Abstract Algebra — Chapter 11 Commutative algebra

Chapter 11
Commutative algebra

In this chapter, all rings are commutative unless otherwise stated.

11.1. Localization of modules

We now turn to localization of modules.

Notation 11.1.1. §

Let $M$ be a module over a commutative ring $R$ , and let $S$ be a multiplicatively closed subset of $R$ . The set of equivalence classes of $S \times M$ under $\sim_{S}$ is denoted $S^{- 1} M$ , and the equivalence class of $(s, m)$ is denoted $s^{- 1} m$ or $\frac{m}{s}$ . We write $\frac{m}{1}$ more simply as $m$ .

Lemma 11.1.2. §

Let $M$ be a module over a commutative ring $R$ , and let $S$ be a multiplicatively closed subset of $R$ . Then the relation $\sim_{S}$ on $S \times M$ defined by $(s, m) \sim_{S} (t, n)$ if there exists $r \in S$ such that $r (𝑠𝑛 - 𝑡𝑚) = 0$ is an equivalence relation.

Proof.

The relation $\sim_{S}$ is clearly reflexive and symmetric, so we only need to check transitivity. For this, let $(s, m) \sim_{S} (s^{'}, m^{'})$ and $(s^{'}, m^{'}) \sim_{S} (s^{″}, m^{″})$ in $S \times M$ . Then there exist $r, r^{'} \in S$ such that $r (s m^{'} - s^{'} m) = r^{'} (s^{'} m^{″} - s^{″} m^{'}) = 0$ . We then have

0 = r r^{'} s^{″} (s m^{'} - s^{'} m) + r r^{'} s (s^{'} m^{″} - s^{″} m^{'}) = r r^{'} s^{'} (s m^{″} - s^{″} m),

so $(s, m) \sim_{S} (s^{″}, m^{″})$ . □

We omit the easy but nonetheless tedious proof of the following.

Proposition 11.1.3. §

Let $M$ be a module over a commutative ring $R$ , and let $S$ be a multiplicatively closed subset of $R$ . The set $S^{- 1} M$ of equivalence classes of $S \times M$ under the equivalence relation $\sim_{S}$ is an $S^{- 1} R$ -module under the operations

\begin{array}{rcl} \frac{m}{s} + \frac{n}{t} = \frac{𝑡𝑚 + 𝑛𝑠}{𝑠𝑡} & and & \frac{a}{s} \cdot \frac{m}{t} = \frac{𝑎𝑚}{𝑠𝑡} \end{array}

for $a \in R$ , $m, n \in M$ , and $s, t \in S$ . There is a canonical map $ι : M \to S^{- 1} M$ of $R$ -modules given by $ι (m) = \frac{m}{1}$ .

Example 11.1.4. §

Let $S$ be a multiplicatively closed subset of a commutative ring $R$ . Then the localization $S^{- 1} R$ of $R$ viewed as a left $R$ -module is just the ring $S^{- 1} R$ viewed as a module over itself.

Example 11.1.5. §

Let $R$ be an integral domain and $S = R - {0}$ . If $M$ is an $R$ -module, then $S^{- 1} M$ is a $Q (R)$ -vector space.

Lemma 11.1.6. §

Let $S$ be a multiplicatively closed subset of a commutative ring $R$ . Let ${M_{i} ∣ i \in I}$ be a collection of $R$ -modules. Then

S^{- 1} (⨁_{i \in I} M_{i}) \to_{}^{\sim} ⨁_{i \in I} S^{- 1} M_{i}

via the canonical map that takes $s^{- 1} {(m_{i})}_{i \in I}$ to ${(s^{- 1} m_{i})}_{i \in I}$ .

Example 11.1.7. §

Let $p$ be a prime number, and let $S_{p}$ be the multiplicatively closed subset of $ℤ$ that is the complement of the prime ideal $(p)$ . For $n \geq 1$ , the localization $S_{p}^{- 1} (ℤ ∕ 𝑛ℤ)$ is isomorphic to $ℤ ∕ p^{k} ℤ$ , where $p^{k}$ is the highest power of $p$ dividing $n$ .

To see this, note that $ℤ ∕ 𝑛ℤ≅ℤ ∕ 𝑚ℤ \times ℤ ∕ p^{k} ℤ$ , where $n = p^{k} m$ , so we have

S_{p}^{- 1} (ℤ ∕ 𝑛ℤ) = S_{p}^{- 1} (ℤ ∕ 𝑚ℤ) \times S_{p}^{- 1} (ℤ ∕ p^{k} ℤ) .

Now, for any $x \in ℤ ∕ 𝑚ℤ$ , we have $x = m^{- 1} (𝑚𝑥) = m^{- 1} 0 = 0$ in $S_{p}^{- 1} (ℤ ∕ 𝑚ℤ)$ . It follows that $S_{p}^{- 1} (ℤ ∕ 𝑚ℤ) = 0$ . On the other hand, if $y \in ℤ ∕ p^{k} ℤ$ and $a \in S_{p}$ are such that $a^{- 1} y = 0$ , then there exists $b \in S_{p}$ such that $𝑏𝑦 = 0$ , which means that $y = 0$ since $b$ is prime to $p$ . Furthermore, any $a^{- 1} y \in S_{p}^{- 1} (ℤ ∕ p^{k} ℤ)$ is in the image of $ℤ ∕ p^{k} ℤ$ as $a \in {(ℤ ∕ p^{k} ℤ)}^{\times}$ . Thus, we have $S_{p}^{- 1} (ℤ ∕ p^{k} ℤ) = ℤ ∕ p^{k} ℤ$ .

