Abstract Algebra — Chapter 8 Topics in ring theory

Chapter 8
Topics in ring theory

8.1. Localization of commutative rings

In this section, we let $R$ be a commutative ring.

Definition 8.1.1. §

A subset $S$ of a commutative ring is multiplicatively closed if it is closed under multiplication, $1 \in S$ , and $0 \notin S$ .

To describe our first class of examples, we first make the following definition.

Definition 8.1.2. §

An element $a$ of a ring is nilpotent if there exists $n \geq 1$ such that $a^{n} = 0$ .

Example 8.1.3. §

Given $a \in R$ , the set ${a^{n} ∣ n \geq 1}$ is multiplicatively closed if and only if $a$ is not nilpotent.

Lemma 8.1.4. §

Let $𝔭$ be a prime ideal of $R$ . Then $S_{𝔭} = R - 𝔭$ is a multiplicatively closed subset of $R$ .

Proof.

If $a, b \in S$ , then since $a, b \notin 𝔭$ and $𝔭$ is prime, we have $𝑎𝑏 \notin 𝔭$ , so $𝑎𝑏 \in S$ . Moreover $0 \notin S$ and $1 \in S$ by definition. □

Next, we consider rings of fractions, or localizations, for arbitrary commutative rings. First, we prove a strengthening of Lemma 3.11.1.

Lemma 8.1.5. §

Let $S$ be a multiplicatively closed subset of $R$ . The relation $\sim$ on $R \times S$ given by $(a, s) \sim (b, t)$ if and only if there exists $r \in S$ such that $𝑟𝑎𝑡 = 𝑟𝑏𝑠$ is an equivalence relation.

Proof.

Let $a, b, c \in R$ and $s, t, u \in S$ . That $\sim$ is reflexive is the fact that $𝑟𝑎𝑠 = 𝑟𝑎𝑠$ for any $r \in S$ , that it is symmetric is the fact that $𝑟𝑎𝑡 = 𝑟𝑏𝑠$ implies $𝑟𝑏𝑠 = 𝑟𝑎𝑡$ . If $q, r \in S$ are such that $𝑟𝑎𝑡 = 𝑟𝑏𝑠$ and $𝑞𝑏𝑢 = 𝑞𝑐𝑡$ , then multiplying the former equality by $𝑞𝑢$ and then applying the latter, we obtain

(𝑟𝑞𝑡) 𝑎𝑢 = q (𝑟𝑎𝑡) u = q (𝑟𝑏𝑠) u = r (𝑞𝑏𝑢) s = r (𝑞𝑐𝑡) s = (𝑟𝑞𝑡) 𝑐𝑠 .

We have $𝑟𝑞𝑡 \in S$ since $S$ is multiplicatively closed, so $(a, s) \sim (c, u)$ . Therefore, $\sim$ is transitive. □

Remark 8.1.6. §

Let $S$ be a multiplicatively closed subset of $R$ . If $S$ contains no zero divisors, then the relation $\sim$ on $R \times S$ is more simply defined by $(a, s) \sim (b, t)$ if and only if $𝑎𝑡 = 𝑏𝑠$ . That is, this implies $𝑟𝑎𝑡 = 𝑟𝑏𝑠$ for all $r \in R$ , and likewise, the latter implies $𝑎𝑡 = 𝑏𝑠$ since $r$ is not a zero divisor.

Definition 8.1.7. §

Let $S$ be a multiplicatively closed subset of $R$ . The equivalence class $\frac{a}{s}$ of a pair $(a, s) \in R \times S$ is called an fraction of $R$ with denominator in $S$ (or $S$ -fraction), and the set of such $S$ -fractions is denoted $S^{- 1} R$ .

Remark 8.1.8. §

Let $S$ be a multiplicatively closed subset of $R$ . By definition, we have $\frac{a}{s} = \frac{𝑎𝑡}{𝑠𝑡}$ for any $a \in R$ and $s, t \in S$ . We denote the fraction $\frac{a}{1}$ more simply by $a$ .

Remark 8.1.9. §

If we were to allow $0 \in S$ , then $S^{- 1} R$ would have just one element $0$ . The condition that $1 \in S$ is not strictly necessary so long as $S$ is nonempty, as we can set $a = \frac{𝑎𝑠}{s}$ for any $s \in S$ anyway.

It is worth describing, for additional clarity, $S^{- 1} R$ in the case that $R$ is a domain, prior to treating the general case. In this case, $S^{- 1} R$ is a subset of the field of fractions $Q (R)$ , as $\sim$ on $R \times S$ is the restriction of the previously defined relation on $R \times (R - {0})$ .

Proposition 8.1.10. §

Let $S$ be a multiplicatively closed subset of a domain $R$ . Then

S^{- 1} R = {\frac{a}{s} ∣ a \in R, s \in S}

is the smallest subring of $Q (R)$ containing $R$ in which every element of $S$ is a unit.

Proof.

We note that $S^{- 1} R \subseteq Q (R)$ is closed under addition and multiplication as $S$ is multiplicatively closed, and it clearly contains additive inverses. It also contains all $r \in R$ as $r = \frac{𝑟𝑠}{s}$ for any $s \in S$ , so $S^{- 1} R$ is a subring of $Q (R)$ containing $R$ . We then have $s \cdot \frac{1}{s} = 1$ in this ring, so $S \subseteq {(S^{- 1} R)}^{\times}$ . Finally, as $\frac{1}{s}$ is the inverse to $s \in S$ in $Q (R)$ , any subring $B$ of $Q (R)$ containing $R$ with $S \subseteq B^{\times}$ contains $a \cdot \frac{1}{s} = \frac{a}{s}$ for all $a \in R$ , $s \in S$ . That is, any such subring $B$ contains $S^{- 1} R$ . □

We now turn to the general case.

Theorem 8.1.11. §

Let $S$ be a multiplicatively closed subset of a commutative ring $R$ . The set $S^{- 1} R$ is a ring under addition and multiplication of fractions:

\begin{array}{rcl} \frac{a}{s} + \frac{b}{t} = \frac{𝑎𝑡 + 𝑏𝑠}{𝑠𝑡} & and & \frac{a}{s} \cdot \frac{b}{t} = \frac{𝑎𝑏}{𝑠𝑡} . \end{array}

Proof.

Suppose that $(a, s) \sim (a^{'}, s^{'})$ , and let $r \in R$ be such that $𝑟𝑎 s^{'} = r a^{'} s$ . Then

r (𝑎𝑡 + 𝑏𝑠) s^{'} t = 𝑟𝑎 s^{'} 𝑡𝑡 + 𝑟𝑏𝑠 s^{'} t = r a^{'} 𝑠𝑡𝑡 + 𝑟𝑏𝑠 s^{'} t = r (a^{'} t + b s^{'}) 𝑠𝑡,

so addition is well-defined, noting its symmetry. Similarly, we have

𝑟𝑎𝑏 s^{'} t = r a^{'} 𝑏𝑠𝑡,

so multiplication is well-defined. By definition, addition and multiplication are commutative, and associativity and distributivity of the two are exactly as in the proof of Theorem 3.11.6. Moreover, for any $s \in S$ , we have

\begin{array}{rcl} \frac{a}{s} + 0 = \frac{a}{s} + \frac{0}{1} = \frac{a}{s} & and & \frac{a}{s} \cdot 1 = \frac{a}{s} \cdot \frac{1}{1} = \frac{a}{s} . \end{array}

Also, we have

\frac{a}{s} + \frac{- a}{s} = \frac{𝑎𝑠 + (- a) s}{s^{2}} = 0,

so $- \frac{a}{s} = \frac{- a}{s}$ . Thus, $S^{- 1} R$ is a ring under addition and multiplication. □

Definition 8.1.12. §

The ring $S^{- 1} R$ consisting of $S$ -fractions for a multiplicatively closed subset $S$ of a commutative ring $R$ is called the ring of $S$ -fractions of $R$ , or the localization of $R$ at $S$ .

Theorem 8.1.13. §

Let $S$ be a multiplicatively closed subset of a commutative ring $R$ .

a.: There is a canonical ring homomorphism $R \to S^{- 1} R$ given by $ϕ_{S} (a) = a$ for all $a \in R$ , and its kernel is ${a \in R ∣ 𝑠𝑎 = 0 for some s \in S}$ . In particular, $ϕ_{S}$ is injective if and only if $S$ contains no zero divisors.
b.: Every element of $S$ maps to a unit in $S^{- 1} R$ under $ϕ_{S}$ . Moreover, $ϕ_{S}$ is universal with this property in the sense that if $Q$ is a commutative ring and $f : R \to Q$ is a homomorphism such that $f (S) \subset Q^{\times}$ , then there is a unique homomorphism $𝜃 : S^{- 1} R \to Q$ such that $f = 𝜃 \circ ϕ_{S}$ .

Proof.

That $ϕ_{S}$ is a homomorphism is simply that

ϕ_{S} (a) ϕ_{S} (b) = \frac{a}{1} \frac{b}{1} = \frac{𝑎𝑏}{1} = ϕ_{S} (𝑎𝑏)

for any $s, t \in S$ . We have $ϕ_{S} (a) = 0$ for $a \in R$ if and only if $(a, 1) \sim (0, 1)$ , and therefore if and only if $𝑟𝑎 = 0$ for some $r \in S$ . Thus, we have part a.

Note that $\frac{1}{s} \in S^{- 1} R$ is clearly a multiplicative inverse of $ϕ_{S} (s) \in S^{- 1} R$ . Define $𝜃 : S^{- 1} R \to Q$ by $𝜃 (\frac{a}{s}) = f {(s)}^{- 1} f (a)$ . It is easily checked to be a homomorphism. It also restricts to $f$ by definition. To see that it is well-defined, note that if $\frac{a}{s} = \frac{b}{t}$ , then $𝑟𝑎𝑡 = 𝑟𝑏𝑠$ for some $r \in R$ , and so

f {(s)}^{- 1} f (a) = f {(𝑟𝑠𝑡)}^{- 1} f (𝑟𝑎𝑡) = f {(𝑟𝑠𝑡)}^{- 1} f (𝑟𝑏𝑠) = f {(t)}^{- 1} f (b) .

Thus, we have part b. □

Remark 8.1.14. §

Theorem 8.1.13 tells us that if $S$ has no zero divisors, then $S^{- 1} R$ is the smallest commutative ring containing $R$ in which every element of $S$ is a unit. In particular, if every element of $S$ already is a unit in $R$ , then $S^{- 1} R = R$ . If $S$ has zero divisors, then the map from $S$ to the unit group of $S^{- 1} R$ is still injective.

Definition 8.1.15. §

The total ring of fractions $Q (R)$ of a commutative ring $R$ is the localization of $R$ at the set of nonzero elements of $R$ that are not zero divisors.

Remark 8.1.16. §

If $R$ is an integral domain, then its total ring of fractions is its field of fractions $Q (R)$ .

Examples 8.1.17. §

Let us examine a few localizations of the ring $R = ℤ \times ℤ$ , employing Theorem 8.1.13 to determine their isomorphism classes.

a.

The total ring of fractions of $R$ is given by inverting the set

{(c, d) ∣ c, d \in ℤ - {0}} .

The homomorphism induced by the inclusion of $ℤ^{2}$ in $ℚ^{2}$ is a ring isomorphism

Q (ℤ \times ℤ) \to_{}^{\sim} ℚ \times ℚ, \frac{(a, b)}{(c, d)} \mapsto (\frac{a}{c}, \frac{b}{d}) .

b.

Take $S = ((ℤ - {0}) \times {0}) \cup {(1, 1)}$ . Then the homomorphism $f : R \to ℚ$ given by $f (a, b) = a$ for $a, b \in ℤ$ induces an isomorphism $S^{- 1} R \to_{}^{\sim} ℚ$ .

We study extensions and contractions of ideals by $ϕ_{S}$ .

Notation 8.1.18. §

For an ideal $I$ of a commutative ring $R$ and a multiplicative set $S$ in $R$ , let $S^{- 1} I$ denote the extension of $I$ through $ϕ_{S}$ .

Proposition 8.1.19. §

Let $R$ be a commutative ring, and let $S$ be a multiplicative subset of $R$ .

a.

For any ideal $I$ of $R$ , we have

S^{- 1} I = {\frac{a}{s} ∣ a \in R, s \in S}

and

ϕ_{S}^{- 1} (S^{- 1} I) = {a \in R ∣ 𝑆𝑎 \cap I \neq \emptyset} .

b.

Every ideal of $S^{- 1} R$ is equal to $S^{- 1} I$ for some ideal $I$ of $R$ .

Proof.

By definition, every element of $S^{- 1} I$ is an $S^{- 1} R$ -linear combination of fractions $\frac{a}{1}$ with $a \in I$ , so an $R$ -linear combination of fractions $\frac{a}{s}$ with $a \in I$ and $s \in S$ . But we can take common denominators and use the fact that $I$ is an $R$ -ideal to write every such fraction as $\frac{b}{t}$ for some $b \in I$ and $t \in S$ .

Let $a \in R$ be such that we have $s \in S$ with $x = 𝑠𝑎 \in I$ . Then $ϕ_{S} (a) = \frac{x}{s} \in S^{- 1} I$ , so $a \in ϕ_{S}^{- 1} (S^{- 1} I)$ . Conversely, if $ϕ_{S} (a) = \frac{x}{s}$ for some $x \in I$ and $s \in S$ , then $𝑟𝑠𝑎 = 𝑟𝑥$ for some $r \in S$ , from which it follows that $𝑆𝑎 \cap I \neq \emptyset$ . This proves part a.

Let $J$ be an ideal of $S^{- 1} R$ , and set $I = ϕ_{S}^{- 1} (J)$ . Then $S^{- 1} I \subseteq J$ by definition. On the other hand, $\frac{a}{s} \in J$ for some $a \in R$ and $s \in S$ , then $\frac{a}{1} \in J$ , so $a \in I$ , and therefore $\frac{a}{s} \in S^{- 1} I$ . Thus $J = S^{- 1} I$ , and we have part b. □

Proposition 8.1.20. §

Let $R$ be a commutative ring, and let $S$ be a multiplicatively closed subset of $R$ . Then extension and contraction give mutually inverse maps between the set of prime ideals of $R$ disjoint from $S$ and the set of prime ideals of $S^{- 1} R$ . Moreover, the extension of any prime ideal of $R$ that intersects $S$ is $S^{- 1} R$ .

Proof.

Let $𝔭$ be a prime ideal of $R$ . Note that if $𝔭$ is not disjoint from $S$ , then $1 = \frac{s}{s} \in S^{- 1} 𝔭$ for any $s \in S \cap 𝔭$ , which means that $S^{- 1} 𝔭 = S^{- 1} R$ .

So, suppose that $𝔭 \cap S = \emptyset$ . We claim that its extension $S^{- 1} 𝔭$ is prime. For this claim, it suffices to check that if $\frac{a}{s} \cdot \frac{b}{t} = \frac{c}{u}$ for some $a, b \in R$ , $c \in 𝔭$ and $s, t, u \in S$ , then either $a \in 𝔭$ or $b \in 𝔭$ . The equality implies that there exists $r \in S$ such that $𝑟𝑎𝑏𝑢 = 𝑟𝑐𝑠𝑡$ , and the right-hand side is in $𝔭$ but $𝑟𝑢 \notin 𝔭$ , so $𝑎𝑏 \in 𝔭$ , and we then need merely note that $𝔭$ is prime.

The ideal

I = ϕ_{S}^{- 1} (S^{- 1} 𝔭) = {a \in R ∣ 𝑆𝑎 \cap 𝔭 \neq \emptyset}

clearly contains $𝔭$ . On the other hand, if $a \in I$ , then there exists $s \in S$ such that $𝑠𝑎 \in 𝔭$ , so $a \in 𝔭$ as $𝔭 \cap S = \emptyset$ . That is, $I = 𝔭$ .

The contraction of a prime ideal is always prime, and from what we have proven every prime ideal of $R$ that is disjoint from $S$ is a contraction by $ϕ_{S}$ of a prime ideal of $S^{- 1} R$ . On the other hand, we have already argued that the extension of a prime ideal of $R$ disjoint from $S$ is a prime ideal of $S^{- 1} R$ . Moreover, any prime ideal of $S^{- 1} R$ is the extension of an ideal of $R$ by Proposition 8.1.19 and then of a prime ideal as it is the extension of its contraction by Proposition 3.7.26. Such a prime ideal must be disjoint from $S$ by the above. Thus, we obtain the inverse bijections as the restriction of the inverse bijections in Proposition 3.7.26. □

Definition 8.1.21. §

Let $R$ be a commutative ring, and let $x \in R$ be a non-nilpotent element. Then the localization of $R$ with respect to $x$ , denoted by $R [x^{- 1}]$ , is the ring $S^{- 1} R$ for $S = {x^{n} ∣ n \geq 0}$ .

Example 8.1.22. §

Let $n \in ℤ$ . Then the ring $ℤ [\frac{1}{n}]$ may be identified with the subset of $ℚ$ consisting of reduced fractions with denominator a product of powers of primes dividing $n$ , or equivalently, with denominator dividing a power of $n$ . The distinct ideals of $ℤ [\frac{1}{n}]$ are generated by nonnegative $a \in ℤ$ with $(a, n) = 1$ .

Example 8.1.23. §

Let $R = ℤ \times ℤ$ and $x = (1, 0)$ , and consider $R [x^{- 1}]$ . Since $x \cdot (0, 1) = 0$ , and $x$ is invertible in $R [x^{- 1}]$ , we have $(0, 1) = 0$ in $R [x^{- 1}]$ . Note also that ${(1, 0)}^{n} (a, 0) = (a, 0)$ , so $\frac{(a, 0)}{{(1, 0)}^{n}} = (a, 0)$ . It follows that the ring homomorphism $ℤ \to R [x^{- 1}]$ given by $a \mapsto (a, 0)$ is an isomorphism.

Definition 8.1.24. §

Let $R$ be a commutative ring, and let $𝔭$ be a prime ideal of $R$ . Then $R_{𝔭} = S_{𝔭}^{- 1} R$ , where $S_{𝔭}^{- 1} = R - 𝔭$ , is referred to as the localization of $R$ at $𝔭$ .

Examples 8.1.25. §

Let $p \in ℤ$ be prime.

a.: The localization $ℤ_{(p)}$ is the subring of $ℚ$ consisting of reduced fractions with denominators not divisible by $p$ .
b.: The ring $ℚ {[x]}_{(x)}$ consists of rational functions with denominator not divisible by $x$ .
c.: The ring $ℤ {[x]}_{(x)}$ equals $ℚ {[x]}_{(x)}$ inside $ℚ (x)$ .
d.: The ring $ℤ {[x]}_{(p, x)}$ is the subring of $ℚ (x)$ of rational functions with denominator having nonzero constant term modulo $p$ .

Example 8.1.26. §

Consider $R = ℤ \times ℤ$ and its prime ideal $𝔭 = 𝑝ℤ \times {0}$ . Then $R_{𝔭} ≅ ℤ_{(p)}$ , where the localization map $ℤ \times ℤ \to ℤ_{(p)}$ sends $(a, b)$ to $a$ .

Definition 8.1.27. §

A commutative ring $R$ is local if it has a unique maximal ideal.

Definition 8.1.28. §

The residue field of a local ring $R$ with maximal ideal $𝔪$ is the field $R ∕ 𝔪$ .

The first part of the following explains something of the meaning of the terminology “localization.”

Proposition 8.1.29. §

Let $𝔭$ be a prime ideal of a commutative ring $R$ .

a.: The ring $R_{𝔭}$ is a local ring with maximal ideal $𝔭 R_{𝔭}$ .
b.: The proper ideals of $R_{𝔭}$ are exactly those of the form $I R_{𝔭}$ for some ideal $I$ of $R$ contained in $𝔭$ .

Proof.