Remark 11.1.8. §

Given a commutative ring $R$ and a multiplicatively closed set $S$ , localization provides a functor $S^{- 1} : R -mod \to S^{- 1} R -mod$ . That is, if $f : M \to N$ is an $R$ -modules homomorphism, then we have an induced $R$ -module homomorphism $S^{- 1} f : S^{- 1} M \to S^{- 1} N$ given by $f (s^{- 1} m) = s^{- 1} f (m)$ .

We may describe the localization of a module over a commutative ring as a tensor product, as follows.

Proposition 11.1.9. §

Let $M$ be a module over a commutative ring $R$ , and let $S$ be a multiplicatively closed subset of $R$ . Then $S^{- 1} 𝑀≅ S^{- 1} R \otimes_{R} M$ as $S^{- 1} R$ -modules.

Proof.

Define a map $𝜃 : S^{- 1} R \times M \to S^{- 1} M$ by $𝜃 (s^{- 1} a, m) = \frac{𝑎𝑚}{s}$ . To see that it is well-defined, note that if $\frac{a}{s} = \frac{b}{t}$ , then we have $r \in S$ with $r (𝑡𝑎 - 𝑠𝑏) = 0$ , and then $r (𝑡𝑎𝑚 - 𝑠𝑏𝑚) = 0$ , so $\frac{𝑎𝑚}{s} = \frac{𝑏𝑚}{t}$ . The map $𝜃$ is easily checked to be left $S^{- 1} R$ -linear, right $R$ -linear, and $R$ -balanced. We then obtain a map of $S^{- 1} R$ -modules $Θ : S^{- 1} R \otimes_{R} M \to S^{- 1} M$ satisfying $Θ (\frac{a}{s} \otimes m) = \frac{𝑎𝑚}{s}$ by the universal property of the tensor product. (That it is an $S^{- 1} R$ -module homomorphism, rather than just an $R$ -module homomorphism, follows directly from the left $S^{- 1} R$ -linearity.) For bijectivity, it suffices to exhibit an inverse function.

Define a function $ψ : S \times M \to S^{- 1} R \otimes_{R} M$ by $ψ (s, m) = s^{- 1} \otimes m$ . If $(s, m) \sim_{S} (t, n)$ , then let $r \in S$ be such that $r (𝑠𝑛 - 𝑡𝑚) = 0$ . We then have

s^{- 1} \otimes m = {(𝑟𝑠𝑡)}^{- 1} \otimes 𝑟𝑡𝑚 = {(𝑟𝑠𝑡)}^{- 1} \otimes 𝑟𝑠𝑛 = t^{- 1} \otimes n,

so we obtain a well-defined map $Ψ : S^{- 1} M \to S^{- 1} R \otimes M$ given by $Ψ (\frac{m}{s}) = s^{- 1} \otimes m$ . (In fact, $Ψ$ is a homomorphism of $S^{- 1} R$ -modules, but it is not necessary to check this to finish the proof, since the inverse of a module isomorphism is one as well.) By definition, $Θ (Ψ (\frac{m}{s})) = \frac{m}{s}$ , and we have

Ψ (Θ (s^{- 1} a \otimes m)) = Ψ (\frac{𝑎𝑚}{s}) = s^{- 1} \otimes 𝑎𝑚 = s^{- 1} a \otimes m .

□

Notation 11.1.10. §

Let $𝔭$ be a prime ideal of a commutative ring $R$ , and let $M$ be an $R$ -module. Then the localization of the $R_{𝔭}$ -module $S_{𝔭}^{- 1} M$ is denoted $M_{𝔭}$ .

Proposition 11.1.11. §

Let $R$ be a commutative ring and $M$ be an $R$ -module. Then the following are equivalent:

i.: $M = 0$ ,
ii.: $M_{𝔭} = 0$ for every prime ideal $𝔭$ of $R$ , and
iii.: $M_{𝔪} = 0$ for every maximal ideal $𝔪$ of $R$ .

Proof.