By Proposition 8.1.19b, every ideal of $R_{𝔭}$ has the form $I R_{𝔭}$ for some ideal $I$ of $R$ . If $a \in R - 𝔭$ , then by definition $a$ is invertible in $R_{𝔭}$ , hence $a R_{𝔭} = R_{𝔭}$ . Thus, $I R_{𝔭} = R_{𝔭}$ for every ideal of $R$ not contained in $𝔭$ . On the other hand $I R_{𝔭} \subseteq 𝔭 R_{𝔭}$ if $I \subseteq 𝔭$ , so $𝔭 R_{𝔭}$ is the unique maximal ideal of $R_{𝔭}$ . □

We note the following lemmas.

Lemma 8.1.30. §

Let $R$ be a local ring and $𝔪$ be its maximal ideal. Then $R^{\times} = R - 𝔪$ .

Proof.

If $(a) \neq R$ , then $a$ is contained in a maximal ideal, which must be $𝔪$ . Conversely, if $a \in R^{\times}$ , then $(a) = R$ , so $a$ is not contained in $𝔪$ . □

Lemma 8.1.31. §

Let $𝔪$ be a maximal ideal of a commutative ring $R$ . Then the canonical ring homomorphism $R ∕ 𝔪 \to R_{𝔪} ∕ 𝔪 R_{𝔪}$ is an isomorphism.

Proof.

Since nonzero maps of fields are injective, it suffices to see that the map is onto. If $r \in R$ and $u \in R - 𝔪$ , then let $v \in R$ be such that $(u + 𝔪) (v + 𝔪) = 1$ . Then $\frac{r}{u} + 𝔪 R_{𝔪}$ is the image of $𝑣𝑟 + 𝔪$ . □

Remark 8.1.32. §

If $𝔭$ is a non-maximal prime ideal in a commutative ring $R$ , then $R ∕ 𝔭 \to R_{𝔭} ∕ 𝔭 R_{𝔭}$ is injective but cannot be an isomorphism, since $R ∕ 𝔭$ is not a field but $R_{𝔭} ∕ 𝔭 R_{𝔭}$ is.

We mention the following result on ideals contained in the complements of multiplicative sets.

Theorem 8.1.33. §

Let $R$ be a commutative ring, let $S$ be a multiplicative set, and let $I$ be an ideal of $R$ disjoint from $S$ . Then there exists an ideal of $R$ containing $I$ and maximal with respect to the property that it is disjoint from $S$ . Moreover, any such ideal is prime.

Proof.

Let $X$ denote the set of ideals of $R$ that contain $I$ and are disjoint from $S$ . Since $I \in X$ , it is nonempty. If $𝒞$ is a chain in $X$ , then the ideal $⋃_{J \in 𝒞} C$ of $R$ is clearly contained in $X$ , hence is an upper bound on $𝒞$ . By Zorn’s lemma, $X$ contains a maximal element $𝔭$ .

If $𝔭$ is not prime, then we may choose $𝑎𝑏 \in 𝔭$ with $a, b \notin 𝔭$ . Set $I = (a) + 𝔭$ and $J = (b) + 𝔭$ so that $𝐼𝐽 \subseteq 𝔭$ . By maximality of $𝔭$ , we have that there exist $c \in I \cap S$ and $d \in J \cap S$ . Since $S$ is a multiplicative set, we then have $𝑐𝑑 \in 𝐼𝐽 \cap S \subseteq 𝔭 \cap S$ , contradicting $𝔭 \in X$ . Thus, $𝔭$ must be prime. □

We next give an interesting characterization of unique factorization domains. First, a definition and a lemma.

Definition 8.1.34. §

A multiplicative subset $S$ of a commutative ring $R$ is saturated if for all $a, b \in R$ with $𝑎𝑏 \in S$ , we have $a, b \in S$ .

Lemma 8.1.35. §

Let $R$ be a domain. The set

S = {a \in R - {0} ∣ a = u p_{1} \dots p_{k} for u \in R^{\times} and p_{i} prime for 1 \leq i \leq k for some k \geq 0}

is a saturated multiplicative set.

Proof.

The set $S$ is clearly multiplicative. We show it is saturated. Let $a, b \in R$ such that $𝑎𝑏 \in S$ . If $𝑎𝑏 \in R^{\times}$ , then $a, b \in R^{\times}$ (as, for instance $a \cdot b {(𝑎𝑏)}^{- 1} = 1$ ). Suppose by induction that if the product element $𝑎𝑏$ can be written as a product of a unit times $k - 1$ prime elements for $k \geq 1$ , the result holds. Write $𝑎𝑏 = p_{1} \dots p_{k}$ with each $p_{i}$ prime. As $p_{k}$ is prime, it divides either $a$ or $b$ . Without loss of generality, we suppose that $p_{k} ∣ a$ , and we write $a = p_{k} a^{'}$ for some $a^{'} \in R$ . Then $a^{'} b = p_{1} \dots p_{k - 1}$ . By induction, we have $a^{'}, b \in S$ . Then $a = a^{'} p_{k} \in S$ as well, as $S$ is multiplicative. □

Theorem 8.1.36. §

A domain $R$ is a UFD if and only if every nonzero prime ideal of $R$ contains a prime element.

Proof.

Suppose first that $R$ is a UFD and $𝔭$ is a prime ideal of $R$ . If $a \in 𝔭 - {0}$ , then we may write $a = π_{1} \dots π_{k}$ with $π_{i} \in R$ irreducible, hence prime, for $1 \leq i \leq k$ . As $𝔭$ is prime, we have $π_{j} \in 𝔭$ for some $1 \leq j \leq k$ .

Conversely, suppose that every nonzero prime ideal of $R$ contains a prime element. Let $S$ be the saturated multipilicative set of Lemma 8.1.35. Suppose that $a \in R - S$ is nonzero. Then for any $r \in R$ , we have $𝑟𝑎 \notin S$ as $S$ is saturated. In other words, $(a)$ is disjoint from $S$ . By Theorem 8.1.33, there exists a prime ideal $𝔭$ containing $(a)$ with $𝔭 \cap S = \emptyset$ . By assumption, $𝔭$ contains a prime element $p$ . But $p \in S$ by definition, a contradiction. Thus, $S = R - {0}$ , and every nonzero element in $R$ factors as a unit times a product of prime elements. By Proposition 5.1.27, we conclude that $R$ is a UFD. □

8.2. Finitely generated and noetherian modules

Definition 8.2.1. §

Let $R$ be a ring. A left $R$ -module $M$ is said to be noetherian if its set of submodules satisfies the ascending chain condition.

Proposition 8.2.2. §

A module $M$ over a ring $R$ is noetherian if and only if every submodule of $M$ is finitely generated over $R$ .

Proof.

If every submodule of $M$ is finitely generated, then the union of every ascending chain ${N_{i} ∣ i \geq 1}$ of submodules of $M$ is finitely generated, and each one of these generators is contained in some $N_{k}$ , so they are all contained in the largest $N_{k}$ among these. Thus, the union is actually equal to $N_{k}$ , so the ACC holds.

On the other hand, if the ACC holds for $M$ and $N$ is an $R$ -submodule of $M$ , then we can pick $n_{1} \in N - {0}$ and then, if it exists, $n_{i + 1} \in N$ with $n \notin N_{i}$ with $N_{i} = \sum_{j = 1}^{i} R n_{i}$ for each $i$ . By definition, $N_{i}$ is properly contained in $N_{i + 1}$ , so by the ACC, eventually we cannot continue the process, which is to say that for some $k$ , we have $N_{k} = N$ , or in other words that $M$ is generated by ${n_{1}, n_{2}, \dots, n_{k}}$ . □

Remark 8.2.3. §

Finitely generated modules need not be noetherian. A ring is left noetherian (i.e., satisfies the ascending chain condition on left ideals) if and only if it is noetherian as a left module over itself. Yet, any ring is finitely generated as a left module over itself, being that it is generated by $1$ .

Lemma 8.2.4. §

Let $R$ be a ring, let $M$ be an $R$ -module, and let $N$ be an $R$ -submodule of $M$ . If $S$ is a generating set of $N$ and $T$ is a subset of $M$ with image generating $M ∕ N$ , then $S \cup T$ generates $M$ .

Proof.

If $m \in M$ , then there exist $n \in N$ , $m_{i} \in T$ , and $b_{i} \in R$ for $1 \leq i \leq j$ for some $j$ such that

m = n + \sum_{i = 1}^{j} b_{i} m_{i},

and then there exist $n_{i} \in S$ and $c_{i} \in R$ for $1 \leq i \leq k$ for some $k$ such that

n = \sum_{i = 1}^{k} c_{i} n_{i},

so $n$ is an $R$ -linear combination of the $m_{i}$ and the $n_{i}$ . That is, $M$ is generated by $S \cup T$ . □

Corollary 8.2.5. §

Let $R$ be a ring, let $M$ be an $R$ -module, and let $N$ be a finitely generated $R$ -submodule of $M$ such that $M ∕ N$ is also finitely generated. Then $M$ is finitely generated.

Lemma 8.2.6. §

Let $R$ be a ring and $N$ be a submodule of an $R$ -module $M$ . Then $M$ is noetherian if and only if both $N$ and $M ∕ N$ are noetherian.

Proof.

If $M$ is noetherian, then $N$ is noetherian by definition. Moreover, the inverse image $P$ of any submodule $O$ of $M ∕ N$ under the quotient map $π : M \to M ∕ N$ is a submodule of $M$ , hence generated by some finite set $S$ . Then $π (S)$ generates $O$ , so we conclude that $M ∕ N$ is noetherian.

If $N$ and $M ∕ N$ are both noetherian and $P$ is a submodule of $M$ , then $P \cap N$ and $P ∕ (P \cap N)$ are finitely generated as submodules of $N$ and $M ∕ N$ , respectively, so $P$ is finitely generated by Corollary 8.2.5. That is, $M$ is noetherian. □

Corollary 8.2.7. §

Finite direct sums of noetherian modules are noetherian.

Proof.

If $M = N \oplus N^{'}$ for $R$ -modules $M$ , $N$ , and $N^{'}$ , then $N^{'} = M ∕ N$ , so by the lemma, $M$ is noetherian if $N$ and $N^{'}$ are. The result then follows by induction on the number of summands. □

Proposition 8.2.8. §

Every finitely generated left module over a left noetherian ring is noetherian.

Proof.

Let $M$ be a finitely generated left module over a noetherian ring $R$ , and let $N$ be a submodule of $M$ . Since $M$ is finitely generated, there is a surjective $R$ -module homomorphism $R^{n} \to M$ for some $n$ . Let $P$ be the inverse image of $N$ in $R^{n}$ . The module $N$ is generated by the image of any set of generators of $P$ under the quotient map $P \to N$ . So, we need only show that any submodule $P$ of $R^{n}$ is finitely generated, which is to say that $R^{n}$ is left noetherian. This is true as $R$ is a left noetherian $R$ -module, and $R^{n}$ is the direct sum of $n$ copies of $R$ . □

We next consider modules that satisfy the descending chain condition.

Definition 8.2.9. §

Let $X$ be a set with a partial ordering $\leq$ . A descending chain on $X$ is an ascending chain with respect to the opposite partial ordering $\geq$ defined by $x \geq y$ if and only if $y \leq x$ for $x, y \in X$ .. We say that $X$ satisfies the descending chain condition, or DCC, if it satisfies the ACC with respect to $\geq$ .

Definition 8.2.10. §

We say that a module over a ring $R$ is artinian if its set of submodules satisfies the descending chain condition.

Example 8.2.11. §

Any finite-dimension vector spaces over a field $F$ is an artinian $F$ -module.

Lemma 8.2.12. §

Let $R$ be a ring and $N$ be a submodule of an $R$ -module $M$ . Then $M$ is artinian if and only if both $N$ and $M ∕ N$ are artinian.

Proof.

That $M$ being artinian implies $N$ and $M ∕ N$ are artinian is straightforward. If $N$ and $M ∕ N$ are artinian and ${(M_{i})}_{i \geq 1}$ is a descending chain in $M$ , then there exists $k \geq 0$ such that $M_{i} \cap N = M_{k} \cap N$ and $(M_{i} + N) ∕ N = (M_{k} + N) ∕ N$ for all $i \geq k$ . But this can only happen if $M_{i} = M_{k}$ for all $i \geq k$ as well: if $m \in M_{k}$ , then $m \in M_{i} + N$ , so $m = m^{'} + n$ for some $m^{'} \in M_{i}$ and $n \in N$ , but then $n = m - m^{'} \in M_{k} \cap N \subseteq M_{i}$ , and therefore $m = m^{'} + n \in M_{i}$ . □

Lemma 8.2.13. §

If $𝔪$ is a maximal ideal in a noetherian ring, then $R ∕ 𝔪^{n}$ is an artinian $R$ -module.

Proof.

Note that $R ∕ 𝔪$ is a field, hence artinian as an $R$ -module. By induction on $n \geq 1$ , we may suppose that $R ∕ 𝔪^{n - 1}$ is artinian as an $R$ -module. By Lemma 8.2.12, it suffices to show that $𝔪^{n - 1} ∕ 𝔪^{n}$ is Artinian over $R$ . Since $R$ is noetherian, $𝔪^{n - 1}$ is a finitely generated $R$ -module, and the images in $𝔪^{n - 1} ∕ 𝔪^{n}$ of any list of generators span it as an $R ∕ 𝔪$ -vector space. Since it is artinian as an $R ∕ 𝔪$ -module, it is also artinian as an $R$ -module. □

Definition 8.2.14. §

The Jacobson radical $J (R)$ of a (possibly noncommutative) ring $R$ is the intersection of all left maximal ideals of $R$ .

The following extends Lemma 8.1.30.

Lemma 8.2.15. §

Let $x \in R$ . Then $x \in J (R)$ if and only if $1 - 𝑟𝑥 \in R^{\times}$ for all $r \in R$ .

Proof.

If $1 - 𝑟𝑥 \notin R^{\times}$ , then there exists a left maximal ideal $𝔪$ containing $1 - 𝑟𝑥$ . Then $𝑟𝑥 \notin 𝔪$ , so $𝑟𝑥 \notin J (R)$ , and therefore $x \notin J (R)$ . Conversely, if $x \notin J (R)$ , then there exists a left maximal ideal $𝔪$ such that $x \notin 𝔪$ . Then there exist $r \in R$ and $y \in 𝔪$ such that $1 = 𝑟𝑥 + y$ . Then $1 - 𝑟𝑥 = y \notin R^{\times}$ . □

Theorem 8.2.16 (Nakayama’s lemma). §

Let $M$ be a finitely generated module over a commutative ring $R$ , and suppose that $J (R) M = M$ . Then $M = 0$ .

Proof.

Let ${m_{1}, m_{2}, \dots, m_{k}}$ be a set of generators of $M$ with $k \geq 1$ . Since $m_{1} \in J (R) M$ , we can find $a_{i} \in J (R)$ for $1 \leq i \leq k$ such that

m_{1} = \sum_{i = 1}^{k} a_{i} m_{i} .

Since $(1 - a_{1}) m_{1}$ is contained in the submodule $M^{'}$ generated by $m_{2}, \dots, m_{k}$ . On the other hand, $1 - a_{1} \in R^{\times}$ by Lemma 8.2.15. But then $m_{1}$ itself is contained in $M^{'}$ , which tells us that $M^{'} = M$ and $k$ is not minimal. That is, the minimal number of generators of $M$ is zero. □

Corollary 8.2.17. §

Let $M$ be a finitely generated module over a local ring $R$ , and suppose that $𝔪𝑀 = M$ , where $𝔪$ is the maximal ideal of $M$ . Then $M = 0$ .

Corollary 8.2.18. §

Let $M$ be a finitely generated module over a local ring $R$ with maximal ideal $𝔪$ , and let $X$ be a set of elements of $M$ such that ${m + 𝔪𝑀 ∣ m \in X}$ generates $M ∕ 𝔪𝑀$ as a vector space over the residue field $R ∕ 𝔪$ . Then $X$ generates $M$ .

Proof.

Let $N$ be the submodule of $M$ generated by $X$ . Then $N + 𝔪𝑀 = M$ , so every element in $M ∕ N$ is the $N$ -coset of some element of $𝔪𝑀$ , which is to say that $𝔪 (M ∕ N) = M ∕ N$ . By Nakayama’s lemma, we have $M ∕ N = 0$ , so $X$ generates $M$ . □

Example 8.2.19. §

Take the set of tuples $(111, 107, 50)$ , $(23, - 17, 41)$ , and $(30, - 8, 104)$ . Suppose that we want to see if they generate the $ℚ$ -vector space $ℚ^{3}$ . It suffices, then, to see that they generate the $ℤ_{(p)}$ -module $ℤ_{(p)}^{3}$ for some prime $p$ . Moreover, the map $𝔽_{p} \to ℤ_{(p)} ∕ p ℤ_{(p)}$ is an isomorphism, so by Corollary 8.2.18, it suffices to see that these tuples generate $𝔽_{p}^{3}$ . Modulo $2$ , they are $(1, 1, 0)$ , $(1, 1, 1)$ , and $(0, 0, 0)$ , so they do not generate $𝔽_{2}^{3}$ . However, modulo $3$ , they are $(0, - 1, - 1)$ , $(- 1, 1, - 1)$ , and $(0, 1, - 1)$ , which do in fact generate $𝔽_{3}^{3}$ , and thus the original tuples generate $ℚ^{3}$ .

8.3. Homomorphism groups

Remark 8.3.1. §

Let $M$ be a left module over an $R$ -algebra $A$ . Then $M$ is an $R$ -module under $r \cdot m = ϕ (r) m$ , where $ϕ : R \to Z (A)$ is given by the structure of $A$ as an $R$ -algebra.

Definition 8.3.2. §

Let $M$ and $N$ be left modules over an $R$ -algebra $A$ . The homomorphism group ${Hom}_{A} (M, N)$ is the $R$ -module of homomorphisms $ϕ : M \to N$ under the usual addition of maps and the scalar multiplication $(r \cdot ϕ) (m) = r \cdot ϕ (m)$ for $r \in R$ and $m \in M$ .

Remark 8.3.3. §

It is traditional to call ${Hom}_{A} (M, N)$ a homomorphism group, even when it has an additional $R$ -module structure (for when we simply take $R = ℤ$ , it is just a $ℤ$ -module, or abelian group).

Example 8.3.4. §

Let $R$ be a commutative ring. Then ${Hom}_{R} (R^{m}, R^{n})$ is a free $R$ -module of rank $𝑚𝑛$ , isomorphic to $M_{𝑛𝑚} (R)$ via $ϕ \mapsto A$ with $ϕ (e_{j}) = \sum_{i = 1}^{n} a_{𝑖𝑗} e_{j}$ .

Example 8.3.5. §

Let $m, n \geq 1$ . Then ${Hom}_{ℤ} (ℤ ∕ 𝑚ℤ, ℤ ∕ 𝑛ℤ) ≅ ℤ ∕ (m, n) ℤ$ . That is, an element this group is completely determined of $ϕ (1)$ , and $ϕ (1)$ has to be an element of order dividing $m$ in $ℤ ∕ 𝑛ℤ$ , so a multiple of $\frac{n}{\gcd (m, n)}$ .

In general, if $M$ and $N$ are $A$ -modules with an additional right module structures that turn them into bimodules, then we can consider transfer these structures to ${Hom}_{A} (M, N)$ , as we briefly explore.

Definition 8.3.6. §

Let $A$ and $B$ be algebras over a commutative ring $R$ . We say that an $A$ - $B$ -bimodule $M$ is $R$ -balanced if $𝑟𝑚 = 𝑚𝑟$ for all $r \in R$ and $m \in M$ .

Examples 8.3.7. §

a.: If $A$ is an $R$ -algebra, then $A$ is an $R$ -balanced $A$ - $A$ -bimodule.
b.: For a commutative ring $R$ , the $M_{m} (R)$ - $M_{n} (R)$ -bimodule $M_{𝑚𝑛} (R)$ is $R$ -balanced.