Clearly, (i) implies (ii) and (ii) implies (iii). Let $m \in M$ be nonzero. Let $I$ be annihilator of $m$ in $R$ , which is to say $I = Ann (𝑅𝑚)$ . Then $I$ is a proper ideal, hence contained in a maximal ideal $𝔪$ of $R$ . If $\frac{r}{u} \in R_{𝔪}$ annihilates $m$ , then $𝑟𝑠𝑚 = 0$ for some $s \in R - 𝔪$ . Thus $𝑟𝑠 \in 𝔪$ , so $r \in 𝔪$ as $𝔪$ is prime. This implies that the annihilator of $m$ in $R_{𝔪}$ in a proper ideal, so $m$ is nonzero in $M_{𝔪}$ . Thus, we have the contrapositive to (iii) implies (i). □

Proposition 11.1.12. §

Let $R$ be a commutative ring and $f : M \to N$ a homomorphism of $R$ -modules. Then the following are equivalent:

i.: $f$ is injective,
ii.: the maps $f_{𝔭} : M_{𝔭} \to N_{𝔭}$ induced by $f$ are injective for all prime ideals $𝔭$ of $R$ , and
iii.: the maps $f_{𝔪} : M_{𝔪} \to N_{𝔪}$ are injective for all maximal ideals $𝔪$ of $R$ .

We have the following.

Proposition 11.1.13. §

Let $𝔪$ be a maximal ideal of a commutative ring $R$ . For $n \geq 0$ , then canonical map

𝔪^{n} ∕ 𝔪^{n + 1} \to 𝔪^{n} R_{𝔪} ∕ 𝔪^{n + 1} R_{𝔪}

is an isomorphism of $R ∕ 𝔪$ -vector spaces.

Proof.

Note that $R_{𝔪} \otimes_{R} 𝔪^{n} ≅ 𝔪^{n} R_{𝔪}$ by Lemma 11.1.9, the isomorphism being induced by the $R$ -linear map $(\frac{a}{s}, b) \mapsto \frac{𝑎𝑏}{s}$ . Therefore, we have an isomorphism

R_{𝔪} \otimes_{R} 𝔪^{n} ∕ 𝔪^{n + 1} \to_{}^{\sim} 𝔪^{n} R_{𝔪} ∕ 𝔪^{n + 1} R_{𝔪}

by right exactness of tensor product with $R_{𝔪}$ .

Now consider the composite isomorphism

\begin{matrix} R ∕ 𝔪 \otimes_{R} 𝔪^{n} ∕ 𝔪^{n + 1} \to_{}^{\sim} 𝔪^{n} ∕ 𝔪^{n + 1} \to 𝔪^{n} R_{𝔪} ∕ 𝔪^{n + 1} R_{𝔪} \\ \to_{}^{\sim} R_{𝔪} \otimes_{R} 𝔪^{n} ∕ 𝔪^{n + 1} \to_{}^{\sim} R_{𝔪} ∕ 𝔪 R_{𝔪} \otimes_{R} 𝔪^{n} ∕ 𝔪^{n + 1} \to_{}^{\sim} R ∕ 𝔪 \otimes_{R} 𝔪^{n} ∕ 𝔪^{n + 1}, \end{matrix}

where in the first and last steps we have used that the maximal ideal annihilates the successive quotients of powers of the maximal ideal in both the case of $R$ and $R_{𝔪}$ and in the last step we have used the $n = 0$ case found in Lemma 8.1.31. An element of the form $1 \otimes x$ with $x \in 𝔪^{n} ∕ 𝔪^{n + 1}$ is sent under the composition to $1 \otimes x$ . The composite is also easily seen to be a map of $R ∕ 𝔪$ -vector spaces, so it is the identity being that it is the identity on simple tensors. In particular, the map $𝔪^{n} ∕ 𝔪^{n + 1} \to 𝔪^{n} R_{𝔪} ∕ 𝔪^{n + 1} R_{𝔪}$ is an isomorphism. □

11.2. Radicals of ideals

Let $R$ be a commutative ring.

Definition 11.2.1. §

The radical $\sqrt{I}$ of an ideal $I$ of $R$ is the set

\sqrt{I} = {a \in R ∣ a^{k} \in I for some k \geq 1} .

Lemma 11.2.2. §

For any ideal $I$ of $R$ , the radical $\sqrt{I}$ of $I$ is an ideal of $R$ .

Proof.

If $r \in R$ and $a \in \sqrt{I}$ , then there exists $k \geq 1$ with $a^{k} \in I$ , and then ${(𝑟𝑎)}^{k} = r^{k} a^{k} \in I$ , so $𝑟𝑎 \in \sqrt{I}$ as well. If we also have $b \in \sqrt{l}$ with $b^{l} \in I$ , then

{(a + b)}^{k + l} = \sum_{i = 0}^{k + l} (\binom{k + l}{i}) a^{i} b^{k + l - i} \in I

since either $i \geq k$ or $k + l - i \geq l$ if $i \in ℤ$ . Thus $a + b \in \sqrt{I}$ . □

The nilradical of $R$ is the radical of the ideal $(0)$ of $R$ .

Definition 11.2.3. §

The nilradical of $R$ is the ideal of nilpotent elements in $R$ .

Example 11.2.4. §

The nilradical of $F [x] ∕ (x^{n})$ for a field $F$ and $n \geq 2$ is generated by $x$ .

Definition 11.2.5. §

A commutative ring is reduced if it has no nonzero nilpotent elements.

Examples 11.2.6. §

a.: Domains are reduced.
b.: The quotient of a commutative ring by its nilradical is reduced.
c.: The ring $ℤ ∕ 6 ℤ$ is reduced, though it is not a domain.

In fact, the following is easily verified.