Proposition 8.3.8. §

Let $A$ , $B$ , and $C$ be $R$ -algebras, let $M$ be an $R$ -balanced $A$ - $B$ -bimodule, and let $N$ be an $R$ -balanced $A$ - $C$ -bimodule. Then ${Hom}_{A} (M, N)$ is an $R$ -balanced $B$ - $C$ -bimodule under the actions given by

\begin{array}{rcl} (b \cdot ϕ) (m) = ϕ (𝑚𝑏) & and & (ϕ \cdot c) (m) = ϕ (m) c \end{array}

for $b \in B$ , $c \in C$ , $m \in M$ , and $ϕ \in {Hom}_{A} (M, N)$ .

Homomorphism groups behave well with respect to direct sums and products, as made precise in the following proposition.

Proposition 8.3.9. §

Let $A$ be an $R$ -algebra.

a.: Let $M$ be a left $A$ -module, and let ${N_{j} ∣ j \in J}$ be a collection of left $A$ -modules. Then there is a canonical isomorphism of left $R$ -modules
${Hom}_{A} (M, \prod_{j \in J} N_{j}) ≅ \prod_{j \in J} {Hom}_{A} (M, N_{j}) .$
b.: Let $N$ be a left $A$ -module, and let ${M_{i} ∣ i \in I}$ be a collection of left $A$ -modules. Then there is a canonical isomorphism of left $R$ -modules
${Hom}_{A} (⨁_{i \in I} M_{i}, N) ≅ \prod_{i \in I} {Hom}_{A} (M_{i}, N) .$

Proof.

Given a collection of $A$ -module homomorphisms $ϕ_{j} : M \to N_{j}$ for $j \in J$ , we define $Φ : M \to \prod_{j \in J} N_{j}$ by $Φ (m) = {(ϕ_{j} (m))}_{j \in J}$ , which is clearly an $A$ -module homomorphism. Conversely, given $Φ$ , we define $ϕ_{j} = π_{j} \circ Φ$ where $π_{j} : \prod_{k \in J} N_{k} \to N_{j}$ is the projection map, and $ϕ_{j}$ is then an $A$ -module homomorphism. The bijection ${(ϕ_{j})}_{j \in J} \mapsto Φ$ is clearly a map of $R$ -modules. Thus, we have part a.

Now, given a collection of $A$ -module homomorphisms $ψ_{i} : M_{i} \to N$ for $i \in I$ , we define $Ψ : \oplus_{i \in I} M_{i} \to N$ by $Ψ ({(m_{i})}_{i \in I}) = \sum_{i \in I} m_{i}$ , which is well-defined as all but finitely many $m_{i} = 0$ by definition of the direct sum. The map $Ψ$ is then an $A$ -module homomorphism. Conversely, given $Ψ$ , we define $ψ_{i} (m) = Ψ (ι_{i} (m))$ , where $ι_{i} : M \to \oplus_{i \in I} M_{i}$ is the inclusion. These are by definition inverse associations, and the bijection ${(ψ_{i})}_{i \in I} \to Ψ$ is again clearly an $R$ -module homomorphism. □

Let us consider the example of a dual vector space.

Definition 8.3.10. §

Let $V$ be a vector space over a field $K$ . The dual vector space is $V^{*} = {Hom}_{K} (V, K)$ .

Remark 8.3.11. §

Note that $𝑉≅ \oplus_{i \in I} K$ for any choice of basis, so

V^{*} ≅ {Hom}_{K} (⨁_{i \in I} K, K) ≅ \prod_{i \in I} {Hom}_{K} (K, K) ≅ \prod_{i \in I} K

by part b of Proposition 8.3.9. That is, $V$ and $V^{*}$ are not in general isomorphic, but they will be so if $V$ is finite-dimensional. However, this isomorphism is not canonical: it depends on a choice of basis, which we next make explicit.

Definition 8.3.12. §

Let $V$ be an $n$ -dimensional vector space over a field $K$ , and let $B = {e_{1}, e_{2}, \dots, e_{n}}$ be a basis of $V$ . The dual basis to $B$ is the basis of $V^{*}$ given by $B^{*} = {f_{1}, f_{2}, \dots, f_{n}}$ , where for $1 \leq i, j \leq n$ , we have

f_{i} (e_{j}) = δ_{𝑖𝑗}

We next consider the double dual $V^{∗∗} = {(V^{*})}^{*}$ of an arbitrary vector space. For a finite-dimensional vector space, it is canonically isomorphic to $V$ .

Proposition 8.3.13. §

Let $V$ be a vector space over a field $K$ . There is a canonical injection $Φ : V \to V^{∗∗}$ of $K$ -vector spaces given by $F (v) (f) = f (v)$ for $v \in V$ and $f \in V^{*}$ . It is an isomorphism if $V$ is finite-dimensional.

Proof.

Let $v \in V$ and $f \in V^{*}$ . First, note that

Φ (v) (f + a f^{'}) = f (v) + 𝑎𝑓 (v^{'}) = Φ (v) (f) + a Φ (v) (f^{'}),

so $Φ (v) \in V^{∗∗}$ . Second, note that

Φ (𝑎𝑣 + v^{'}) (f) = f (𝑎𝑣 + v^{'}) = 𝑎𝑓 (v) + f (v^{'}) = a \cdot Φ (v) (f) + Φ (v^{'}) (f),

so $Φ$ is a $K$ -linear transformation. Third, note that if $Φ (v) = 0$ , then $Φ (v) (f) = f (v) = 0$ for all $f \in V^{*}$ . If $v \neq 0$ , we can extend ${v}$ to a basis $B$ of $V$ and define $f \in V^{*}$ by $f (v) = 1$ and $f (w) = 0$ for all $w \in B - {v}$ . Thus, the fact that $f (v) = 0$ for all $f \in V^{*}$ implies that $v = 0$ , so $Φ$ is injective.

Now, suppose that $V$ is $n$ -dimensional, let ${e_{1}, e_{2}, \dots, e_{n}}$ be a basis, and let ${f_{1}, f_{2}, \dots, f_{n}}$ be its dual basis in $V^{*}$ . If $φ \in V^{∗∗}$ , then set $c_{j} = φ (f_{j})$ for each $1 \leq j \leq n$ . Then

Φ (\sum_{i = 1}^{n} φ (f_{i}) e_{i}) (f_{j}) = f_{j} (\sum_{i = 1}^{n} φ (f_{i}) e_{i}) = φ (f_{j})

for all $j$ , so $φ \in Φ (V)$ . That is, $Φ$ is an isomorphism. □

8.4. Tensor products

Definition 8.4.1. §

Let $A$ be ring, let $M$ be a right $A$ -module, and let $N$ be a left $A$ -module. The tensor product $M \otimes_{A} N$ of $M$ and $N$ over $A$ is the abelian group that is the quotient of the free abelian group with basis $M \times N$ by its subgroup generated by

i.: $(m + m^{'}, n) - (m, n) - (m^{'}, n)$ for all $m, m^{'} \in M$ and $n \in N$ ,
ii.: $(m, n + n^{'}) - (m, n) - (m, n^{'})$ for all $m \in M$ and $n, n^{'} \in N$ , and
iii.: $(𝑚𝑎, n) - (m, 𝑎𝑛)$ for all $m \in M$ , $n \in N$ , and $a \in A$ .

The image of $(m, n)$ in $M \otimes_{A} N$ is denoted $m \otimes n$ .

Definition 8.4.2. §

Let $A$ be ring, let $M$ be a right $A$ -module, and let $N$ be a left $A$ -module. An element of $M \otimes_{A} N$ of the form $m \otimes n$ for some $m \in M$ and $n \in N$ is called a simple tensor.

Remark 8.4.3. §

Any tensor product $M \otimes_{A} N$ is generated as an abelian group by simple tensors $m \otimes n$ for $m \in M$ and $n \in N$ . It is not in general equal to the set of such tensors.

Proposition 8.4.4. §

Let $A$ be an algebra over a commutative ring $R$ , let $M$ be a right $A$ -module, and let $N$ be a left $A$ -module. The tensor product $M \otimes_{A} N$ is an $R$ -module under the unique action that satisfies

r (m \otimes n) = 𝑚𝑟 \otimes n = m \otimes 𝑟𝑛 .

for all $r \in R$ , $m \in M$ , and $n \in N$ .

Proof.

If we consider the free abelian group on $M \times N$ as an $R$ -module via $r (m, n) = (𝑚𝑟, n)$ for $r \in R$ , $m \in M$ , and $n \in N$ , then the elements providing the relations in Definition 8.4.1 define an $R$ -submodule. Therefore, the quotient becomes an $R$ -module under this action. □

Remark 8.4.5. §

If $A$ is an $R$ -algebra, the tensor product $M \otimes_{A} N$ is isomorphic to the quotient of the free $R$ -module on $M \times N$ by the submodule generated by the elements of Definition 8.4.1, along with the elements $r (m, n) - (𝑚𝑟, n)$ for $r \in R$ , $m \in M$ , and $n \in N$ .

Definition 8.4.6. §

a.

Let $L$ , $M$ , and $N$ be abelian groups. A map $ϕ : M \times N \to L$ is said to be bilinear if

\begin{array}{rcl} ϕ (m + m^{'}, n) = ϕ (m, n) + ϕ (m^{'}, n) & and & ϕ (m, n + n^{'}) = ϕ (m, n) + ϕ (m, n^{'}) . \end{array}

for all $m, m^{'} \in M$ and $n, n^{'} \in N$ . Here, the first equality (for all $m$ , $m^{'}$ , and $n$ ) is referred to as left linearity (or linearity in the first variable) and the second as right linearity.

b.

Let $L$ , $M$ , and $N$ be left modules over a commutative ring $R$ . A bilinear map $ϕ : M \times N \to L$ satisfying

𝑟𝜙 (m, n) = ϕ (𝑟𝑚, n) = ϕ (m, 𝑟𝑛)

for all $r \in R$ , $m \in M$ , and $n \in N$ , then $ϕ$ is said to be $R$ -bilinear.

Definition 8.4.7. §

Let $A$ be a ring, let $M$ be a right $A$ -module, and let $N$ be a left $A$ -module. A function $ϕ : M \times N \to L$ is said to be $A$ -balanced if $ϕ (𝑚𝑎, n) = ϕ (m, 𝑎𝑛)$ for all $a \in A$ .

Remark 8.4.8. §

Let $A$ be an algebra over a commutative ring $R$ , let $M$ be a right $A$ -module, and let $N$ be a left $A$ -module. The tensor product $M \otimes_{A} N$ is endowed with an $A$ -balanced $R$ -bilinear map

ι_{M, N} : M \times N \to M \otimes_{A} N, ϕ (m, n) = m \otimes n,

as seen directly from the relations defining $M \otimes_{A} N$ .

The tensor product enjoys a universal property, exhibited in the following proposition.

Proposition 8.4.9. §

Let $A$ be an algebra over a commutative ring $R$ , let $M$ be a right $A$ -module, let $N$ be a left $A$ -module, and let $L$ be an $R$ -module. Let $ϕ : M \times N \to L$ be $R$ -bilinear and $A$ -balanced. Then there exists a unique $R$ -module homomorphism $Φ : M \otimes_{A} N \to L$ such that $Φ (m \otimes n) = ϕ (m, n)$ for all $m \in M$ and $n \in N$ .

Proof.

We use the alternate construction of $M \otimes_{A} N$ of Remark 8.4.5. The map $ϕ$ induces a unique $R$ -module homomorphism

F : ⨁_{(m, n) \in M \times N} R (m, n) \to L, F ((m, n)) = ϕ (m, n),

since the direct sum is free. The $R$ -bilinearity of $ϕ$ tells us that the elements $(m + m^{'}, n) - (m, n) - (m^{'}, n)$ , $(m, n + n^{'}) - (m, n) - (m, n^{'})$ , and $r (m, n) - (𝑚𝑟, n)$ lie its the kernel. The fact that $ϕ$ is $A$ -balanced similarly tells us that the elements $(𝑚𝑎, n) - (m, 𝑎𝑛)$ are contained in its kernel. The first homomorphism theorem then provides an $R$ -module homomorphism $Φ : M \otimes_{A} N \to L$ with $Φ (m \otimes n) = ϕ (m, n)$ for all $m \in M$ and $n \in N$ .

If $Ψ : M \otimes_{A} N \to L$ is an $R$ -module homomorphism also satisfying $Ψ \circ ι_{M, N} = ϕ$ , then $Ψ (m \otimes n) = ϕ (m, n) = Φ (m \otimes n)$ for all $m \in M$ and $n \in N$ , but the symbols $m \otimes n$ generate $M \otimes_{A} N$ as an $R$ -module, since the tensor product is defined as the quotient of the free $R$ -module on $M \times N$ . Therefore, we must have $Φ = Ψ$ . □

Remark 8.4.10. §

The defining property of the map $Φ : M \otimes_{A} N \to L$ of Proposition 8.4.9 is stated more succinctly as $Φ \circ ι_{M, N} = ϕ$ .

The reader may check the following, which gives the uniqueness of the tensor product up to unique isomorphism as a module satisfying the universal property of the tensor product.

Proposition 8.4.11. §

Let $A$ be an algebra over a commutative ring $R$ , let $M$ be a right $A$ -module, and let $N$ be a left $A$ -module. Let $P$ be an $R$ -module, and let $λ : M \times N \to P$ be an $R$ -bilinear map such that for any $R$ -bilinear, $A$ -balanced map $ϕ : M \times N \to L$ , there exists a unique $R$ -module homomorphism $Φ : P \to L$ such that $Φ \circ λ = ϕ$ . Then there is a unique isomorphism $ψ : P \to_{}^{\sim} M \otimes_{A} N$ such that $ψ \circ λ = ι_{M, N}$ .

Remark 8.4.12. §

Let $A$ be an algebra over a commutative ring $R$ , let $M$ be a right $A$ -module, and let $N$ be a left $A$ -module. For any $m \in M$ and $n \in N$ , we have $m \otimes 0 = 0 = 0 \otimes n$ . For the first equality, note that $m \otimes 0 = 0 (m \otimes 0) = 0$ .

We give an example by way of a proposition.

Proposition 8.4.13. §

Let $m, n \geq 1$ . Then $(ℤ ∕ 𝑚ℤ) \otimes_{ℤ} (ℤ ∕ 𝑛ℤ) ≅ℤ ∕ (m, n) ℤ$ .

Proof.

Let $d = \gcd (m, n)$ , and write $d = 𝑎𝑚 + 𝑏𝑛$ for some $a, b \in ℤ$ . Note that $x \otimes y = 𝑥𝑦 (1 \otimes 1)$ , so $T = (ℤ ∕ 𝑚ℤ) \otimes_{ℤ} (ℤ ∕ 𝑛ℤ)$ is cyclic, and moreover,

d (1 \otimes 1) = (𝑎𝑚 + 𝑏𝑛) (1 \otimes 1) = a (m \otimes 1) + b (1 \otimes n) = 0,

so the order of $T$ divides $d$ .

We can define a bilinear map $ϕ : ℤ ∕ 𝑚ℤ \times ℤ ∕ 𝑛ℤ \to ℤ ∕ 𝑑ℤ$ by $ϕ (x, y) = 𝑥𝑦 mod d$ for $x \in ℤ ∕ 𝑚ℤ$ and $y \in ℤ ∕ 𝑛ℤ$ . We then have a homomorphism $Φ : T \to ℤ ∕ (m, n) ℤ$ with $Φ (1 \otimes 1) = 1$ , and it is therefore surjective. This forces $| T | = d$ and $Φ$ to be an isomorphism, as desired. □

Proposition 8.4.14. §

Let $A$ be an $R$ -algebra, let $M$ be a right $A$ -module, and let ${N_{i} ∣ i \in I}$ be a collection of left $A$ -modules. Then

M \otimes_{A} (⨁_{i \in I} N_{i}) ≅ ⨁_{i \in I} (M \otimes_{A} N_{i}) .

Proof.

First, define an $R$ -bilinear, $A$ -balanced map

ϕ : M \times (⨁_{i \in I} N_{i}) \to ⨁_{i \in I} (M \otimes_{A} N_{i}), ϕ (m, \sum_{i \in I} n_{i}) = \sum_{i \in I} m \otimes n_{i} .

This induces an $R$ -module homomorphism

Φ : M \otimes_{A} (⨁_{i \in I} N_{i}) ≅ ⨁_{i \in I} (M \otimes_{A} N_{i})

with $Φ (m \otimes n_{i}) = m \otimes n_{i}$ for $m \in M$ and $n_{i} \in I$ for some $i \in I$ .

Next, define $R$ -bilinear, $A$ -balanced maps

ψ_{i} : M \times N_{i} \to M \otimes_{A} (⨁_{i \in I} N_{i}), ψ_{i} (m, n_{i}) = m \otimes n_{i} .

The collection ${(ψ_{i})}_{i \in I}$ gives rise to a unique $R$ -module homomorphism

Ψ : ⨁_{i \in I} (M \otimes_{A} N_{i}) \to M \otimes_{A} (⨁_{i \in I} N_{i}),

satisfying $Ψ (m \otimes n_{i}) = m \otimes n_{i}$ for $m$ and $n_{i}$ as above by Proposition 8.3.9b. By definition, the maps $Φ$ and $Ψ$ are inverse to each other. □

Proposition 8.4.15. §

Let $M$ and $N$ be modules over a commutative ring $R$ . Then there is a unique isomorphism of $R$ -modules

M \otimes_{R} N \to_{}^{\sim} N \otimes_{R} M, m \otimes n \mapsto n \otimes m .

Proof.

Consider the $R$ -bilinear map $ϕ : M \times N \to N \otimes_{R} M$ given by $ϕ (m, n) = n \otimes m$ . It induces an $R$ -module homomorphism $Φ : M \otimes_{R} N \to N \otimes_{R} M$ satisfying $Φ (m \otimes n) = n \otimes m$ by the universal property of the tensor product. It then has inverse the similarly defined map $Ψ : N \otimes_{R} M \to M \otimes_{R} N$ with $Ψ (n \otimes m) = m \otimes n$ . □

Remark 8.4.16. §

We can allow a tensor product over an arbitrary $R$ -algebra $A$ in Proposition 8.4.15, but we obtain $M \otimes_{A} 𝑁≅𝑁 \otimes_{A^{op}} M$ as $R$ -modules (noting that $A^{op}$ has the same $R$ -module structure as $A$ ).

Example 8.4.17. §

Let $R$ be a commutative ring. The tensor product $R^{m} \otimes_{R} R^{n}$ is a free $R$ -module of rank $𝑚𝑛$ with basis ${e_{i} \otimes e_{j} ∣ 1 \leq i \leq m, 1 \leq j \leq n}$ . This follows immediately from Proposition 8.4.14 (and Proposition 8.4.15) and the fact that $R \otimes_{R} 𝑅≅𝑅$ . Here, the latter isomorphism is induced the $R$ -bilinear map $(x, y) \mapsto 𝑥𝑦$ , its inverse being the map $R \to R \otimes_{R} R$ with $x \mapsto x \otimes 1$ .

Proposition 8.4.18. §

Let $A$ and $B$ be $R$ -algebras. Let $L$ be a right $A$ -module, let $M$ be an $R$ -balanced $A$ - $B$ -bimodule, and let $N$ be a left $B$ -module. Then there is a unique isomorphism of $R$ -modules

(L \otimes_{A} M) \otimes_{B} N \to_{}^{\sim} L \otimes_{A} (M \otimes_{B} N), (l \otimes m) \otimes n \mapsto l \otimes (m \otimes n) .

Proof.

Let $ϕ : (L \otimes_{A} M) \times N \to P$ be an $R$ -bilinear, $B$ -balanced map to some $R$ -module $P$ . This gives rise to a map

ψ = ϕ \circ (ι_{L, M} \times {id}_{N}) : L \times M \times N \to P

which is $R$ -linear in each variable separately and satisfies

\begin{array}{rcl} ψ (𝑙𝑎, m, n) = ψ (l, 𝑎𝑚, n) & and & ψ (l, 𝑚𝑏, n) = ψ (l, m, 𝑏𝑛) \end{array}

for all $a \in A$ , $b \in B$ , $l \in L$ , $m \in M$ , and $n \in N$ . In particular, for each $l \in L$ , we obtain an $R$ -module homomorphism $Ψ_{l} : M \otimes_{B} N \to P$ with $Ψ_{l} (m \otimes n) = ψ (l, m, n)$ by the fact that $ψ_{l} : M \times N \to P$ with $ψ_{l} (m, n) = ψ (l, m, n)$ is $R$ -bilinear and $B$ -balanced. We then obtain an $R$ -bilinear, $A$ -balanced map

𝜃 : L \times (M \otimes_{B} N) \to P, 𝜃 (l, m \otimes n) = ψ_{l} (m \otimes n) = ψ (l, m, n)

which in turn induces an $R$ -module homomorphism

Θ : L \otimes_{A} (M \otimes_{B} N) \to P, Θ (l \otimes (m \otimes n)) = ψ (l, m, n) .

Since the elements $l \otimes (m \otimes n)$ generate $L \otimes_{A} (M \otimes_{B} N)$ , this is the unique homomorphism that agrees with $ψ$ on these simple tensors. Since $ψ (l, m, n) = ϕ (l \otimes m, n)$ , the $R$ -module $L \otimes_{A} (M \otimes_{B} N)$ satisfies the universal property of the tensor product $L \otimes_{A} (M \otimes_{B} N)$ , hence is canonically isomorphic to it via the indicated map, as in Proposition 8.4.11. □

Lemma 8.4.19. §

Let $A$ be an $R$ -algebra. Let $M$ and $M^{'}$ be right $A$ -modules, and let $N$ and $N^{'}$ be left $A$ -bmodules. Let $ϕ : M \to M^{'}$ and $ψ : N \to N^{'}$ be homomorphisms of left and right $A$ -modules, respectively. Then there exists a homomorphism of $R$ -modules

ϕ \otimes ψ : M \otimes_{A} N \to M^{'} \otimes_{A} N^{'}, (ϕ \otimes ψ) (m \otimes n) = ϕ (m) \otimes ψ (n) .

Proof.

The map $𝜃 : M \times N \to M^{'} \times N^{'}$ with $𝜃 (m, n) = ϕ (m) \otimes ψ (n)$ is immediately seen to be $R$ -bilinear, and it is $A$ -balanced since

ϕ (𝑚𝑎) \otimes ψ (n) = ϕ (m) a \otimes ψ (n) = ϕ (m) \otimes 𝑎𝜓 (n) = ϕ (m) \otimes ψ (𝑎𝑛) .

Thus, it induces an $R$ -module homomorphism $M \otimes_{A} N \to M^{'} \otimes_{A} N^{'}$ with the desired property. □

We can also form the tensor product of $R$ -algebras.

Proposition 8.4.20. §

Let $A$ and $B$ be algebras over a commutative ring $R$ . The tensor product $A \otimes_{R} B$ is an $R$ -algebra under the unique multiplication satisfying

(a \otimes b) (a^{'} \otimes b^{'}) = a a^{'} \otimes b b^{'} .

for $a, a^{'} \in A$ and $b, b^{'} \in B$ .

Proof.

First, we should check the desired multiplication on $A \otimes_{R} B$ is well-defined. To start, given $a \in A$ and $b \in B$ , we claim that the map $A \times B \to A \otimes_{R} B$ given by $(a^{'}, b^{'}) \mapsto a a^{'} \otimes b b^{'}$ is $R$ -bilinear (and therefore $R$ -balanced). To see this, we merely note that $a (r a^{'}) \otimes b b^{'} = r (a a^{'} \otimes b b^{'})$ and $a (a^{'} + a^{″}) \otimes b b^{'} = a a^{'} \otimes b b^{'} + a a^{″} \otimes b b^{'}$ . Therefore, we obtain a well-defined map

ψ : A \times B \to {End}_{R} (A \otimes_{R} B), ψ (a, b) (a^{'} \otimes b^{'}) = a a^{'} \otimes b b^{'} .

Note also that $ψ$ is $R$ -bilinear as well, so we obtain an $R$ -module homomorphism $A \otimes_{R} B \to {End}_{R} (A \otimes_{R} B)$ , which we may rewrite then as a well-defined operation

(A \otimes_{R} B) \times (A \otimes_{R} B) \to A \otimes_{R} B, (a \otimes b, a^{'} \otimes b^{'}) \mapsto a a^{'} \otimes b b^{'} .

This operation is $R$ -bilinear by what we have said. As it clearly satisfies $(1 \otimes 1) (a^{'} \otimes b^{'}) = a^{'} \otimes b^{'}$ , so we need only observe its associativity to finish the proof of the result. This can be checked on simple tensors, for which it is in an immediate consequence of the associativity of the operations on $A$ and $B$ . □

Proposition 8.4.21. §

Let $A$ and $B$ be algebras over a commutative ring $R$ . An abelian group $M$ that is a left $A$ -module and a right $B$ -module is an $R$ -balanced $A$ - $B$ -bimodule if and only if it is an $A \otimes_{R} B^{op}$ -module under the action $(a \otimes b) m = (𝑎𝑚) b$ .

Proof.

Let $M$ be an $R$ -balanced $A$ - $B$ -bimodule. We endow it with an $A \times B^{op}$ -action by $(a, b) m = 𝑎𝑚𝑏$ . This is action is $R$ -bilinear, so it factors through an action of $A \otimes_{R} B^{op}$ that clearly satisfies $(1 \otimes 1) m = m$ and $(a \otimes b) (m + m^{'}) = (a \otimes b) m + (a \otimes b) m^{'}$ and therefore makes $M$ into an $A \otimes_{R} B^{op}$ -module.

Conversely, if $M$ is an $A \otimes_{R} B^{op}$ -module, it is in particular an $R$ -balanced $A$ - $B$ -bimodule via the actions $𝑎𝑚 = (a \otimes 1) m$ and $𝑚𝑏 = (1 \otimes b) m$ , as the reader may quickly verify. □

When $M$ and $N$ have $R$ -balanced bimodule structures, we can also attain a bimodule structure on their tensor product.

Proposition 8.4.22. §

Let $A$ , $B$ , and $C$ be $R$ -algebras over a commutative ring $R$ . Let $M$ be an $A$ - $B$ -bimodule and $N$ be an $R$ -balanced $B$ - $C$ -bimodule. Then $M \otimes_{B} N$ is an $R$ -balanced $A$ - $C$ -bimodule with respect to actions satisfying

\begin{array}{rcl} a (m \otimes n) = (𝑎𝑚) \otimes n & and & (m \otimes n) c = m \otimes (𝑛𝑐) \end{array}

for all $a \in A$ , $c \in C$ , $m \in M$ , and $n \in N$ .

Proof.

For $a \in A$ and $c \in C$ , we can define an $R$ -bilinear map

ϕ : M \times N \to M \otimes_{B} N, (m, n) \mapsto (𝑎𝑚) \otimes (𝑛𝑐),

noting that

ϕ_{a, c} (𝑟𝑚 + m^{'}, n) = (a (𝑚𝑟 + m^{'})) \otimes (𝑛𝑐) = r (𝑎𝑚 \otimes 𝑛𝑐) + a m^{'} \otimes 𝑛𝑐 = 𝑟𝜙 (m, n) + ϕ (m, n^{'})

and similarly for the second variable. We thus have an induced map

Φ_{a, c} : M \otimes_{B} N \to M \otimes_{B} N, m \otimes n \mapsto (𝑎𝑚) \otimes_{B} (𝑛𝑐)

of $R$ -modules. The map

A \times C^{op} \to {End}_{R} (M \otimes_{B} N), (a, c) \to Φ_{a, c}

then defines an $R$ -algebra homomorphism. In other words, this gives $M \otimes_{B} N$ the structure of a left $A \otimes_{R} C^{op}$ -module. □

We give an application.

Proposition 8.4.23. §

Let $A$ be a ring, let $M$ be a left $A$ -module, and let $I$ be a two-sided ideal of $A$ . Then there is an isomorphism of left $A$ -modules

M ∕ 𝐼𝑀 \to_{}^{\sim} A ∕ I \otimes_{A} M, m + 𝐼𝑀 \mapsto 1 \otimes m .

Proof.

Note that $A ∕ I$ is an $A$ - $A$ -bimodule, so $A ∕ I \otimes_{A} M$ has the structure of an $A$ -module by Proposition 8.4.22. In one direction, we can define an $A$ -module homomorphism $Ψ : M \to A ∕ I \otimes_{A} M$ by $Ψ (m) = 1 \otimes m$ . In the other, we can define an left $A$ -linear, $A$ -balanced map $ϕ : A ∕ I \times M \to M ∕ 𝐼𝑀$ by $ϕ (a + I, m) = 𝑎𝑚 + 𝐼𝑀$ , which induces an $A$ -module homomorphism $Φ : A ∕ I \otimes_{A} M \to M ∕ 𝐼𝑀$ which is clearly inverse to $Ψ$ . □

We have the following direct corollary.

Corollary 8.4.24. §

Let $A$ be a ring, and let $M$ be a left $A$ -module. Then $𝑀≅𝐴 \otimes_{A} M$ as $A$ -modules.

Remark 8.4.25. §

Note that Proposition 8.4.23 requires $I$ to be a two-sided ideal, though the definition of $M ∕ 𝐼𝑀$ only requires $I$ to be a left ideal. That is, we need a right $A$ -action on $A ∕ I$ in order to define $A ∕ I \otimes_{A} M$ . We cannot take $M \otimes_{A} A ∕ I$ either, as $M$ is a left $A$ -module.

Corollary 8.4.26. §

Let $M$ and $N$ be free modules over a commutative ring $R$ with bases $B$ and $C$ , respectively. Then $M \otimes_{R} N$ is a free $R$ -module with basis ${m \otimes n ∣ m \in B, n \in C}$ .

Proof.

Note that $𝑀≅ \oplus_{m \in B} R$ and $𝑁≅ \oplus_{n \in C} R$ via the isomorphisms given by the bases $B$ and $C$ . Since tensor products and direct sums commute, and $R \otimes_{R} 𝑅≅𝑅$ by Corollary 8.4.24, we have isomorphisms

M \otimes_{R} 𝑁≅ ⨁_{m \in B} R \otimes_{R} N ≅ ⨁_{m \in B} ⨁_{n \in C} R \otimes_{R} R ≅ ⨁_{(m, n) \in B \times C} R

such that the composite identifies $m \otimes n$ with $1$ in the $(m, n)$ -coordinate of the direct sum. In particular, ${m \otimes n ∣ m \in B, n \in C}$ maps to the standard basis of the direct sum, so is a basis of $M \otimes_{R} N$ . □

The following corollary is also useful.

Corollary 8.4.27. §

Let $R$ be a commutative ring, let $I$ be an ideal of $R$ , let $M$ and $N$ be left $R$ -modules. Suppose that $𝐼𝑀 = 0$ , so $M$ can also be viewed as an $R ∕ I$ -module. Then there is an isomorphism of $R$ -modules

M \otimes_{R} N \to_{}^{\sim} M \otimes_{R ∕ I} N ∕ 𝐼𝑁, m \otimes n \mapsto m \otimes (n + 𝐼𝑁) .

Proof.

By Proposition 8.4.23, we have $𝑀≅𝑅 ∕ I \otimes_{R} M$ and $N ∕ 𝐼𝑁≅𝑅 ∕ I \otimes_{R} N$ , so by associativity and commutativity of tensor products over $R$ , we have

M \otimes_{R} 𝑁≅𝑅 ∕ I \otimes_{R} M \otimes_{R} 𝑁≅𝑀 \otimes_{R} N ∕ 𝐼𝑁

via the $R$ -modules homomorphism which sends $m \otimes n$ to $m \otimes (n + 𝐼𝑁)$ for $m \in M$ and $n \in N$ . Finally, note that the canonical maps from $M \times N$ to simple tensor products in $M \otimes_{R} N ∕ 𝐼𝑁$ and $M \otimes_{R ∕ I} N ∕ 𝐼𝑁$ are both $R ∕ I$ -bilinear, and therefore induce inverse maps between $M \otimes_{R} N ∕ 𝐼𝑁$ and $M \otimes_{R ∕ I} N ∕ 𝐼𝑁$ . □

Here is an interesting comparison of tensor products and homomorphism groups in the case of vector spaces.

Lemma 8.4.28. §

Let $V$ and $W$ be finite-dimensional vector spaces over a field $F$ . Then we have an $F$ -linear isomorphism

Ψ : V^{*} \otimes_{F} W \to_{}^{\sim} {Hom}_{F} (V, W), Ψ (ϕ \otimes w) (v) = ϕ (v) w

for $ϕ \in V^{*}$ , $v \in V$ , and $w \in W$ .

Proof.

One checks directly that the map $ψ : V^{*} \times W \to {Hom}_{F} (V, W)$ with $ψ (ϕ, w) (v) = ϕ (v) w$ is $F$ -bilinear, thus induces a map on the tensor product. Let $B$ be a basis of $V$ and $C$ be a basis of $W$ . For each $v \in B$ , and $φ \in {Hom}_{F} (V, W)$ , write

φ (v) = \sum_{w \in C} a_{v, w} w .

Define $ϕ_{w} \in V^{*}$ for $w \in C$ by $ϕ_{w} (v) = a_{v, w}$ for $v \in B$ . We can then define $Θ : {Hom}_{F} (V, W) \to V^{*} \otimes_{F} W$ by

Θ (φ) = \sum_{w \in C} ϕ_{w} \otimes w .

By definition, $Ψ (Θ (φ)) (v) = ϕ (v)$ and

Θ (Ψ (ϕ \otimes w)) = Θ (v \mapsto ϕ (v) w) = ϕ \otimes w .

□

Theorem 8.4.29. §

Let $A$ and $B$ be algebras over a commutative ring $R$ . Let $M$ be an $R$ -balanced $A$ - $B$ -bimodule, let $N$ be a left $B$ module, and let $L$ be a left $A$ -module. Then there is an isomorphism of $R$ -modules

Ξ : {Hom}_{A} (M \otimes_{B} N, L) \to_{}^{\sim} {Hom}_{B} (N, {Hom}_{A} (M, L))

given by

Ξ (f) (n) (m) = f (m \otimes n)

for all $f \in {Hom}_{A} (M \otimes_{B} N, L)$ , $m \in M$ and $n \in N$ .

Proof.

First, define $Ξ$ as in the statement of the theorem. Note that

\begin{matrix} Ξ (f) (n) (𝑎𝑚 + m^{'}) = f ((𝑎𝑚 + m^{'}) \otimes n) = f (a (m \otimes n) + (m^{'} \otimes n)) \\ = 𝑎𝑓 (m \otimes n) + f (m^{'} \otimes n) = a Ξ (f) (n) (m) + Ξ (f) (n) (m^{'}), \end{matrix}

so $Ξ (f) (n)$ is a homomorphism of $A$ -modules. Moreover,

\begin{matrix} Ξ (f) (b n^{'} + n^{″}) (m) = f (m \otimes (b n^{'} + n^{″})) = f (𝑚𝑏 \otimes n^{'}) + f (m \otimes n^{″}) \\ = Ξ (f) (𝑚𝑏) (n^{'}) + Ξ (f) (m) (n^{″}) = (b Ξ (f)) (m) (n^{'}) + Ξ (f) (m) (n^{″}), \end{matrix}

so $Ξ (f)$ is a homomorphism of $B$ -modules. Thus, $Ξ$ is well-defined. In addition,

\begin{matrix} Ξ (𝑟𝑓) (n) (m) = (𝑟𝑓) (m \otimes n) = f (r (m \otimes n)) = f (𝑚𝑟 \otimes n) \\ = f (m \otimes 𝑟𝑛) = Ξ (f) (𝑟𝑛) (m) = (r Ξ (f)) (n) (m) \end{matrix}

so $Ξ (𝑟𝑓) = r Ξ (f)$ , and since $Ξ$ is also clearly a homomorphism of abelian groups, $Ξ$ is a homomorphism of $R$ -modules.

To finish the proof, we must exhibit an inverse to $Ξ$ . For this, suppose we are given $λ \in {Hom}_{B} (N, {Hom}_{A} (M, L))$ and define $ϕ : M \times N \to L$ by $ϕ (m, n) = λ (n) (m)$ . This map satisfies

\begin{array}{rcl} ϕ (𝑟𝑚 + m^{'}, n) = λ (n) (𝑟𝑚 + m^{'}) = 𝑟𝜆 (n) (m) + λ (n) (m^{'}) = 𝑟𝜙 (m, n) + ϕ (m^{'}, n), \\ ϕ (m, 𝑟𝑛 + n^{'}) = (𝑟𝜆 (n)) (m) + λ (n^{'}) (m) = 𝑟𝜆 (n) (m) + λ (n^{'}) (m) = 𝑟𝜙 (m, n) + ϕ (m, n^{'}), \\ ϕ (𝑚𝑏, n) = λ (𝑚𝑏) (n) = (λ (m) b) (n) = λ (m) (𝑏𝑛) = ϕ (m, 𝑏𝑛), \\ ϕ (𝑎𝑚, n) = λ (𝑎𝑚) (n) = (𝑎𝜆 (m)) (n) = 𝑎𝜆 (m) (n) = 𝑎𝜙 (m, n) \end{array}

for all $m, m^{'} \in M$ , $n, n^{'} \in N$ , $r \in R$ , $a \in A$ , and $b \in B$ . Thus, $ϕ$ induces a unique map $Φ : M \otimes_{B} N \to L$ of $A$ -modules with $Φ (m \otimes n) = λ (n) (m)$ . The map $λ \to Φ$ is then by definition inverse to $Ξ$ , which tells us that $Ξ$ is a bijection, hence an isomorphism. □

Remark 8.4.30. §

If we suppose in Theorem 8.4.29 that $N$ is an $R$ -balanced $B$ - $C$ -bimodule and $L$ is an $R$ -balanced $A$ - $D$ -bimodule for $R$ -algebras $C$ and $D$ , then the isomorphism $Ξ$ is one of $R$ -balanced $C$ - $D$ -bimodules.

8.5. Exterior powers

In this section, $R$ will denote a commutative ring.

Definition 8.5.1. §

Let $M$ be an $R$ -module. For a nonnegative integer $k$ , the $k$ th tensor power $M^{\otimes k}$ of $M$ over $R$ is the tensor product $M \otimes_{R} M \otimes_{R} \dots \otimes_{R} M$ of $k$ copies of $M$ if $k$ is positive and $R$ if $k = 0$ .

Definition 8.5.2. §

Let $M_{1}, M_{2}, \dots, M_{k}$ and $N$ be $R$ -modules for some $k \geq 1$ . A map $f : M_{1} \times M_{2} \times \dots \times M_{k} \to N$ is said to be $R$ -multilinear if it is $R$ -linear in each of its $k$ variables, which is to say that

\begin{matrix} f (m_{1}, m_{2}, \dots, m_{i - 1}, r m_{i} + m_{i}^{'}, m_{i + 1}, \dots, m_{k}) \\ = f (m_{1}, m_{2}, \dots, m_{i - 1}, r m_{i}, m_{i + 1}, \dots, m_{k}) + f (m_{1}, m_{2}, \dots, m_{i - 1}, m_{i}^{'}, m_{i + 1}, \dots, m_{k}) \end{matrix}

for $r \in R$ and all $m_{j}$ and $m_{j}^{'} \in M_{j}$ for $1 \leq j \leq k$ .

The reader will quickly check the following.

Proposition 8.5.3. §

Let $M_{1}, M_{2}, \dots, M_{k}$ and $N$ be $R$ -modules for some $k \geq 1$ . For an $R$ -multilinear map $𝜃 : \prod_{i = 1}^{k} M_{i} \to N$ , there exists a unique $R$ -module homomorphism

Θ : M_{1} \otimes_{R} M_{2} \otimes_{R} \dots \otimes_{R} M_{k} \to N

such that $Θ (m_{1} \otimes m_{2} \otimes \dots \otimes m_{k}) = 𝜃 (m_{1}, m_{2}, \dots, m_{k})$ for all $m_{i} \in M_{i}$ with $1 \leq i \leq k$ .