Lemma 11.2.7. §

If $π : R \to R ∕ I$ is the projection of $R$ onto its quotient by an ideal $I$ , then $π (\sqrt{I})$ is the nilradical of $R ∕ I$ .

Definition 11.2.8. §

An ideal is radical, or a semiprime ideal, if it is its own radical.

Examples 11.2.9. §

a.

Prime ideals are radical. That is, if $a^{n} \in 𝔭$ for some $n \geq 1$ , then $a \in 𝔭$ by primality.

b.

Let $F$ be a field, $f_{1}, \dots, f_{r} \in F [x]$ be irreducible, and $k_{1}, \dots, k_{r}$ be positive integers. Then

\sqrt{(f_{1}^{k_{1}} \dots f_{r}^{k_{r}})} = (f_{1} \dots f_{r}) .

Thus, the nonzero radical ideals of $F [x]$ are exactly the ideals generated by products of distinct irreducible elements.

c.

Radicals of ideals are radical.

Proposition 11.2.10. §

Let $I$ be a proper ideal of $R$ . Then $\sqrt{I}$ is the intersection of all prime ideals of $R$ containing $I$ .

Proof.

If $𝔭$ is a prime ideal containing $I$ , then $𝔭 = \sqrt{𝔭} \supseteq \sqrt{I}$ . Thus, $\sqrt{I}$ is contained in the intersection of all prime ideals $J$ containing $I$ .

Suppose that $a \in J - \sqrt{I}$ . Then $S = {a^{n} ∣ n \geq 0}$ is a multiplicative set disjoint from $I$ . By Theorem 8.1.33, there exists a prime ideal $𝔭$ containing $I$ and disjoint from $S$ . Then $a$ cannot be in the intersection $J$ , which is a contradiction. Thus, we have that $J = \sqrt{I}$ . □

Proposition 11.2.11. §

Let $R$ be noetherian and $I$ be an ideal of $R$ . Then there exists $n \geq 1$ such that ${(\sqrt{I})}^{n} \subseteq I$ .

Proof.

Let $a_{1}, \dots, a_{m} \in R$ be such that $\sqrt{I} = (a_{1}, \dots, a_{m})$ . For $1 \leq i \leq m$ , let $k_{i} \geq 1$ be such that $a_{i}^{k_{i}} \in I$ , and let $k = \max {k_{i} ∣ 1 \leq i \leq m}$ . For any $x = \sum_{i = 1}^{m} r_{i} a_{i} \in \sqrt{I}$ , we have

x^{𝑘𝑚} \in ({a_{1}^{i_{1}} \dots a_{m}^{i_{m}} ∣ i_{j} \geq 0 for all j with i_{1} + \dots + i_{m} = 𝑘𝑚}) \subseteq (a_{1}^{k}, \dots, a_{m}^{k}) \subseteq I .

Thus, we may take $n = 𝑘𝑚$ in the statement. □

Definition 11.2.12. §

An ideal $I$ of $R$ is nilpotent if there exists $n \geq 1$ such that $I^{n} = 0$ .

Corollary 11.2.13. §

The nilradical of a noetherian commutative ring is nilpotent.

It is easy to see why this can fail in a noncommutative ring.

Example 11.2.14. §

Consider the polynomial ring $R = F [x_{1}, x_{2}, \dots]$ in countably many variables over a field $F$ and its ideal $I = (x_{k}^{k} ∣ k \geq 1)$ . Its radical is $\sqrt{I} = (x_{k} ∣ k \geq 1)$ but no power of $\sqrt{I}$ is contained in $I$ .

The taking of radicals behaves well with respect to localization.

Lemma 11.2.15. §

Let $S$ be a multiplicatively closed subset of $R$ , and let $I$ be an ideal of $R$ . Then $S^{- 1} \sqrt{I} = \sqrt{S^{- 1} I}$ .

11.3. Primary decomposition

Definition 11.3.1. §

A proper ideal $𝔮$ of $R$ is primary if for any $a, b \in R$ with $𝑎𝑏 \in 𝔮$ , one has either $a \in 𝔮$ or $b^{n} \in 𝔮$ for some $n \geq 1$ .

That is, an ideal $𝔮$ is primary if whenever $𝑎𝑏 \in 𝔮$ , either $a \in 𝔮$ or $b \in \sqrt{𝔮}$ . Of course, prime ideals are primary. The following is just a rephrasing of the definition of primary.

Lemma 11.3.2. §

A proper ideal $𝔮$ of $R$ is primary if and only if every zero divisor in $R ∕ 𝔮$ is nilpotent.

The following is a key property of primary ideals.

Proposition 11.3.3. §

The radical of any primary ideal is a prime ideal.

Proof.

Let $𝔮$ be a primary ideal of $R$ . If $a, b \in R$ with $𝑎𝑏 \in \sqrt{𝔮}$ , then $a^{k} b^{k} \in 𝔮$ for some $k \geq 1$ , and therefore either $a^{k} \in 𝔮$ or there exists $n \geq 1$ such that $b^{𝑘𝑛} \in 𝔮$ . In the first, case $a \in \sqrt{𝔮}$ , and in the second, $b \in \sqrt{𝔮}$ , so $\sqrt{𝔮}$ is prime. □

In particular, the radical of a primary ideal $𝔮$ is the smallest prime ideal of $R$ containing $𝔮$ , given that it is also the intersection of all prime ideals containing $𝔮$ .

Definition 11.3.4. §

The radical $𝔭$ of a primary ideal $𝔮$ of $R$ is called the associated prime to $𝔮$ , and we say that $𝔮$ is $𝔭$ -primary.

Examples 11.3.5. §

Let $F$ be a field.

a.: The ideal $(x^{2}, y)$ of $F [x, y]$ is primary since $F [x, y] ∕ (x^{2}, y) ≅𝐹 [x] ∕ (x^{2})$ , and every zero divisor in the latter ring is nilpotent. Its associated prime is $(x, y)$ .
b.: Consider $R = F [x, y, z] ∕ (𝑥𝑦 - z^{2})$ and its ideal $𝔭 = (x, z)$ , which is prime since $R ∕ 𝔭≅𝐹 [y]$ . We have $𝑥𝑦 \in 𝔭^{2}$ , but $x \notin 𝔭^{2}$ and $y \notin \sqrt{𝔭^{2}} = 𝔭$ . Thus $𝔭^{2}$ is not primary, even though $𝔭$ is prime.

Lemma 11.3.6. §

If $I$ is an ideal of $R$ such that $\sqrt{I}$ is maximal, then $I$ is primary. In particular, any power of a maximal ideal $𝔪$ is primary with associated prime $𝔪$ .

Proof.

Suppose that $𝔪 = \sqrt{I}$ is maximal. The image of $𝔪$ in $R ∕ I$ is the nilradical of $R ∕ I$ , which means that the nilradical is the only prime ideal of $R ∕ I$ . In particular, $R ∕ I$ is local, and every element of $R ∕ I$ that is not nilpotent is a unit. In particular, every zero divisor of $R ∕ I$ is nilpotent. Thus, $I$ is $𝔪$ -primary. □

The following is easily checked.

Lemma 11.3.7. §

A finite intersection of primary ideals with the same associated prime is primary.

Definition 11.3.8. §

Let $I$ be an ideal of $R$ .

a.: A primary decomposition of $I$ is a finite collection ${𝔮_{1}, \dots, 𝔮_{k}}$ of primary ideals of $R$ such that $I = ⋂_{i = 1}^{k} 𝔮_{i}$ .
b.: We say that $I$ is decomposable if it has a primary decomposition.
c.: A primary decomposition ${𝔮_{1}, \dots, 𝔮_{k}}$ of $I$ is minimal if the radicals $\sqrt{𝔮_{i}}$ are all distinct and no proper subset of the primary decomposition is also a primary decomposition of $I$ .

Every ideal with a primary decomposition has a minimal such decomposition.

Lemma 11.3.9. §

Let $I$ be a decomposable ideal of $R$ . Then $I$ has a minimal primary decomposition.

Proof.

From this decomposition, we may first remove one at a time any primary ideals that contain the intersection of the others. By Lemma 11.3.7, we may then replace the subcollection of those ideals in the decomposition with the same associated prime by the single primary ideal that is its intersection. The resulting collection is minimal. □

Example 11.3.10. §

Let $F$ be a field. The ideal $(x y^{2})$ of $F [x, y]$ has a minimal primary decomposition $(x y^{2}) = (x) \cap (y^{2})$ , and the associated primes of these primary ideals are $(x)$ and $(y)$ .

Definition 11.3.11. §

A proper ideal $I$ of $R$ is irreducible if for any ideals $𝔞$ and $𝔟$ of $R$ with $I = 𝔞 \cap 𝔟$ , either $I = 𝔞$ or $I = 𝔟$ .

Proposition 11.3.12. §

Let $R$ be noetherian. Then every irreducible ideal of $R$ is primary.

Proof.

Let $I$ be an irreducible ideal of $R$ , and let $a, b \in R$ with $𝑎𝑏 \in I$ but $b \notin I$ . For each $n \geq 1$ , let $J_{n} = {r \in R ∣ a^{n} r \in I}$ , and note that $J_{n}$ is an ideal of $R$ . Then $J_{n}$ form an ascending chain of ideals containing $I$ , and since $R$ is noetherian, this chain is eventually constant, say $J_{n} = J_{n + 1}$ for all $n \geq N$ with $N \geq 1$ . Consider the ideals $𝔞 = (a^{N}) + I$ and $𝔟 = (b) + I$ containing $I$ . We claim that $𝔞 \cap 𝔟 = I$ . Let $c \in 𝔞 \cap 𝔟$ . Then $c = a^{N} r + q$ for some $r \in R$ and $q \in I$ . Since $c \in (b) + I$ , we have $𝑎𝑐 \in (𝑎𝑏) + I = I$ . In other words, $a^{N + 1} r + 𝑞𝑎 \in I$ , so $a^{N + 1} r \in I$ , so $r \in J_{N + 1} = J_{N}$ . Therefore $a^{N} r \in I$ as well, so $c \in I$ , and the claim holds. Since $I$ is irreducible and $b \notin I$ , we must have $I = 𝔞$ , which means that $a^{N} \in I$ . Therefore, $I$ is primary. □

Examples 11.3.13. §

Let $F$ be a field, and consider the ring $F [x, y]$ .