Definition 8.5.4. §

Let $M$ be a module over a commutative ring $R$ . For a nonnegative integer $k$ , the $k$ th exterior power $\land^{k} M$ is the quotient of $M^{\otimes k}$ by the $R$ -submodule generated by the elements of the form $m_{1} \otimes m_{2} \otimes \dots \otimes m_{k}$ , where $m_{i} = m_{j}$ for some $1 \leq i < j \leq k$ . The image of a tensor $m_{1} \otimes m_{2} \otimes \dots \otimes m_{k}$ in $\land^{k} M$ is denoted $m_{1} \land m_{2} \land \dots \land m_{k}$ .

Remark 8.5.5. §

The $k$ th exterior power of a module $M$ is often referred to as the wedge product of $M$ with itself $k$ times.

Definition 8.5.6. §

Let $M$ and $N$ be abelian groups. A multilinear map $f : M^{k} \to N$ is said to be alternating if

f (m_{1}, m_{2}, \dots, m_{k}) = 0

for any $m_{j} \in M$ for $1 \leq j \leq k$ such that $m_{i} = m_{i + 1}$ for some $1 \leq i \leq k - 1$ .

Remark 8.5.7. §

There is an alternating, $R$ -bilinear map $κ : M^{k} \to \land^{k} M$ for any $k \geq 0$ such that $κ (m_{1}, m_{2}, \dots, m_{k}) = m_{1} \land m_{2} \land \dots \land m_{k}$ for all $m_{j} \in M$ for $1 \leq j \leq k$

Proposition 8.5.8. §

Let $M$ and $N$ be $R$ -modules, and let $ψ : M^{k} \to N$ be $R$ -multilinear and alternating. Then there exists a unique $R$ -module homomorphism $Ψ : \land^{k} M \to N$ such that

Ψ (m_{1} \land m_{2} \land \dots \land m_{k}) = ψ (m_{1}, m_{2}, \dots, m_{k})

for all $m_{i} \in M$ for $1 \leq i \leq k$ .

Proof.

Since $ψ$ is $R$ -multilinear, there exists by Proposition 8.5.3 a unique $R$ -module homomorphism $Θ : M^{\otimes k} \to N$ with $Θ (m_{1} \otimes m_{2} \otimes \dots \otimes m_{k}) = ψ (m_{1}, m_{2}, \dots, m_{k})$ for all $m_{i} \in M$ for $1 \leq i \leq k$ . If $m_{i} = m_{i + 1}$ for some $1 \leq i \leq k - 1$ , then

Θ (m_{1} \otimes m_{2} \otimes \dots \otimes m_{k}) = ψ (m_{1}, m_{2}, \dots, m_{k}) = 0

as $ψ$ is alternating, so $Θ$ factors through the desired map $Ψ : \land^{k} M \to N$ .

If $Ψ^{'} : \land^{k} M \to N$ also has the property of the proposition, then we may compose $Ψ$ with the quotient map $κ : M^{\otimes k} \to N$ to obtain a map $Θ^{'} = Ψ \circ κ : M^{\otimes k} \to N$ that satisfies the universal property of Proposition 8.5.3, hence is equal to $Θ$ . This then forces the equality $Ψ^{'} = Ψ$ for the induced maps on the exterior product. □

We leave the following to the reader.

Lemma 8.5.9. §

Let $φ : M \to N$ be a homomorphism of $R$ -modules. Then for any $k \geq 0$ , there exists a homomorphism $\land^{k} φ : ⋀^{k} M \to ⋀^{k} N$ satisfying

(⋀^{k} φ) (m_{1} \land m_{2} \land \dots \land m_{k}) = φ (m_{1}) \land φ (m_{2}) \land \dots \land φ (m_{k}) .

Lemma 8.5.10. §

Let $M$ be an $R$ -module. Then we have

m_{1} \land m_{2} \land \dots \land m_{k} = - m_{τ_{i} (1)} \land m_{τ_{i} (2)} \land \dots \land m_{τ_{i} (k)},

where $τ_{i} = (i i + 1) \in S_{k}$ , for all $1 \leq i \leq k$ and all $m_{j} \in M$ for $1 \leq j \leq k$ .

Proof.

The proof in the general case amounts to the following calculation in the case $k = 2$ . For any $m, n \in M$ , we have

0 = (m + n) \land (m + n) = m \land m + m \land n + n \land m + n \land n = m \land n + n \land m,

so $m \land n = - n \land m$ . □

Remark 8.5.11. §

The property that $m \land n = - n \land m$ for all $m, n \in M$ tells us directly that $m \land m = - m \land m$ , and so $2 m \land m = 0$ , by taking $m = n$ . In other words, if $2$ is invertible in $R$ , the submodule of $M \otimes_{R} M$ generated by tensors of the form $m \otimes n + n \otimes m$ contains the tensors of the form $m \otimes m$ .

Theorem 8.5.12. §

Let $M$ be a free $R$ -module of rank $n$ . Then the $k$ th exterior power $\land^{k} M$ of $M$ over $R$ is a free $R$ -module of rank $(\binom{n}{k})$ for any $k \in ℤ$ , where we take $(\binom{n}{k}) = 0$ for $k > n$ .

Proof.

Let $m_{1}, \dots, m_{n}$ be a basis of $M$ . The $0$ th exterior power is just $R$ , so the result holds for $k = 0$ . For $k \geq 1$ , we know that $M^{\otimes k}$ is $R$ -free with a basis of elements of the form $m_{i_{1}} \otimes m_{i_{2}} \otimes \dots \otimes m_{i_{k}}$ with $1 \leq i_{j} \leq n$ for each $1 \leq j \leq k$ . Since we can switch the orders of the terms of elements of $\land^{k} M$ with only a change of sign, we have that $\land^{k} M$ is generated by the $m_{i_{1}} \land m_{i_{2}} \land \dots \land m_{i_{k}}$ with $1 \leq i_{1} \leq i_{2} \leq \dots \leq i_{k} \leq n$ . But by definition of the exterior product, those elements with $i_{j} = i_{j + 1}$ for some $j$ are $0$ , so it is generated by those with $1 \leq i_{1} < i_{2} < \dots < i_{k} \leq n$ . The number of such elements is $(\binom{n}{k})$ .

It remains only to see $R$ -linear independence. For this, fix $1 \leq i_{1}^{'} < i_{2}^{'} < \dots < i_{k}^{'} \leq n$ , and define $f : M^{k} \to R$ as the unique $R$ -multilinear map satisfying that $f (m_{i_{1}}, m_{i_{2}}, \dots, m_{i_{k}})$ equals $0$ unless ${i_{1}, \dots, i_{k}} = {i_{1}^{'}, \dots, i_{k}^{'}}$ , in which case it is $sign (σ)$ for $σ \in S_{n}$ such that $σ (i_{j}) = i_{j}^{'}$ for $1 \leq j \leq k$ and $σ$ fixes every other element of ${1, 2, \dots, n}$ . (Recall from Proposition 4.12.1 that the sign map can be defined independently of the definition of the determinant, so as to avoid circularity in our argument.) That this map is alternating can be easily checked: let $m = \sum_{i = 1}^{n} r_{i} m_{i} \in M$ , and consider

f (\dots, m, m, \dots) = \sum_{i = 1}^{n} \sum_{j = 1}^{n} r_{i} r_{j} f (\dots, m_{i}, m_{j}, \dots) .

We then note that

f (\dots, m_{i}, m_{j}, \dots) + f (\dots, m_{j}, m_{i}, \dots) = 0

for $i \neq j$ and $f (\dots, m_{i}, m_{i}, \dots) = 0$ for all $i$ to see that the sum is trivial. The map $f$ then induces an element $F \in {Hom}_{R} (\land^{k} M, R)$ . Given some nontrivial $R$ -linear combination $x$ in $\land^{k} M$ of the generators $m_{r_{1}} \land m_{r_{2}} \land \dots \land m_{r_{k}}$ with $r_{1} < r_{2} < \dots < r_{k}$ , the value $F (x)$ is also the coefficient of $m_{i_{1}} \land m_{i_{2}} \land \dots \land m_{i_{k}}$ in the linear combination $x$ . So, if $x = 0$ , then the linear combination must be the zero linear combination, which verifies $R$ -linear independence. □

Corollary 8.5.13. §

The $R$ -module $\land^{n} R^{n}$ is one-dimensional with basis vector $e_{1} \land e_{2} \land \dots \land e_{n}$ , where ${e_{1}, e_{2}, \dots, e_{n}}$ is the standard basis of $R^{n}$ such that $e_{i} = {(δ_{i, j})}_{j}$ with $δ_{i, j} \in {0, 1}$ equal to $1$ if and only if $i = j$ .

8.6. Graded rings

Definition 8.6.1. §

A graded ring $A$ is a ring determined by a sequence of abelian groups $A_{i}$ for $i \geq 0$ and biadditive maps $ϕ_{i, j} : A_{i} \times A_{j} \to A_{i + j}$ for $i, j \geq 0$ satisfying

ϕ_{i + j, k} (ϕ_{i, j} (r_{i}, r_{j}), r_{k}) = ϕ_{i, j + k} (r_{i}, ϕ_{j, k} (r_{j}, r_{k}))

for $r_{i} \in A_{i}$ , $r_{j} \in A_{j}$ , $r_{k} \in A_{k}$ and $i, j, k \geq 0$ and such that $A_{0}$ is a ring with multiplication $ϕ_{0, 0}$ , where the additive group of $A$ is $\oplus_{i = 0}^{\infty} A_{i}$ and the multiplication on $A$ is given by

(\sum_{i = 0}^{\infty} r_{i}) \cdot (\sum_{i = 0}^{\infty} s_{i}) = \sum_{k = 0}^{\infty} \sum_{i = 0}^{k} ϕ_{i, k - i} (r_{i}, s_{k - i}),

where the sums are finite and $r_{i}, s_{i} \in A$ for all $i$ . The group $A_{i}$ is called the degree $i$ part, or $i$ th graded piece, of $A$ , and an element of $A_{i}$ is said to be homogeneous of degree $i$ .

Definition 8.6.2. §

A graded algebra over a commutative ring $R$ is an $R$ -algebra $A$ that is a graded ring with structure map $R \to A_{0} \cap Z (A)$ .

Definition 8.6.3. §

For a commutative ring $R$ , an homomorphism of graded $R$ -algebras $ψ : A \to B$ is a homomorphism of rings such that $ψ (A_{i}) \subseteq B_{i}$ for each $i \geq 0$ .

Clearly if $R$ has a grading, then $R$ is a graded ring with respect to the resulting subgroups and maps.

Definition 8.6.4. §

A grading on a ring $R$ is a sequence of additive subgroups $R_{i}$ with $i \geq 0$ such that $R_{0}$ is a subring and $R = \oplus_{i = 0}^{\infty} R_{i}$ such that the multiplication on $R$ restricts to maps $ϕ_{i, j} : R_{i} \times R_{j} \to R_{i + j}$ for all $i, j \geq 0$ . We say that $R$ is graded by the $R_{i}$ .

Example 8.6.5. §

Any commutative or noncommutative polynomial ring $R$ on a set $X$ has a grading under which the $n$ th graded piece is the $R$ -span of of the words in $X$ of length $n$ . In fact, there are many possible gradings by assigning arbitrary choices of positive degrees to the different elements of $X$ .

Example 8.6.6. §

Given a ring $A$ and an ideal $I$ , we may form the graded ring ${gr}_{I} A = \oplus_{n = 0}^{\infty} I^{n} ∕ I^{n + 1}$ , where the maps $I^{i} ∕ I^{i + 1} \times I^{j} ∕ I^{j + 1} \to I^{i + j} ∕ I^{i + j + 1}$ are given by $(x + I^{i + 1}, y + I^{j + 1}) \mapsto 𝑥𝑦 + I^{i + j + 1}$ . If $A$ is an $R$ -algebra, then ${gr}_{I} A$ is a graded $R$ -algebra via the map $R \to A ∕ I$ .

We can form an algebra out of the tensor powers of a module.

Definition 8.6.7. §

For a commutative ring $R$ and $n \geq 0$ , the $n$ th tensor power of an $R$ -module $M$ is $T^{n} (M) = M^{\otimes n} = M \otimes_{R} \dots \otimes_{R} M$ , the $n$ -fold $R$ -tensor product of $M$ with itself, which is taken to be $R$ if $n = 0$ .

Definition 8.6.8. §

For a commutative ring $R$ and nonzero $R$ -module $M$ , the tensor algebra $T_{R} (M)$ of $M$ is the graded $R$ -algebra with $i$ th graded piece $T^{i} (M)$ together with the unique $R$ -bilinear maps $ϕ_{i, j} : T^{i} (M) \times T^{j} (M) \to T^{i + j} (M)$ satisfying

ϕ_{i, j} (m_{1} \otimes \dots \otimes m_{i}, n_{1} \otimes \dots n_{j}) = m_{1} \otimes \dots \otimes m_{i} \otimes n_{1} \otimes \dots n_{j},

where the $R$ -algebra structure map is the identity $R \to T^{0} (M)$

Example 8.6.9. §

The $R$ -tensor algebra of $R$ is isomorphic to $R [x]$ as a graded $R$ -algebra. That is, we have an isomorphism $ψ : R [x] \to T_{R} (R)$ of graded $R$ -algebras uniquely determined by $ψ (x) = 1 \in T^{1} (R)$ . More generally, the $R$ -tensor algebra of $R^{n}$ is isomorphic to $R ⟨ x_{1}, \dots, x_{n} ⟩$ as a graded algebra (where the $x_{i}$ have degree $1$ ).

Definition 8.6.10. §

A graded ideal of a graded ring is an ideal that has a homogeneous generating set.

The reader can verify the following.

Lemma 8.6.11. §

An ideal $I$ of a graded ring $A$ is homogeneous if and only if $I = \oplus_{n = 0}^{\infty} I_{n}$ , where $I_{n} = A_{n} \cap I$ for all $n \geq 0$ .

Lemma 8.6.12. §

The quotient of a graded $R$ -algebra $A$ by a homogeneous ideal $I$ is a graded $R$ -algebra with $i$ th graded piece $A_{i} ∕ (A_{i} \cap I)$ , where $A_{i}$ is the $i$ th graded piece of $A$ .

Definition 8.6.13. §

Let $R$ be a commutative ring and $M$ be an $R$ -module.

a.: The symmetric algebra $S_{R} (M)$ on a $R$ -module $M$ is the quotient of $T_{R} (M)$ by the homogeneous ideal generated by the elements $m \otimes n - n \otimes m$ with $m, n \in R$ .
b.: The $n$ th graded piece $S^{n} (M)$ of $S_{R} (M)$ is called the $n$ th symmetric power of $M$ .

Notation 8.6.14. §

For $M$ an $R$ -module and $x, y \in T_{R} (M)$ , the image of their product $x \otimes y$ in $S_{R} (M)$ is denoted $x \cdot y$ .

Example 8.6.15. §

The symmetric algebra $S_{R} (R^{n})$ is isomorphic to $R [x_{1}, \dots, x_{n}]$ .

Definition 8.6.16. §

Let $R$ be a commutative ring and $M$ be an $R$ -module. The exterior algebra $\land_{R} M$ on $M$ is the quotient of $T_{R} (M)$ by the homogeneous ideal generated by the elements $m \otimes m$ with $m \in R$ .

Notation 8.6.17. §

For $M$ an $R$ -module and $x, y \in T_{R} (M)$ , the image of their product $x \otimes y$ in $S_{R} (M)$ is denoted $x \land y$ .

Lemma 8.6.18. §

The multiplication on $\land_{R} M$ for an $R$ -module $M$ satisfies $x \land x = 0$ and $x \land y = - y \land x$ for all $x, y \in \land_{R} M$ .

Proof.

For any $x, y \in T_{R} (M)$ , we have

(x + y) \otimes (x + y) = x \otimes x + x \otimes y + y \otimes x + y \otimes y,

which reduces the problem to proving that $x \otimes x$ lies in the homogeneous ideal $I$ generated by the $m \otimes m$ for $m \in R$ . By the distributive property of multiplication, the result is further reduced to the case of simple tensors. For $m_{1}, \dots, m_{n} \in R$ , we claim that

m_{1} \otimes \dots \otimes m_{n} \otimes m_{1} \otimes \dots \otimes m_{n} \in I,

and for this it suffices to show that

m_{1} \otimes (m_{2} \otimes \dots \otimes m_{n}) \otimes m_{1} \in I,

This is clear if $n = 1$ . For $n \geq 2$ , from the case $n = 1$ it follows that $m_{1} \otimes m_{2} - m_{2} \otimes m_{1} \in I$ , which reduces us to showing that

m_{1} \otimes (m_{3} \otimes \dots \otimes m_{n}) \otimes m_{1} \in I,

which now follows by induction. □

The reader can now verify the following.

Lemma 8.6.19. §

For an $R$ -module $M$ and $n \geq 0$ , the $n$ th graded piece of $\land_{R} M$ is isomorphic to $\land^{n} M$ under the $R$ -linear map that takes the image of $m_{1} \otimes \dots \otimes m_{n}$ to $m_{1} \land \dots \land m_{n}$ for $m_{1}, \dots, m_{n} \in M$ .

8.7. Determinants

In this section, $R$ denotes a commutative ring.

Definition 8.7.1. §

Let $n \geq 1$ .

a.

The determinant $\det (A)$ of a matrix $A \in M_{n} (R)$ with columns $v_{1}, \dots, v_{n} \in R^{n}$ is the unique element of $R$ such that

v_{1} \land v_{2} \land \dots \land v_{n} = \det (A) \cdot e_{1} \land e_{2} \land \dots \land e_{n},

where $e_{i}$ denotes the $i$ th element in the standard basis of $R^{n}$ .

b.

The determinant map

\det : M_{n} (R) \to R

is the map that takes a matrix to its determinant.

Remark 8.7.2. §

The determinant map is an alternating, multilinear map if we view $M_{n} (R)$ as $\land^{n} R^{n}$ by taking a matrix to the wedge product $v_{1} \land v_{2} \land \dots \land v_{n}$ of its columns $v_{1}, v_{2}, \dots, v_{n}$ .

Proposition 8.7.3. §

The determinant map $\det : M_{n} (R) \to R$ satisfies

\det (A) = \sum_{σ \in S_{n}} sign (σ) a_{1 σ (1)} a_{2 σ (2)} \dots a_{𝑛𝜎 (n)}

for any $A = (a_{𝑖𝑗}) \in M_{n} (R)$ .

Proof.

Let $v_{1}, v_{2}, \dots, v_{n}$ denote the columns of $A = (a_{𝑖𝑗})$ . Then $v_{j} = \sum_{i = 1}^{n} a_{𝑖𝑗} e_{i}$ . We have

v_{1} \land v_{2} \land \dots \land v_{n} = \sum_{i_{1} = 1}^{n} \sum_{i_{2} = 1}^{n} \dots \sum_{i_{n} = 1}^{n} a_{i_{1} 1} a_{i_{2} 2} \dots a_{i_{n} n} \cdot e_{i_{1}} \land e_{i_{2}} \land \dots \land e_{i_{n}},

but note that all the terms such that the $j_{1}, j_{2}, \dots, j_{n}$ are not all distinct are zero. The remaining nonzero terms correspond to permutations $σ \in S_{n}$ with $σ (j) = i_{j}$ for each $1 \leq j \leq n$ . We then have

\begin{array}{l} v_{1} \land v_{2} \land \dots \land v_{n} & = \sum_{σ \in S_{n}} a_{σ (1) 1} a_{σ (2) 2} \dots a_{σ (n) n} \cdot e_{σ (1)} \land e_{σ (2)} \land \dots \land e_{σ (n)} \\ = \sum_{σ \in S_{n}} sign (σ) a_{σ (1) 1} a_{σ (2) 2} \dots a_{σ (n) n} \cdot e_{1} \land e_{2} \land \dots \land e_{n} \\ = \sum_{σ \in S_{n}} sign (σ) a_{1 σ (1)} a_{2 σ (2)} \dots a_{𝑛𝜎 (n)} \cdot e_{1} \land e_{2} \land \dots \land e_{n}, \end{array}

the latter step coming from rearranging the terms and replacing $σ$ by $σ^{- 1}$ . □

Lemma 8.7.4. §

Let $A, B \in M_{n} (R)$ for some $n \geq 1$ . Then

\det (𝐴𝐵) = \det (A) \det (B) .