The ideal ${(x, y)}^{2}$ is $(x, y)$ -primary, but ${(x, y)}^{2} = (x^{2}, y) \cap (x, y^{2})$ , so ${(x, y)}^{2}$ is not irreducible. Thus, the converse to Proposition 11.3.12 does not hold.
The ideal $(x^{2}, y)$ is irreducible, and therefore primary, but it is not prime.

Proposition 11.3.14. §

Let $R$ be noetherian. The every proper ideal of $R$ is a finite intersection of irreducible ideals.

Proof.

Let $X$ be the set of proper ideals of $R$ that cannot be written as a finite intersection of irreducible ideals of $R$ . Since $R$ is noetherian, either $X$ is empty or $X$ has a maximal element $𝔪$ . Since $𝔪 \in X$ , it is not irreducible, so there exist ideals $𝔞$ and $𝔟$ properly containing $𝔪$ with $𝔪 = 𝔞 \cap 𝔟$ . Since $𝔪$ is maximal in $X$ , both of $𝔞$ and $𝔟$ can be written as a finite intersection of irreducible ideals, so $𝔪$ may be as well, which contradicts the existence of $𝔪$ . Therefore $X$ is empty, as desired. □

Combining Propositions 11.3.12 and 11.3.14, we have the following.

Theorem 11.3.15 (Primary decomposition theorem). §

Every proper ideal of a noetherian commutative ring $R$ is decomposable.

Definition 11.3.16. §

For ideals $I$ and $J$ of a commutative ring $R$ , the ideal quotient of $I$ by $J$ is $(I : J) = {r \in R ∣ 𝑟𝐽 \subseteq I}$ .

Notation 11.3.17. §

For an ideal $I$ of $R$ and $a \in I$ , we set $I_{a} = (I : (a))$ .

Note that ideal quotients are ideals.

Lemma 11.3.18. §

Let $𝔮$ be a $𝔭$ -primary ideal of $R$ . For $a \in R$ , we have $𝔮_{a} = R$ if $a \in 𝔮$ and $𝔮_{a}$ is $𝔭$ -primary otherwise.

Proof.

If $a \in 𝔮$ , then $1 \in 𝔮_{a}$ , so we are done. If $a \notin 𝔮$ , let $b, c \in R$ with $𝑏𝑐 \in 𝔮_{a}$ and $b \notin 𝔮_{a}$ . Then $𝑎𝑏𝑐 \in 𝔮$ but $𝑎𝑏 \notin 𝔮$ , so $b \notin 𝔮$ , and $c \in \sqrt{𝔮}$ as $𝔮$ is primary. Hence $𝔮_{a}$ is primary. Moreover, we have

\sqrt{𝔮_{a}} = {r \in R ∣ r^{n} a \in 𝔮 for some n \geq 1} = \sqrt{𝔮} = 𝔭 .

□

Corollary 11.3.19. §

Let $I$ be a decomposable ideal of $R$ . Let ${𝔮_{1}, \dots, 𝔮_{n}}$ be a minimal primary decomposition of $I$ , and let $𝔭_{i} = \sqrt{𝔮_{i}}$ for each $i$ . For $a \in R$ , we have

\sqrt{I_{a}} = ⋂_{\begin{array}{c} i = 1 \\ a \notin 𝔮_{i} \end{array}}^{n} 𝔭_{i} .

In particular, for each $i$ , there exists an $a \in R$ such that $\sqrt{I_{a}} = 𝔭_{i}$ .

Proof.

For $a \in R$ , we have $I_{a} = ⋂_{i = 1}^{n} {(𝔮_{i})}_{a}$ . By Lemma 11.3.18, we have

\sqrt{I_{a}} = ⋂_{i = 1}^{n} \sqrt{{(𝔮_{i})}_{a}} = ⋂_{\begin{array}{c} i = 1 \\ a \notin 𝔮_{i} \end{array}}^{n} 𝔭_{i} .

We may choose $a \in 𝔮_{j}$ for $j \neq i$ such that $a \notin 𝔮_{i}$ by the minimality of the decomposition. For such an $a$ , we then have $\sqrt{I_{a}} = 𝔭_{i}$ . □

In fact, for noetherian $R$ , we have the following refinement of the last statement of Corollary 11.3.19.

Lemma 11.3.20. §

Suppose that $R$ is noetherian, and let $I$ be a proper ideal of $R$ . Let $𝔭$ be an associated prime of $I$ . Then there exists $a \in R$ such that $I_{a} = 𝔭_{i}$ .

Proof.