Proof.

Let $v_{i}$ be the $i$ th column of $A$ , let $w_{i}$ be the $i$ th column of $B$ , and let $z_{i}$ be the $i$ th column of $𝐴𝐵$ . Then $A w_{i} = z_{i}$ for all $i$ . By Lemma 8.5.9, we then obtain

z_{1} \land z_{2} \land \dots \land z_{n} = \det (A) w_{1} \land w_{2} \land \dots \land w_{n} .

Since

w_{1} \land w_{2} \land \dots \land w_{n} = \det (B) e_{1} \land e_{2} \land \dots \land e_{n},

the result holds. □

Definition 8.7.5. §

Let $R$ be a ring. Two matrices $A$ and $A^{'}$ in $M_{n} (R)$ for some $n \geq 1$ are called similar if there exists a matrix $Q \in {GL}_{n} (R)$ such that $A^{'} = Q^{- 1} 𝐴𝑄$ .

Remark 8.7.6. §

Let $T : R^{n} \to R^{n}$ be a linear transformation represented by the matrix $A$ with respect to the standard basis of $R^{n}$ . If $A^{'} = Q^{- 1} 𝐴𝑄$ for $Q \in {GL}_{n} (R)$ , then $A^{'}$ represents $T$ with respect to the basis $B = {v_{1}, \dots, v_{n}}$ with $v_{j} = \sum_{i = 1}^{n} q_{𝑖𝑗} e_{i}$ for $1 \leq j \leq n$ . Conversely, any two matrices that each represent $T$ with respect to some basis are similar.

Lemma 8.7.4 has the following corollary.

Corollary 8.7.7. §

Let $R$ be a commutative ring.

a.: For any $A \in {GL}_{n} (R)$ , we have $\det (A) \det (A^{- 1}) = 1$ .
b.: Let $A$ and $B$ be similar matrices in $M_{n} (R)$ . Then $\det (A) = \det (B)$ .

Lemma 8.7.8. §

Let $A \in M_{n} (R)$ , and let $A^{T}$ denote its transpose. Then $\det (A^{T}) = \det (A)$ .

Proof.

Write $A = (a_{i, j})$ . By Proposition 8.7.3, we have

\det (A^{T}) = \sum_{σ \in S_{n}} sign (σ) a_{σ (1) 1} a_{σ (2) 2} \dots a_{σ (n) n},

but $a_{σ (j) j} = a_{σ (j) σ^{- 1} (σ (j))}$ , so

a_{σ (1) 1} a_{σ (2) 2} \dots a_{σ (n) n} = a_{1 σ^{- 1} (1)} a_{2 σ^{- 1} (2)} \dots a_{n σ^{- 1} (n)} .

Noting that $sign (σ) = sign (σ^{- 1})$ for all $σ \in S_{n}$ , we have

\det (A^{T}) = \sum_{σ \in S_{n}} sign (σ^{- 1}) a_{1 σ^{- 1} (1)} a_{2 σ^{- 1} (2)} \dots a_{n σ^{- 1} (n)} = \det (A) .

□

We also have the following standard properties of the determinant.

Lemma 8.7.9. §

Let $A \in M_{n} (R)$ .

a.: Let $B$ be a matrix obtained by switching either two rows or two columns of $A$ . Then $\det (B) = - \det (A)$ .
b.: Let $C$ be a matrix obtained by adding an $R$ -multiple of one row (resp., column) of $A$ to another row (resp., column). Then $\det (C) = \det (A)$ .
c.: Let $D$ be a matrix obtained by multiplying one row or column of $A$ by some $c \in R$ . Then $\det (D) = c \det (A)$ .

Proof.

By Lemma 8.7.8, it suffices to prove these for columns. Part a follows from the more general fact that

v_{σ (1)} \land v_{σ (2)} \land \dots \land v_{σ (n)} = sign (σ) v_{1} \land v_{2} \land \dots \land v_{n},

and part b follows from

\begin{array}{l} v_{1} \land \dots \land v_{i} \land \dots & \land (v_{j} + r v_{i}) \land \dots \land v_{n} \\ = (v_{1} \land \dots \land v_{i} \land \dots \land v_{j} \land \dots \land v_{n}) + r (v_{1} \land \dots \land v_{i} \land \dots \land v_{i} \land \dots \land v_{n}) \\ = v_{1} \land \dots \land v_{i} \land \dots \land v_{j} \land \dots \land v_{n} . \end{array}

Part c follows from the multilinearity of the exterior product. □

Lemma 8.7.10. §

Let $A \in M_{n} (R)$ be a block diagonal matrix with $A \in M_{n_{i}} (R)$ for $1 \leq i \leq m$ and some $m \geq 1$ . Then $\det (A) = \prod_{i = 1}^{m} \det (A_{i})$ .

Proof.

We have

A e_{1} \land A e_{2} \land \dots \land A e_{n} = \det (A_{1}) (e_{1} \land \dots \land e_{n_{i}}) \land \dots \land \det (A_{m}) (e_{n - n_{m} + 1} \land \dots \land e_{n}),

as required. □

Definition 8.7.11. §

For $A \in M_{n} (R)$ and $1 \leq i, j \leq n$ , the $(i, j)$ -minor of $A$ is the matrix $A_{𝑖𝑗} \in M_{n - 1} (R)$ obtained by removing the $i$ th row and $j$ th column from $A$ . The $(i, j)$ -cofactor of $A$ is ${(- 1)}^{i + j} \det (A_{𝑖𝑗})$ .

Proposition 8.7.12 (Cofactor expansion). §

Let $A = (a_{𝑖𝑗}) \in M_{n} (R)$ . Then for any $i$ with $1 \leq i \leq n$ , we have

\det (A) = \sum_{j = 1}^{n} {(- 1)}^{i + j} a_{𝑖𝑗} \det (A_{𝑖𝑗}),

and for any $j$ with $1 \leq j \leq n$ , we have

\det (A) = \sum_{i = 1}^{n} {(- 1)}^{i + j} a_{𝑖𝑗} \det (A_{𝑖𝑗}) .

Proof.

The first follows from the second by taking the transpose. So, fix $j$ . Denote the $i$ th column of $A$ by $v_{i}$ . Set

w_{k}^{(i)} = v_{k} - a_{𝑖𝑘} e_{i}

for each $k$ , which is the column vector given by replacing the $i$ th entry of $v_{k}$ by zero. We may then view $w_{1}^{(i)}, \dots, w_{j - 1}^{(i)}, w_{j}^{(i)}, \dots, w_{n}^{(i)}$ as the column vectors of the minor $A_{𝑖𝑗}$ in the ordered basis $e_{1}, \dots, e_{i - 1}, e_{i + 1}, \dots, e_{n}$ . In particular, we have

w_{1}^{(i)} \land \dots \land w_{j - 1}^{(i)} \land w_{j + 1}^{(i)} \land \dots \land w_{n}^{(i)} = \det (A) e_{1} \land \dots \land e_{i - 1} \land e_{i + 1} \land \dots \land e_{n} .

We then have

\begin{array}{l} v_{1} \land v_{2} \land \dots \land v_{n} & = {(- 1)}^{j - 1} v_{i} \land v_{1} \land \dots \land v_{j - 1} \land v_{j + 1} \land \dots \land v_{n} \\ = {(- 1)}^{j - 1} \sum_{i = 1}^{n} a_{𝑖𝑗} e_{i} \land v_{1} \land \dots \land v_{j - 1} \land v_{j + 1} \land \dots \land v_{n} \\ = {(- 1)}^{j - 1} \sum_{i = 1}^{n} a_{𝑖𝑗} e_{i} \land w_{1}^{(i)} \land \dots \land w_{j - 1}^{(i)} \land w_{j + 1}^{(i)} \land \dots \land w_{n}^{(i)} \\ = {(- 1)}^{j - 1} \sum_{i = 1}^{n} a_{𝑖𝑗} \det (A_{𝑖𝑗}) e_{i} \land e_{1} \land \dots \land e_{i - 1} \land e_{i + 1} \land \dots \land e_{n} \\ = {(- 1)}^{i + j} \sum_{i = 1}^{n} a_{𝑖𝑗} \det (A_{𝑖𝑗}) e_{1} \land e_{2} \land \dots \land e_{n}, \end{array}

where in the third equality we have applied Lemma 8.7.9(b). □

Definition 8.7.13. §

Let $A \in M_{n} (R)$ . The adjoint matrix to $A$ is the matrix with $(i, j)$ -entry ${(- 1)}^{i + j} \det (A_{𝑗𝑖})$ .

Theorem 8.7.14. §

Let $A \in M_{n} (R)$ , and let $B$ be its adjoint matrix. Then $𝐴𝐵 = \det (A) I_{n}$ .

Proof.

The $(i, j)$ -entry of $𝐴𝐵$ is $\sum_{k = 1}^{n} {(- 1)}^{k + j} a_{𝑖𝑘} \det (A_{𝑗𝑘})$ . If $i = j$ , this is just $\det (A)$ by Proposition 8.7.12. If $i \neq j$ , then the same proposition tells us that this equals the determinant of a matrix which has the $i$ th row of the matrix obtained by replacing the $j$ th row of $A$ by the $i$ th row of $A$ . Since this matrix has two rows which are the same, its determinant is $0$ . □

Corollary 8.7.15. §

A matrix $A \in M_{n} (R)$ is invertible if and only if $\det (A) \in R^{\times}$ , in which case its inverse is $A^{- 1} = \det {(A)}^{- 1} B$ , where $B$ is the adjoint matrix to $A$ .

As any two similar matrices in $M_{n} (R)$ have the same determinant and any two matrices representing a linear transformation are similar, the following is well-defined.

Definition 8.7.16. §

Let $V$ be a free $R$ -module of finite rank. The determinant of an $R$ -module homomorphism $T : V \to V$ is the determinant of a matrix representing $T$ with respect to an $R$ -basis of $V$ .

Remark 8.7.17. §

Let $T : V \to V$ be a homomorphism of free $R$ -modules, and let $A$ be an $R$ -algebra. Then we have an $A$ -module homomorphism ${id}_{A} \otimes T : A \otimes_{F} V \to A \otimes_{F} V$ , which we usually denote more simply by $T$ . It satisfies $T (a \otimes v) = a \otimes T (v)$ for any $a \in A$ and $v \in V$ .

Definition 8.7.18. §

a.: The characteristic polynomial of a matrix $A \in M_{n} (R)$ is $c_{A} (x) = \det (𝑥𝐼 - A)$ .
b.: The characteristic polynomial of an $R$ -module homomorphism $T : V \to V$ with $V$ a free $R$ -module of finite rank is $c_{T} (x) = \det (x id - T)$ , where $id$ denotes the identity map on $F [x] \otimes_{F} V$ .

Definition 8.7.19. §

The trace of a matrix $A = (a_{𝑖𝑗}) \in M_{n} (R)$ is

tr (A) = \sum_{i = 1}^{n} a_{𝑖𝑖} \in R .

The trace is a homomorphism of additive groups.

Lemma 8.7.20. §

If $A, B \in M_{n} (R)$ , then $tr (A + B) = tr (A) + tr (B)$ .

Lemma 8.7.21. §

Let $A \in M_{n} (R)$ . The constant coefficient of $c_{A} (x)$ is ${(- 1)}^{n} \det (A)$ , and the coefficient of $x^{n - 1}$ is $- tr (A)$ .

Proof.

We have $c_{A} (0) = \det (- A) = {(- 1)}^{n} \det (A)$ . The second part is an easy consequence of the permutation formula for the determinant applied to $x I_{n} - A$ , from which it is seen that only the term corresponding to the identity of $S_{n}$ has degree at least $n - 1$ . This term is equal to $(x - a_{11}) (x - a_{22}) \dots (x - a_{𝑛𝑛})$ , and its $x^{n - 1}$ -coefficient is $- tr (A)$ . □

Corollary 8.7.22. §

If $A$ and $B$ are similar matrices in $M_{n} (R)$ , then $tr (A) = tr (B)$ .

The reader may also verify the following directly.

Lemma 8.7.23. §

Let $A, B \in M_{n} (R)$ . Then $tr (𝐴𝐵) = tr (𝐵𝐴)$ .

8.8. Torsion and rank

Definition 8.8.1. §

Let $M$ be a module over an integral domain $R$ . We say that $m \in M$ is an $R$ -torsion element if there exists a nonzero element $a \in R$ with $𝑎𝑚 = 0$ .

Definition 8.8.2. §

Let $M$ be a module over an integral domain $R$ . Then $M$ is said to be a torsion module if all of its nonzero elements are $R$ -torsion elements.

Lemma 8.8.3. §

Let $M$ be an module over an integral domain $R$ . The set of $R$ -torsion elements of $M$ is an $R$ -submodule of $M$ .

Proof.

If $m \in M$ and $r \in R - {0}$ are such that $𝑟𝑚 = 0$ , then clearly $r (𝑎𝑚) = 0$ for all $a \in R$ as well. If moreover $m^{'} \in M$ and $r^{'} \in R - {0}$ with $r^{'} m^{'} = 0$ , then $r r^{'} (m + m^{'}) = r^{'} (𝑟𝑚) + r (r^{'} m^{'}) = 0$ , and $r r^{'} \in R - {0}$ as $R$ is a domain. Thus, the set of $R$ -torsion elements in $M$ is indeed a submodule. □

Definition 8.8.4. §

Let $M$ be a module over an integral domain $R$ . The $R$ -torsion submodule $M_{tor}$ of $M$ is the set of $R$ -torsion elements of $M$ .

Definition 8.8.5. §

Let $R$ be a ring and $M$ a left $R$ -module. The annihilator of $M$ in $R$ is the left ideal

Ann (M) = {r \in R ∣ 𝑟𝑚 = 0 for all m \in M}

of $R$ .

The reader will easily verify the following.

Lemma 8.8.6. §

The annihilator $Ann (M)$ of a left $R$ -module over a ring $R$ is a two-sided ideal of $R$ .

Definition 8.8.7. §

Let $R$ be a ring and $M$ a left $R$ -module. We say that an $R$ -module $M$ is faithful if $Ann (M) = 0$ .

Remark 8.8.8. §

Let $R$ be an integral domain and $M$ an $R$ -module. If $Ann (M) \neq 0$ , then $M$ is $R$ -torsion since any nonzero $r \in Ann (M)$ satisfies $𝑟𝑚 = 0$ for all $m \in M$ .

Lemma 8.8.9. §

Let $R$ be an integral domain, and let $M$ be a finitely generated $R$ -module. Then $Ann (M) \neq 0$ if and only if $M$ is $R$ -torsion.

Proof.

We may suppose that $M$ is $R$ -torsion. Let $m_{1}, \dots, m_{n}$ generate $M$ , and let $r_{1}, \dots, r_{n} \in R - {0}$ be such that $r_{i} m_{i} = 0$ for $1 \leq i \leq n$ . Then $r_{1} r_{2} \dots r_{n}$ is a nonzero element of $Ann (M)$ . □

Example 8.8.10. §

The abelian group $\oplus_{n = 1}^{\infty} ℤ ∕ 𝑛ℤ$ is both faithful and torsion as a $ℤ$ -module.

Let us introduce a general notion of rank for modules over integral domains.

Definition 8.8.11. §

Let $R$ be an integral domain, and let $M$ be an $R$ -module. The rank of $M$ over $R$ , or $R$ -rank of $M$ , is the largest nonnegative integer $n = {rank}_{R} M$ such that $M$ contains $n$ elements that are linearly independent over $R$ , if it exists. If ${rank}_{R} M$ exists, then $R$ is said to have finite rank, and otherwise it has infinite rank.

For free modules over integral domains, this agrees with the notion of rank defined above. We can give an alternative characterization of the rank. For this, we introduce the following lemma.

Lemma 8.8.12. §

Let $ι : M \to Q (R) \otimes_{R} M$ be the $R$ -module homomorphism defined by $ι (m) = 1 \otimes m$ for $m \in M$ . Then $\ker ι = M_{tor}$ .

Proof.

By Proposition 11.1.9, the module $Q (R) \otimes_{R} M$ is canonically isomorphic to the localization of $M$ by $S = R - {0}$ . The map $ι$ becomes identified with the map $M \to S^{- 1} M$ given by $m \mapsto \frac{m}{1}$ . The definition of $S^{- 1} M$ tells us that $\frac{m}{1} = 0$ if and only if there exists $r \in R - {0}$ such that $𝑟𝑚 = 0$ , which is to say $m \in M_{tor}$ . □

Proposition 8.8.13. §

Let $R$ be an integral domain, and let $M$ be an $R$ -module. Then $M$ has finite rank over $R$ if and only if $Q (R) \otimes_{R} M$ is finite-dimensional over $Q (R)$ , in which case

{rank}_{R} M = \dim_{Q (R)} Q (R) \otimes_{R} M .

Proof.

First, suppose that $m_{1}, m_{2}, \dots, m_{n}$ are $n$ elements of $M$ . First, suppose that the elements $m_{i}$ are $R$ -linearly dependent. Let $a_{i} \in R$ not all $0$ be such that $\sum_{i = 1}^{n} a_{i} m_{i} = 0$ . Then

\sum_{i = 1}^{n} a_{i} (1 \otimes m_{i}) = \sum_{i = 1}^{n} a_{i} \otimes m_{i} = \sum_{i = 1}^{n} 1 \otimes a_{i} m_{i} = 0,

so the elements $1 \otimes m_{i}$ are $Q (R)$ -linearly dependent.

Conversely, suppose that the elements $1 \otimes m_{i}$ are $Q (R)$ -linearly dependent. Let $α_{i} \in Q (R)$ with $\sum_{i = 1}^{n} α_{i} \otimes m_{i} = 0$ and not all $α_{i} = 0$ . Let $d \in R$ be such that $a_{i} = d α_{i} \in R$ for all $i$ , and set $m = \sum_{i = 1}^{n} a_{i} m_{i}$ . We then have

1 \otimes m = \sum_{i = 1}^{n} a_{i} \otimes m_{i} = d \sum_{i = 1}^{n} α_{i} \otimes m_{i} = 0,

so there exists $r \in R - {0}$ with $𝑟𝑚 = 0$ by Lemma 8.8.12. That is, the $m_{i}$ are $R$ -linearly dependent.

In particular, if $M$ has no maximal $Q (R)$ -linearly independent subset, then neither does $Q (R) \otimes_{R} M$ , so $R$ has finite rank if and only if $Q (R) \otimes_{R} M$ is infinite dimensional.

If $M$ has a finite linear independent set $X = {m_{1}, \dots, m_{n}}$ of maximal order $n$ , then the set of $1 \otimes m_{i}$ is a $Q (R)$ -linearly independent subset of $Q (R) \otimes_{R} M$ of the same order. We have seen that for any $y \in Q (R) \otimes_{R} M$ , there exists $c \in R$ such that $𝑐𝑦$ has the form $1 \otimes n$ for some $n \in M$ , and we have then seen that $y$ and the elements $1 \otimes m_{i}$ are linearly dependent, as $X$ is maximal. Since ${1 \otimes m_{i} ∣ 1 \leq i \leq n}$ can be extended to a basis of $Q (R) \otimes_{R} M$ , it must already then be a basis, so $n = \dim_{Q (R)} Q (R) \otimes_{R} M$ . □

Example 8.8.14. §

Consider any nonzero ideal $I$ of a domain $R$ . The usual method shows the existence of a $Q (R)$ -linear map $Φ : Q (R) \otimes_{R} I \to Q (R)$ satisfying $Φ (f \otimes x) = 𝑥𝑓$ for $x \in I$ and $f \in Q (R)$ . This map is clearly onto since $Φ (\frac{1}{x} \otimes x) = 1$ for any nonzero $x \in I$ , so

{rank}_{R} I = \dim_{Q (R)} Q (R) \otimes_{R} I \geq 1 .

On the other hand, if $x \in I$ is nonzero, then any simple tensor $g \otimes y \in ℚ (R) \otimes_{R} I$ can be rewritten as $g \otimes y = \frac{𝑔𝑦}{x} \otimes x$ . The $Q (R)$ -linear transformation $ψ : Q (R) \to Q (R) \otimes_{R} I$ given by $ψ (f) \otimes f \otimes x$ is therefore onto, so ${rank}_{R} I \leq 1$ . Thus, $I$ has $R$ -rank $1$ , but it is a free $R$ -module if and only if it is principal.

Though a bit off of the topic of this section, we also note the following consequence of Nakayama’s lemma for free modules, which we shall employ later. In the case of local domains, it is a considerable strengthening of the statement that the rank of a free module is the dimension of its reduction modulo a maximal ideal.

Lemma 8.8.15. §

Let $M$ be a finitely generated free module over a local ring $R$ with maximal ideal $𝔪$ , and let $X$ be a subset of $M$ . If the image of $X$ in $M ∕ 𝔪𝑀$ is $R$ -linearly independent, then $X$ is $R$ -linearly independent and can be extended to a basis of $M$ .

Proof.

Let $\bar{X}$ denote the image of $X$ in $M ∕ 𝔪𝑀$ . Extend $\bar{X}$ to a basis $\bar{B}$ of $M ∕ 𝔪𝑀$ , and let $B \subset M$ be a lift of $\bar{B}$ to $M$ with $X \subseteq B$ . Then $B$ spans $M$ by Corollary 8.2.18. To see that it is linearly independent, suppose that $B$ has $n$ elements $m_{1}, \dots, m_{n}$ and consider the sum $\sum_{i = 1}^{n} a_{i} m_{i}$ for some $a_{i} \in R$ . Suppose that not all $a_{i}$ are zero, and let $k \geq 0$ be minimal such that $a_{i} \in 𝔪^{k}$ for all $i$ . Note that the map

𝔪^{k} ∕ 𝔪^{k + 1} \otimes_{R} M \to 𝔪^{k} M ∕ 𝔪^{k + 1} M

induced by the $R$ -action on $M$ is an isomorphism by the freeness of $M$ , since tensor products commute with direct sums and it is clearly true for $M = R$ .

By Corollary 8.4.27, we also have an isomorphism

𝔪^{k} ∕ 𝔪^{k + 1} \otimes_{R} M \to_{}^{\sim} 𝔪^{k} ∕ 𝔪^{k + 1} \otimes_{R ∕ 𝔪} M ∕ 𝔪𝑀

via the map induced by the identity in the first variable and the quotient map in the second. But $\sum_{i = 1}^{n} a_{i} \otimes m_{i} \neq 0$ in the right-hand side since $\bar{B}$ is a basis of $M ∕ 𝔪𝑀$ and the tensor product is of $R ∕ 𝔪$ -vector spaces. Therefore $\sum_{i = 1}^{n} a_{i} m_{i} \neq 0$ . In other words $B$ is a basis of $M$ , and $X$ is $R$ -linearly independent. □

8.9. Modules over PIDs

Lemma 8.9.1. §

Let $R$ be a PID. Any finitely generated $R$ -submodule of $Q (R)$ is cyclic.

Proof.

Let $M$ be an $R$ -module generated by some subset ${α_{1}, \dots, α_{n}}$ of $Q (R)$ . Let $d \in R - {0}$ be such that $d α_{i} \in R$ for all $i$ . Then $d : M \to 𝑑𝑀$ is an isomorphism, and $𝑑𝑀$ is an ideal of $R$ , hence principal. That is, $𝑑𝑀$ is cyclic as an $R$ -module, so $M$ is as well. □

Proposition 8.9.2. §

Let $R$ be a PID. Let $V$ be an $n$ -dimensional $Q (R)$ -vector space, and let $M$ be a finitely generated $R$ -submodule of $V$ . Then there exists a basis ${v_{1}, v_{2}, \dots, v_{n}}$ of $V$ and $k \leq n$ such that $M$ is a free $R$ -module with $R$ -basis ${v_{1}, v_{2}, \dots, v_{k}}$ .

Proof.

We suppose without loss of generality that $M$ is nonzero. Pick a nonzero element $m_{1} \in M$ . Recall that $R$ is noetherian as it is a PID. Then $Q (R) m_{1}$ is a $1$ -dimensional $Q (R)$ -vector space, and $M$ is noetherian being that it is $R$ -finitely generated, so $M \cap Q (R) m_{1}$ is $R$ -finitely generated. Since $Q (R) m_{1}$ is a $1$ -dimensional $Q (R)$ -vector space with $R$ -submodule $M \cap Q (R) m_{1}$ , Lemma 8.9.1 tells us that $M \cap Q (R) m_{1} = R v_{1}$ for some $v_{1} \in Q (R) m_{1}$ . Set $\bar{M} = M ∕ R v_{1}$ . This is an $R$ -submodule of the $(n - 1)$ -dimensional vector space $\bar{V} = V ∕ Q (R) v_{1}$ , since if $x \in M$ is such that $x + R v_{1}$ is in the kernel of $\bar{M} \to \bar{V}$ , then there exists $α \in Q (R)$ such that $x = α v_{1}$ . But then $x \in M \cap Q (R) m_{1}$ , which is to say $x \in R v_{1}$ .

Now, by induction on $n$ , there exist $v_{2}, \dots, v_{k} \in M$ for some $k \geq 1$ such that $v_{2} + R v_{1}, \dots, v_{k} + R v_{1}$ form an $R$ -basis of $\bar{M}$ . Then $v_{1}, v_{2}, \dots, v_{k} \in M$ generate $M$ by Lemma 8.2.4, and we claim they are $R$ -linearly independent. That is, if $\sum_{i = 1}^{k} c_{i} v_{i} = 0$ for some $c_{i} \in R$ , then

\sum_{i = 2}^{k} c_{i} (v_{i} + R m_{1}) = 0,

and so $c_{i} = 0$ for $2 \leq i \leq k$ . As $v_{1} \neq 0$ , this forces $c_{1} = 0$ as well. To finish, we merely extend ${v_{1}, \dots, v_{k}}$ to a $Q (R)$ -basis ${v_{1}, \dots, v_{n}}$ of $V$ , noting that an $R$ -linearly independent subset of $V$ is also $Q (R)$ -linearly independent. □

Corollary 8.9.3. §

Every finitely generated, torsion-free module over a principal ideal domain is free.

Proof.

Let $M$ be a finitely generated, torsion-free module over a PID $R$ . We have seen in Lemma 8.8.12 that the canonical map $M \to Q (R) \otimes_{R} M$ is injective, in that $M_{tor} = 0$ . The result is then immediate from Proposition 8.9.2, as $Q (R) \otimes_{R} M$ is a finite-dimensional $Q (R)$ -vector space. □

Corollary 8.9.4. §

Any submodule of a free module of rank $n$ over a principal domain is free of rank at most $n$ .

Proof.

Let $R$ be a PID. Let $M$ be a free $R$ -module of rank $n$ , and let $N$ be an $R$ -submodule of $M$ . Then $N \to Q (R) \otimes_{R} M$ is injective, so we can apply Proposition 8.9.2. □

Proposition 8.9.5. §

Let $M$ be a finitely generated module over a principal ideal domain $R$ . Then $𝑀≅ R^{r} \oplus M_{tor}$ , where $r$ is the rank of $M$ .

Proof.

Note that $M ∕ M_{tor}$ is free of finite rank by Corollary 8.9.3, so isomorphic to $R^{r}$ for some $r$ . By Proposition 5.7.27, we have that $𝑀≅ M_{tor} \oplus R^{r}$ . □

Every $ℤ$ -submodule of $ℤ$ is free, but $0$ and $ℤ$ are the only direct summands of $ℤ$ . On the other hand, if we consider $ℤ ∕ 𝑛ℤ$ as a $(ℤ ∕ 𝑛ℤ)$ -module for some $n \geq 1$ , then $0$ and $ℤ ∕ 𝑛ℤ$ are its only free subbodules, and of course they are direct summands. The following lemma implies (via an application of the Chinese remainder theorem) that free submodules of finitely generated modules over quotients of PIDs by nonzero elements are always direct summands.

Lemma 8.9.6. §

Let $R$ be a principal ideal domain, let $π \in R$ be an irreducible element. Let $k \geq 1$ and set $\bar{R} = R ∕ (π^{k})$ . Then any free $\bar{R}$ -submodule $F$ of a finitely generated $\bar{R}$ -module $M$ is a direct summand of $M$ . If $F$ is maximal, then $𝑀≅𝐹 \oplus C$ , where $π^{k - 1} C = 0$ .

Proof.

We work by induction on $k$ . Let $M$ be an $\bar{R}$ -module and $F$ a free submodule of $M$ . If $k = 1$ , then $M$ is a finite-dimensional $\bar{R}$ -vector space. Then $F$ is a direct summand of $M$ , since any basis of it extends to a basis of $M$ .

Now take $k \geq 2$ . Suppose first that $A$ is a maximal free $\bar{R}$ -submodule of $M$ . Consider the subgroup

N = {m \in M ∣ π^{k - 1} m = 0} .

We have $𝜋𝐴 \subseteq N$ , which is an $R ∕ (π^{k - 1})$ -module. By induction on $k$ , the free $R ∕ (π^{k - 1})$ -module $𝜋𝐴$ is a direct summand of $N$ . We have $N = 𝜋𝐴 \oplus C$ for some $\bar{R}$ -submodule $C$ of $A$ .

Note that $N$ contains $𝜋𝑀$ , and therefore any $R ∕ (π)$ -linearly independent subset of $M ∕ N$ lifts an $R ∕ (π)$ -linearly independent subset of $M ∕ 𝜋𝑀$ . Note that $\bar{R}$ is a local ring with maximal ideal generated by $π$ . Any set of representatives in $M$ of an $R ∕ (π)$ -basis of $M ∕ N$ is then $\bar{R}$ -linearly indepedent by Lemma 8.8.15.

The canonical map $f : A ∕ 𝜋𝐴 \to M ∕ N$ is injective as $A \cap N = {a \in A ∣ π^{k - 1} A = 0} = 𝜋𝐴$ by the freeness of $A$ . In particular, we have $A \cap C = 0$ . The map $f$ is also surjective: if it were not, then by what we have shown, we could extend the image of a basis of $A$ to a basis of $M ∕ N$ and lift to obtain a free $\bar{R}$ -submodule of $M$ properly containing $A$ . We therefore have $A + N = M$ , so $A + C = M$ . Thus $M = A \oplus C$ .

It remains to show that an arbitrary free $\bar{R}$ -module $F$ is a direct summand of $M$ . It suffices to show that $F$ is a direct summand of a maximal free $\bar{R}$ -submodule $A$ . For this, note that the map $F ∕ 𝜋𝐹 \to A ∕ 𝜋𝐴$ is injective, since if $a \in 𝜋𝐴 \cap F$ , then $π^{k - 1} a = 0$ , so $a \in 𝜋𝐹$ by the freeness of $F$ . We may then choose a set $X$ that is a basis of a complement of $F ∕ 𝜋𝐹$ in $A ∕ 𝜋𝐴$ . Again by Lemma 8.8.15, any lift of $X$ to a linearly independent subset of $A$ will span an $\bar{R}$ -complement to $F$ . Thus $F$ is a direct summand of $A$ . □

We are now ready to prove the structure theorem for finitely generated modules over principal ideal domains.

Theorem 8.9.7 (Structure theorem for finitely generated modules over PIDs). §

Let $R$ be a PID, and let $M$ be a finitely generated $R$ -module.

a.

There exist unique nonnegative integers $r$ and $k$ and nonzero proper principal ideals $I_{1} \subseteq I_{2} \subseteq \dots \subseteq I_{k}$ of $R$ such that

𝑀≅ R^{r} \oplus R ∕ I_{1} \oplus R ∕ I_{2} \oplus \dots \oplus R ∕ I_{k} .

b.

There exist unique nonnegative integers $r$ and $l$ , and for $1 \leq i \leq l$ , distinct nonzero prime ideals $𝔭_{i}$ of $R$ and positive integers $v_{i, 1} \geq v_{i, 2} \geq \dots \geq v_{i, m_{i}}$ for some $m_{i} \geq 1$ such that

𝑀≅ R^{r} \oplus ⨁_{i = 1}^{l} ⨁_{j = 1}^{m_{i}} R ∕ 𝔭_{i}^{v_{i, j}} .

(8.9.1)

Moreover, $r$ and $l$ are unique, and the tuple ${(𝔭_{i}, {(v_{i, j})}_{j})}_{i}$ is unique up to ordering in $i$ .

Proof.

By Proposition 8.9.5, it suffices to consider the case that $M$ is torsion. Note that the uniqueness of $r$ in parts (a) and (b) follows from the fact that $r = \dim_{Q (R)} Q (R) \otimes_{R} M = {rank}_{R} M$ . So, let $M$ be a finitely generated torsion $R$ -module.

We first demonstrate the existence of a decomposition as in part b. Let $c \in R$ be a generator of the annihilator $Ann (M)$ of $M$ . Since we have unique factorization in $R$ , we may write $c = u π_{1}^{k_{1}} \dots π_{l}^{k_{l}}$ with $u \in R^{\times}$ and with $π_{1}, \dots, π_{r}$ distinct irreducible elements of $M$ and $k_{1}, \dots, k_{;}$ positive integers for some $k \geq 0$ . By the Chinese remainder theorem, we have an isomorphism

R ∕ (c) ≅ \prod_{i = 1}^{l} R ∕ (π_{i}^{k_{i}})

of rings, which in turn provides a direct sum decomposition

M = M ∕ 𝑐𝑀≅𝑀 \otimes_{R} R ∕ (c) ≅ ⨁_{i = 1}^{l} M \otimes_{R} R ∕ (π_{i}^{k_{i}}) ≅ ⨁_{i = 1}^{l} M ∕ π_{i}^{k_{i}} M .

In particular, we are reduced to the case that $M$ is a module over the local ring $R ∕ (π^{k})$ for some irreducible element $π$ of $R$ .

For the moment, suppose that $M$ is a nonzero finitely generated $R ∕ (π^{k})$ -module for some $k \geq 1$ . Note that if $k = 1$ , then $M$ is simply a finite dimensional $R ∕ (π)$ -vector space, so a choice of basis gives a direct sum decomposition $𝑀≅ \oplus_{i = 1}^{m} R ∕ (π)$ for some $m$ . For general $k$ , let $F$ be a maximal free $R ∕ (π^{k})$ -submodule of $M$ . By Lemma 8.9.6, we have $M = F \oplus C$ , where $C$ is a finitely generated $R ∕ (π^{k - 1})$ -module. By induction on $k$ , this gives the decomposition of part b.

We next prove the existence in part a using the decomposition in part b. Let us take $π_{i}$ to be an irreducible element generating $𝔭_{i}$ for each $i$ . For $j \geq 1$ , set $b_{j} = π_{1}^{v_{1, j}} π_{2}^{v_{2, j}} \dots π_{l}^{v_{l, j}}$ , where we take $v_{i, j} = 0$ for $j > m_{i}$ . By construction, we have that $b_{j + 1} ∣ b_{j}$ for each $j \geq 1$ . Set $I_{j} = (b_{j})$ , and let $k$ be maximal such that $I_{k} \neq R$ or zero if all $I_{k} = R$ . Applying the Chinese remainder theorem again to see that

⨁_{i = 1}^{l} R ∕ π_{i}^{v_{i, j}} ≅ R ∕ I_{j}

we obtain a decomposition as in part a.

Next, we exhibit uniqueness. If $𝔭 = (π)$ is a nonzero prime ideal of $R$ and $v \geq 0$ , then the map $R ∕ 𝔭 \to 𝔭^{v} ∕ 𝔭^{v + 1}$ induced by multiplication by $π^{v}$ is an isomorphism. Let $N = R ∕ 𝔮^{w}$ for a nonzero prime ideal $𝔮$ and $w \geq 0$ . Note that $𝔭^{v} + 𝔮^{w} = R$ if $𝔭 \neq 𝔮$ , being that $𝔭$ and $𝔮$ are generated by coprime elements. We therefore have

𝔭^{v} N ∕ 𝔭^{v + 1} 𝑁≅ 𝔭^{v} ∕ ((𝔭^{v + 1} + 𝔮^{w}) \cap 𝔭^{v}) ≅ {\begin{matrix} R ∕ 𝔭 & if 𝔭 = 𝔮 and v < w \\ 0 & if 𝔭 \neq 𝔮 or v \geq w . \end{matrix}

For any $v \geq 1$ , we then have that

𝔭_{i}^{v} M ∕ 𝔭_{i}^{v + 1} 𝑀≅ {(R ∕ 𝔭_{i})}^{u},

where $u$ is the number of $j$ such that $v < v_{i, j}$ . Thus, the $𝔭_{i}$ and $v_{i, j}$ in any decomposition of $M$ as in part b are the same.

Finally, we reduce the uniqueness in part a to the known uniqueness of part b. Given any decomposition $𝑀≅𝑅 ∕ I_{1} \oplus \dots \oplus R ∕ I_{k}$ as in part a, we can again obtain a decomposition of $M$ into $R$ modulo power of prime ideals, applying CRT to expand out each $R ∕ I_{j}$ . Since $I_{j} \supseteq I_{j + 1}$ , this decomposition then satisfies $I_{j} = 𝔭_{1}^{v_{1, j}} 𝔭_{2}^{v_{2, j}} \dots 𝔭_{l}^{v_{l, j}}$ with $v_{i, j} \geq v_{i, j + 1}$ for each $j \geq 0$ . Thus, the decomposition is as in part b, and by its uniqueness, we obtain the uniqueness of the decomposition in part a. □

Remark 8.9.8. §

The structure theorem for finitely generated abelian groups is the special case of the structure theorem for finitely generated modules over a PID for the PID $ℤ$ .

Definition 8.9.9. §

Let $R$ be a PID, and let $M$ be a finitely generated $R$ -module.

a.: The ideals $I_{1}, I_{2}, \dots, I_{k}$ associated to $M$ by Theorem 8.9.7a are called the invariant factors of $M$ .
b.: The prime powers $𝔭_{i}^{v_{i, j}}$ associated to $M$ by Theorem 8.9.7b are called the elementary divisors of $M$ .

8.10. Canonical forms

Definition 8.10.1. §

Let $V$ be a vector space over a field $F$ , and let $T : V \to V$ be an $F$ -linear transformation.

a.

An eigenvector $v$ of $T$ with eigenvalue $λ \in F$ is an element $v \in V - {0}$ such that $T v = 𝜆𝑣$ .

b.

An element $λ \in F$ is called an eigenvalue of $T$ if there exists an eigenvector in $V$ with eigenvalue $λ$ .

c.

The eigenspace of $F$ for $λ \in F$ of $T$ is the nonzero subspace

E_{λ} (T) = {v \in V ∣ T (v) = 𝜆𝑣}

of $V$ .

Note that $E_{λ} (T) = \ker (T - λ {id}_{V})$ .

Lemma 8.10.2. §

Let $V$ be a vector space over a field $F$ . The following are equivalent for an $F$ -linear transformation $T : V \to V$ and $λ \in F$ :

i.: $E_{λ} (T) \neq 0$ ,
ii.: $λ$ is an eigenvalue of $T$ , and
iii.: $c_{T} (λ) = 0$ .

Proof.

The first two are equivalent by definition. Moreover, $T - λ {id}_{V}$ has a nonzero kernel if and only if $c_{T} (λ) = \det (λ {id}_{V} - T) = 0$ . □

Terminology 8.10.3. §

We may speak of eigenvectors, eigenvalues, and eigenspaces $E_{λ} (A)$ of a matrix in $A \in M_{n} (F)$ , taking them to be the corresponding objects for the linear transformation $T : F^{n} \to F^{n}$ that $A$ represents.