Let $𝔮$ be a $𝔭$ -primary ideal in a minimal primary decomposition of $I$ , and let $J$ be the intersection of the other primary ideals set decomposition. Consider the nonempty set $X = {I_{a} ∣ a \in J, a \notin 𝔮}$ , and note that by Corollary 11.3.19, we have $\sqrt{I_{a}} = 𝔭$ for each $I_{a} \in X$ . As $R$ is noetherian, the set $X$ contains a maximal element $I_{b}$ for some $b \in J$ , $b \notin 𝔮$ . It will suffice to show that $I_{b}$ is prime, since in that case $I_{b}$ is radical with radical $𝔭$ .

Now suppose that $c, d \in R$ are such that $𝑐𝑑 \in I_{b}$ but $c \notin I_{b}$ , which is to say $𝑏𝑐𝑑 \in I$ but $𝑏𝑐 \notin I$ . We know that $𝑏𝑐 \in J$ since $b \in J$ , so $𝑏𝑐 \notin 𝔮$ . But then $I_{𝑏𝑐} \in X$ contains the maximal element $I_{b}$ , so $I_{𝑏𝑐} = I_{b}$ , and therefore $d \in I_{b}$ . That is, $I_{b}$ is prime. □

This lemma has the following interesting consequence.

Proposition 11.3.21. §

Let $R$ be noetherian, and let $I$ be proper. An element $x \in R$ reduces to a zero divisor in $R ∕ I$ if and only if it is contained in an associated prime of $I$ .

Proof.

Let $Z$ be the set of elements of $R$ that reduce to zero divisors modulo $I$ . Then $Z = ⋃_{a \in R - I} I_{a}$ . Let $𝔭$ be an associated primes of $I$ . By Lemma 11.3.20, we may choose $b \in R$ with $I_{b} = 𝔭$ Since $b \notin I$ as $I_{b} \neq R$ , have $𝔭 \subseteq Z$ .

Next take $r \in Z$ , and define $X_{r} = {I_{a} ∣ a \notin I, r \in I_{a}}$ , which is nonempty by definition of $Z$ . Since $R$ is noetherian, it has a maximal element $I_{b}$ for some $b \notin I$ . It suffices to show that $I_{b}$ is prime, as it contains $r$ . For this, let $c, d \in R$ with $𝑐𝑑 \in I_{b}$ but $c \notin I_{b}$ . Then $𝑏𝑐𝑑 \in I$ but $𝑏𝑐 \notin I$ . Since $r \in I_{𝑏𝑐}$ , we have $I_{𝑏𝑐} \in X_{r}$ , and it contains $I_{b}$ , so $I_{𝑏𝑐} = I_{b}$ by maximality of $I_{b}$ . As $d \in I_{𝑏𝑐}$ , we then have $d \in I_{b}$ . Thus, $I_{b}$ is prime. □

We now consider uniqueness of primary decompositions, given the existence of one.

Theorem 11.3.22. §

Let $I$ be a decomposable ideal of $R$ . The set of associated primes to the primary ideals in a minimal primary decomposition of $I$ is uniquely determined by $I$ .

Proof.

Let ${𝔮_{1}, \dots, 𝔮_{n}}$ be a minimal primary decomposition of $I$ , with $𝔭_{i} = \sqrt{𝔮_{i}}$ for each $i$ . By Corollary 11.3.19, we may choose $a$ such that $\sqrt{I_{a}} = 𝔭_{i}$ for any $i$ . On the other hand, for any $a \in R$ such that $\sqrt{I_{a}}$ is prime, part b of Lemma 3.10.19 and Corollary 11.3.19 tell us that $\sqrt{I_{a}} = 𝔭_{i}$ for some $i$ (with $a \notin 𝔮_{i}$ ). Thus, the set of associated primes ${𝔭_{1}, \dots, 𝔭_{n}}$ is uniquely determined by $I$ . □

Definition 11.3.23. §

Let $I$ be a decomposable ideal of $R$ . A prime ideal is called an associated prime of $I$ if it is the associated prime of an element of a minimal primary decomposition of $I$ .

Definition 11.3.24. §

Let $I$ be a proper ideal of $R$ . An isolated prime of $I$ is a minimal element in the set of prime ideals of $R$ containing $I$ , ordered by inclusion.

Remark 11.3.25. §

If $I$ is a proper ideal of a noetherian ring $R$ , then every prime ideal of $R ∕ I$ is contained in a minimal prime, and the minimal primes of $R ∕ I$ are exactly the images of the isolated primes of $I$ .

Proposition 11.3.26. §

Let $I$ be a decomposable ideal of $R$ . A prime ideal of $R$ is an isolated prime of $I$ if and only if it is a minimal element under inclusion in the set of associated primes of $I$ .

Proof.

Let $I = ⋂_{i = 1}^{n} 𝔮_{i}$ with each $𝔮_{i}$ primary, and let $𝔭_{i}$ be the associated prime of $𝔮_{i}$ . If $𝔭$ is a prime ideal of $R$ containing $I$ , then

𝔭 = \sqrt{𝔭} \supseteq \sqrt{I} = ⋂_{i = 1}^{n} \sqrt{𝔮_{i}} = ⋂_{i = 1}^{n} 𝔭_{i} .