The following is the key to the application of the structure theorem for modules over PIDs to linear algebra.

Notation 8.10.4. §

If $T : V \to V$ is a linear transformation and $f = \sum_{i = 1}^{k} c_{i} x^{i} \in F [x]$ , then we set

f (T) = \sum_{i = 1}^{n} c_{i} T^{i} : V \to V,

where $T^{i} : V \to V$ denotes the $i$ -fold composition of $T$ with itself.

Remark 8.10.5. §

If $T$ is represented by a matrix $A$ and $f = \sum_{i = 1}^{k} c_{i} x^{i} \in F [x]$ , then $f (T)$ is represented by

f (A) = \sum_{i = 1}^{k} c_{i} A^{i} \in M_{n} (F) .

Definition 8.10.6. §

Let $T : V \to V$ be an $F$ -linear endomorphism of an $F$ -vector space $V$ . The $F [x]$ -module structure endowed on $V$ by $T$ is that which satisfies $f (x) \cdot v = f (T) v$ for all $T \in F [x]$ .

This construction gives us one way to define the minimal polynomial of a linear transformation.

Definition 8.10.7. §

a.: Let $V$ be a finite-dimensional $F$ -vector space. The minimal polynomial $m_{T} (x)$ of a linear transformation $T : V \to V$ is the unique monic generator of the annihilator $Ann (V)$ under the $F [x]$ -module structure on $V$ induced by $T$ .
b.: The minimal polynomial $m_{A} (x)$ is the minimal polynomial of the linear transformation $T : F^{n} \to F^{n}$ that $A$ represents with respect to the standard basis of $F^{n}$ .

Lemma 8.10.8. §

The minimal polynomial of an endomorphism $T$ of a finite-dimensional vector space $V$ divides the characteristic polynomial of $T$ .

Proof.

Let $v \in V$ . Then $\det (𝑥𝐼 - T) \cdot v = \det (T - T) \cdot v = 0$ by definition, so $c_{T} (x) \in Ann (V)$ . □

Lemma 8.10.9. §

If $A$ and $B$ are similar matrices in $M_{n} (F)$ , then $c_{A} (x) = c_{B} (x)$ and $m_{A} (x) = m_{B} (x)$ .

Proof.

Suppose $Q \in {GL}_{n} (F)$ is such that $B = 𝑄𝐴 Q^{- 1}$ . Then $𝑥𝐼 - B = Q (𝑥𝐼 - A) Q^{- 1}$ , so

c_{B} (x) = \det (𝑥𝐼 - B) = \det (𝑥𝐼 - A) = c_{A} (x) .

Moreover, if $g \in F [x]$ is such that $g (A) v = 0$ for all $v \in F^{n}$ , then

g (B) 𝑄𝑣 = 𝑄𝑔 (A) v = 0

for all $v \in F^{n}$ , so $g (B)$ annihilates $F^{n}$ as well. By symmetry, we have $m_{A} (x) = m_{B} (x)$ . □

Lemma 8.10.10. §

Let $A$ be a block diagonal matrix with blocks $A_{i} \in M_{n_{i}} (F)$ for $1 \leq i \leq m$ and some $m \geq 1$ . Then

c_{A} (x) = \prod_{i = 1}^{m} c_{A_{i}} (x),

while $m_{A} (x)$ is the least common multiple of the $m_{A_{i}} (x)$ with $1 \leq i \leq m$ .

Suppose that we endow a finite-dimensional $F$ -vector space $V$ with the structure of an $F [x]$ -module through a linear transformation $T : V \to V$ . Since $F [x]$ is a PID, the structure theorem for modules over a PID tells us that there exists an $F [x]$ -module isomorphism

𝑉≅ ⨁_{i = 1}^{m} F [x] ∕ (f_{i}),

where $m \geq 0$ and the $f_{i} \in F [x]$ are monic, nonconstant polynomials for $1 \leq i \leq m$ such that $f_{i + 1} ∣ f_{i}$ for $1 \leq i < m$ .

Lemma 8.10.11. §

Let $f = \sum_{i = 0}^{n} c_{i} x^{i} \in F [x]$ be a monic polynomial of degree $n \geq 1$ . With respect to the ordered basis ${1, x, \dots, x^{n - 1}}$ of $V = F [x] ∕ (f)$ as an $F$ -vector space, the linear transformation given by multiplication by $x$ on $V$ is represented by the matrix

A_{f} = (\begin{matrix} 0 & 0 & \dots & 0 & - c_{0} \\ 1 & 0 & 0 & - c_{1} \\ 0 & 1 & ⋱ & ⋮ & ⋮ \\ ⋮ & ⋱ & ⋱ & 0 & - c_{n - 2} \\ 0 & \dots & 0 & 1 & - c_{n - 1} \end{matrix}) .

(This matrix is taken to be $(- c_{0})$ if $n = 1$ .)

Proof.

Let $T_{f} : V \to V$ be the linear transformation given by left multiplication by $x$ . Note that $T_{f} (x^{i}) = x^{i + 1}$ for $0 \leq i \leq n - 2$ and

T_{f} (x^{n - 1}) = x^{n} = - \sum_{i = 0}^{n - 1} c_{i} x^{i} .

Thus, if $A_{f} = (a_{i, j})$ , we have $a_{i + 1, i} = 1$ for $1 \leq i \leq n - 1$ and $a_{i, n} = - c_{i - 1}$ for $1 \leq i \leq n$ , and all other entries are zero. □

Definition 8.10.12. §

For any monic, nonconstant $f \in F [x]$ , the matrix $A_{f}$ of Lemma 8.10.11 is known as the companion matrix to $f$ .

Lemma 8.10.13. §

If $f \in F [x]$ is nonconstant and monic, then $c_{A_{f}} (x) = m_{A_{f}} (x) = f$ .

Proof.

The case $n = 1$ is clear. Write $f = 𝑥𝑔 + c_{0}$ for some $g \in F [x]$ . By induction of the degree of $f$ , we have

c_{A_{f}} = \det (𝑥𝐼 - A_{f}) = x \det (𝑥𝐼 - A_{g}) + {(- 1)}^{n - 1} c_{0} \det (- I_{n - 1}) = 𝑥𝑔 + c_{0} = f .

□

We can make the following definition as a consequence of the structure theorem for $F [x]$ -modules.

Definition 8.10.14. §

Let $V$ be a finite-dimensional $F$ -vector space and $T : V \to V$ an $F$ -linear transformation. Write $𝑉≅ \oplus_{i = 1}^{m} F [x] ∕ (f_{i})$ for some $m \geq 0$ and monic $f_{i} \in F [x]$ with $f_{i} ∣ f_{i + 1}$ for all $i < m$ . The rational canonical form of $T$ is the block-diagonal matrix

(\begin{matrix} A_{f_{1}} \\ A_{f_{2}} \\ ⋱ \\ A_{f_{m}} \end{matrix}),

where $A_{f_{i}}$ is the companion matrix of $f_{i}$ .

Remark 8.10.15. §

The rational canonical form represents $T$ with respect to the basis of $V$ determined by taking the image under the isomorphism $\oplus_{i = 1}^{m} F [x] ∕ (f_{i}) \to_{}^{\sim} V$ of the ordered basis of the direct sum given by concatenating the bases ${1, x, \dots, x^{\deg (f_{i}) - 1}}$ of the $i$ th summands in order of increasing $i$ .

We also note the following.

Remark 8.10.16. §

By definition of rational canonical form, a matrix in rational canonical form in one field is already in rational canonical form in any extension field.

Definition 8.10.17. §

The rational canonical form of a matrix $A \in M_{n} (F)$ is the rational canonical form of the linear transformation that $A$ represents with respect to the standard basis of $F^{n}$ .

Remark 8.10.18. §

By Remark 8.10.15 and the change of basis theorem, $A$ is similar to its rational canonical form. Moreover, two matrices are similar if and only if they have the same rational canonical form, since similar $n$ -by- $n$ matrices give rise to isomorphic $F [x]$ -module structures on $F^{n}$ and conversely.

Definition 8.10.19. §

The invariant factors of $A \in M_{n} (F)$ are the monic generators in $F [x]$ of the invariant factors of $F^{n}$ viewed as an $F [x]$ -module via the linear transformation represented by $A$ with respect to the standard basis.

As a simple consequence of Lemmas 8.10.13 and 8.10.10, we have the following.

Lemma 8.10.20. §

Let $f_{1}, f_{2}, \dots, f_{m}$ be the invariant factors of a matrix $A$ (with $f_{i} ∣ f_{i + 1}$ for $i < m$ ). Then $m_{A} (x) = f_{m} (x)$ and $c_{A} (x) = \prod_{i = 1}^{m} f_{i} (x)$ .

As a consequence, the irreducible divisors of $c_{A} (x)$ and $m_{A} (x)$ are the same. In particular, we have:

Corollary 8.10.21. §

Let $A \in M_{n} (F)$ and $λ \in F$ . The following are equivalent:

i.: An element $λ \in F$ is an eigenvalue of $A \in M_{n} (F)$ .
ii.: The polynomial $x - λ$ divides the minimal polynomial $m_{A} (x)$ .
iii.: The polynomial $x - λ$ divides the characteristic polynomial $c_{A} (x)$ .

This lemma is at times enough to calculate the rational canonical form of a matrix.

Examples 8.10.22. §

Let $A \in M_{n} (F)$ .

a.

If $f = c_{A} (x)$ is a product of distinct monic, irreducible polynomials, then the rational canonical form of $A$ is $A_{f}$ .

b.

If $f = m_{A} (x)$ has degree $n$ , then the rational canonical form of $A$ is $A_{f}$ .

c.

If $f = c_{A} (x) = m_{A} {(x)}^{d}$ and $g = m_{A} (x)$ is irreducible of degree $\frac{n}{d}$ , then $A$ has $d$ invariant factors of the form $A_{g}$ .

d.

Note that

\begin{array}{rcl} (\begin{matrix} 0 \\ 0 \\ 0 & 0 \\ 1 & 0 \end{matrix}) & and & (\begin{matrix} 0 & 0 \\ 1 & 0 \\ 0 & 0 \\ 1 & 0 \end{matrix}) \end{array}

are both $4$ -by- $4$ matrices in rational canonical form with characteristic polynomial $x^{4}$ and minimal polynomial $x^{2}$ .

Recall that we have a second decomposition of $V$ for the $F [x]$ -module structure given by $T$ . That is, there exist distinct monic, irreducible polynomials $p_{1} (x), p_{2} (x), \dots, p_{l} (x)$ and positive integers $v_{i, j}$ for $1 \leq j \leq m_{i}$ for some $m_{i} \geq 1$ for $1 \leq i \leq l$ such that

𝑉≅ ⨁_{i = 1}^{l} ⨁_{j = 1}^{m_{i}} F [x] ∕ (p_{i} {(x)}^{v_{i, j}}) .

If the field $F$ contains all the roots of $c_{A} (x)$ , then it contains all the roots of the $p_{i}$ , so being irreducible, these polynomials must be linear. This occurs, for instance, if $F$ is algebraically closed. Let us assume this is the case and write $p_{i} (x) = x - λ_{i}$ for some $λ_{i} \in F$ .

Lemma 8.10.23. §

Let $V = F [x] ∕ ({(x - λ)}^{n})$ for some $λ \in F$ and $n \geq 0$ . The linear transformation given by multiplication by $x$ on $V$ is represented by the matrix

J_{λ, n} = (\begin{matrix} λ & 1 \\ λ & ⋱ \\ ⋱ & 1 \\ λ \end{matrix})

with respect to the ordered basis ${{(x - λ)}^{n - 1}, {(x - λ)}^{n - 2}, \dots, x - λ, 1}$ of $V$ .

Proof.

The linear transformation $T : V \to V$ that is multiplication by $x$ satisfies

T ({(x - λ)}^{j}) = {(x - λ)}^{j + 1} + λ {(x - λ)}^{j}

for all $j$ , with ${(x - λ)}^{j} = 0$ in $V$ for $j \geq n$ . The result follows. □

Definition 8.10.24. §

A matrix $J_{λ, n}$ of the form Lemma 8.10.23 is called a Jordan block of dimension $n$ for $λ$ .

Definition 8.10.25. §

Let $V$ be a finite-dimensional $F$ -vector space, and let $T : V \to V$ an $F$ -linear transformation such that $c_{T} (x)$ splits in $F$ . Write $𝑉≅ \oplus_{i = 1}^{m} F [x] ∕ ({(x - λ_{i})}^{n_{i}})$ for some $λ_{i} \in F$ and $n_{i} \geq 1$ for $1 \leq i \leq m$ and some $m \geq 1$ . The Jordan canonical form of $T$ is a block-diagonal matrix

(\begin{matrix} J_{λ_{1}, n_{1}} \\ J_{λ_{2}, n_{2}} \\ ⋱ \\ J_{λ_{m}, n_{m}} \end{matrix})

where $J_{λ_{i}, n_{i}}$ is the Jordan block of dimension $n_{i}$ for $λ_{i}$ .

Terminology 8.10.26. §

If the characteristic polynomial of $c_{T} (x)$ splits in $F$ , we say that $T$ has a Jordan canonical form over $F$ .

The Jordan canonical form is unique up to ordering of the Jordan blocks.

Definition 8.10.27. §

The Jordan canonical form of a matrix $A \in M_{n} (F)$ is the Jordan canonical form of the linear transformation that $A$ represents with respect to the standard basis of $F^{n}$ .

Remark 8.10.28. §

Every (square) matrix has a rational canonical form, while every matrix over an algebraically closed field has a Jordan canonical form.

Proposition 8.10.29. §

Suppose that $T : V \to V$ has a Jordan canonical form over $F$ . Then $λ \in F$ is an eigenvalue of $T$ if and only if it is a diagonal entry of the Jordan canonical form of $T$ .

Proof.

Consider the isomorphism $𝑉≅ \oplus_{i = 1}^{m} F [x] ∕ ({(x - λ_{i})}^{n_{i}})$ The image $v \in V$ of ${(x - λ_{i})}^{n_{i - 1}}$ in the $i$ th term of the right-hand side of the above isomorphism is an eigenvector with eigenvalue $λ_{i}$ . On the other hand, if $λ \neq λ_{i}$ , then $(x - λ) f \neq 0$ for nonzero $f \in F [x] ∕ ({(x - λ_{i})}^{n_{i}})$ , so $λ$ is not an eigenvalue of $T$ . □

Examples 8.10.30. §

Let $A \in M_{n} (F)$ , and suppose that the characteristic polynomial of $A$ splits in $F$ .

a.: If $c_{A} (x)$ is a product of distinct linear factors, then the Jordan canonical form of $A$ is diagonal with entries the distinct eigenvalues of $A$ .
b.: If $m_{A} (x)$ is a product of distinct linear factors, then the Jordan canonical form of $A$ is diagonal with entries that are all distinct eigenvalues of $A$ .
c.: If $m_{A} (x)$ has degree $n$ , then the rational canonical form of $A$ is block-diagonal with Jordan blocks $J_{λ_{i}, n_{i}}$ , where the $λ_{i}$ are all distinct.
d.: If $c_{A} (x) = {(x - λ)}^{n}$ for some $λ \in F$ and $m_{A} (x) = x - λ$ , then $A = J_{λ, n}$ .

Definition 8.10.31. §

Suppose that $T : V \to V$ . The generalized eigenspace of $λ \in F$ for $T$ is the $F [x]$ -submodule

{v \in V ∣ {(T - λ)}^{n} (v) = 0 for some n \geq 0}

of $V$ .

Remark 8.10.32. §

The generalized eigenspace of $λ \in F$ under $T : V \to V$ contains the eigenspace $E_{λ} (T)$ of $λ$ .

Example 8.10.33. §

The generalized eigenspace of

A = (\begin{matrix} J_{λ_{1}, n_{1}} \\ J_{λ_{2}, n_{2}} \\ ⋱ \\ J_{λ_{m}, n_{m}} \end{matrix}) \in M_{n} (F)

is the span of the elements $e_{i}$ of the standard basis of $F^{n}$ for which the $i$ th diagonal entry of $A$ is $λ_{i}$ .

The following is an easy consequence of the example just given.

Proposition 8.10.34. §

Let $T : V \to V$ be a linear transformation. Then $V$ is the direct sum of its nontrivial generalized eigenspaces if and only if $c_{T} (x)$ splits in $F$ .

We provide one example of how to obtain the rational and Jordan canonical forms of a matrix.

Example 8.10.35. §

Let

A = (\begin{matrix} 2 & - 2 & 8 \\ 0 & 3 & - 5 \\ 0 & 0 & 2 \end{matrix}) \in M_{3} (ℚ) .

We have

c_{A} (x) = \det (\begin{matrix} x - 2 & 2 & - 8 \\ 0 & x - 3 & 5 \\ 0 & 0 & x - 2 \end{matrix}) = {(x - 2)}^{2} (x - 3) .

So, $ℚ^{3}$ is the direct sum of its generalized eigenspaces for $2$ and $3$ . We compute that

\begin{array}{rcl} A - 2 = (\begin{matrix} 0 & - 2 & 8 \\ 0 & 1 & - 5 \\ 0 & 0 & 0 \end{matrix}) & and & {(A - 2)}^{2} = (\begin{matrix} 0 & - 2 & 10 \\ 0 & 1 & - 5 \\ 0 & 0 & 0 \end{matrix}), \end{array}

so $\ker (A - 2) = ⟨ e_{1} ⟩$ and $\ker ({(A - 2)}^{2}) = ⟨ e_{1}, 5 e_{2} + e_{3} ⟩$ , where we use angle brackets to denote the $ℚ$ -span. Note that $(A - 2) (5 e_{2} + e_{3}) = - 2 e_{1}$ . Similarly, we compute $\ker (A - 3) = ⟨ 2 e_{1} - e_{2} ⟩$ . The Jordan canonical form of the matrix $A$ is then

(\begin{matrix} 2 & 1 \\ 2 \\ 3 \end{matrix})

with respect to the basis $⟨ - 2 e_{1}, 5 e_{2} + e_{3}, 2 e_{1} - e_{2} ⟩$ .

By the Chinese remainder theorem, we have $F^{3} ≅𝐹 [x] ∕ ({(x - 2)}^{2} (x - 3))$ under the $F [x]$ -module structure on $F^{3}$ induced by $A$ . Note that

{(x - 2)}^{2} (x - 3) = (x^{2} - 4 x + 4) (x - 3) = x^{3} - 7 x^{2} + 16 x - 12 .

To find a basis of the rational canonical form

(\begin{matrix} 0 & 0 & 12 \\ 1 & 0 & - 16 \\ 0 & 1 & 7 \end{matrix})

of $A$ , we pick a vector $v$ that generates $F^{3}$ as an $F [x]$ -module, and then $A$ is in rational canonical form with respect to the basis ${v, 𝐴𝑣, A^{2} v}$ . To find $v$ , note that

(A - 2) (A - 3) = (\begin{matrix} 0 & - 2 & 8 \\ 0 & 1 & - 5 \\ 0 & 0 & 0 \end{matrix}) (\begin{matrix} - 1 & - 2 & 8 \\ 0 & 0 & - 5 \\ 0 & 0 & - 1 \end{matrix}) = (\begin{matrix} 0 & 0 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}),

so $v = e_{3}$ works, and one possible basis is ${e_{3}, 8 e_{1} - 5 e_{2} + e_{3}, 42 e_{1} - 25 e_{2} + 4 e_{3}}$ .

Definition 8.10.36. §

We say that a matrix $A \in F^{n}$ is diagonalizable if it is similar to a diagonal matrix. A linear transformation $T : V \to V$ is diagonalizable if and only if $T$ is representable by a diagonal matrix with respect to some basis of $V$ .

Clearly, a linear transformation $T : V \to V$ is diagonalizable if and only if $V$ is the direct sum of its distinct eigenspaces. The following is then a special case of Proposition 8.10.34.

Proposition 8.10.37. §

A linear transformation $T : V \to V$ is diagonalizable if and only if $m_{T} (x)$ splits in $F$ .