By part b of Lemma 3.10.19, we have that $𝔭$ contains some $𝔭_{i}$ for some $i$ , so it contains some minimal prime in the set of associated primes of $I$ . □

Remark 11.3.27. §

If $(0)$ is decomposable, then $R$ has finitely many minimal primes. These are exactly the isolated primes in a minimal primary decomposition of $(0)$ .

Definition 11.3.28. §

Let $I$ be a decomposable ideal of $R$ . An embedded prime of $I$ is an associated prime of $I$ that is not isolated.

Example 11.3.29. §

Let $F$ be a field. The ideal $I = (𝑥𝑦, y^{2})$ of $F [x, y]$ has a minimal primary decomposition $I = {(x, y)}^{2} \cap (y)$ , so it has associated primes $(x, y)$ and $(y)$ . The ideal $(y)$ is the unique isolated prime of $I$ . Note that $I$ also has the primary decomposition $I = (x, y^{2}) \cap (y)$ .

In fact, the following uniqueness result also holds.

Proposition 11.3.30. §

Let $I$ be a decomposable ideal of $R$ . Let $𝔭_{1}, \dots, 𝔭_{n}$ be distinct isolated primes of $I$ . Let $Q$ be a minimal primary decomposition of $I$ , and let $𝔮_{i} \in Q$ be $𝔭_{i}$ -primary for each $1 \leq i \leq n$ . Then the ideal $⋂_{i = 1}^{n} 𝔮_{i}$ is independent of the choice of $Q$ .

Proof.

Consider the multiplicative set $S = R - ⋃_{i = 1}^{n} 𝔭_{i}$ . For any associated prime ideal $𝔭$ of $I$ , the intersection $S \cap 𝔭$ is empty if and only if $𝔭 = 𝔭_{i}$ for some $i$ by Proposition 8.1.20. Given a primary decomposition $Q$ , suppose that $𝔮 \in Q$ is $𝔭$ -primary for some prime $𝔭$ . If $𝔭 = 𝔭_{i}$ for some $i$ , then $S^{- 1} 𝔭$ is a prime of $S^{- 1} R$ and $S^{- 1} 𝔮_{i}$ is $S^{- 1} 𝔭$ -primary, and otherwise we have $S^{- 1} 𝔭 = S^{- 1} R$ . We then have

S^{- 1} I = ⋂_{𝔮 \in Q} S^{- 1} 𝔮 = ⋂_{i = 1}^{n} S^{- 1} 𝔮_{i},

and the contraction of $S^{- 1} I$ to $R$ is the intersection of the contractions of the primary ideals $S^{- 1} 𝔮_{i}$ . By Lemma 8.1.19(b), the contraction of $S^{- 1} 𝔮_{i}$ is the ideal of $a \in 𝔮_{i}$ such that there exists $r \in S$ with $𝑟𝑎 \in 𝔮_{i}$ . Since $S$ is contained in the complement of $𝔭_{i}$ , we have $r \notin 𝔮_{i}$ , which as $𝔮_{i}$ is primary forces such an $a$ to lie in $𝔮_{i}$ . In other words, $⋂_{i = 1}^{n} 𝔮_{i}$ is the contraction of $S^{- 1} I$ , and this is independent of the choice of $Q$ . □

Corollary 11.3.31. §

If $𝔭$ is any isolated prime of a decomposable ideal $I$ , then the unique $𝔭$ -primary ideal in any minimal primary decomposition of $I$ is independent of the choice of decomposition.

11.4. Integral extensions

Definition 11.4.1. §

We say that $B ∕ A$ is an extension of commutative rings if $A$ and $B$ are commutative rings such that $A$ is a subring of $B$ .

Definition 11.4.2. §

Let $B ∕ A$ be an extension of commutative rings. We say that $β \in B$ is integral over $A$ if $β$ is the root of a monic polynomial in $A [x]$ .

Examples 11.4.3. §

a.: Every element $a \in A$ is integral over $A$ , in that $a$ is a root of $x - a$ .
b.: If $L ∕ K$ is a field extension and $α \in L$ is algebraic over $K$ , then $α$ is integral over $K$ , being a root of its minimal polynomial, which is monic.
c.: If $L ∕ K$ is a field extension and $α \in L$ is transcendental over $K$ , then $α$ is not integral over $K$ .
d.: The element $\sqrt{2}$ of $ℚ (\sqrt{2})$ is integral over $ℤ$ , as it is a root of $x^{2} - 2$ .
e.: The element $α = \frac{1 - \sqrt{5}}{2}$ of $ℚ (\sqrt{5})$ is integral over $ℤ$ , as it is a root of $x^{2} - x - 1$ .

Proposition 11.4.4. §

Let $B ∕ A$ be an extension of commutative rings. For $β \in B$ , the following conditions are equivalent:

i.: the element $β$ is integral over $A$ ,
ii.: there exists $n \geq 0$ such that ${1, β, \dots, β^{n}}$ generates $A [β]$ as an $A$ -module,
iii.: the ring $A [β]$ is a finitely generated $A$ -module, and
iv.: there exists a faithful $A [β]$ -submodule of $B$ that is finitely generated over $A$ .

Proof.

Suppose that (i) holds. Then $β$ is a root of a monic polynomial $g \in A [x]$ . Given any $f \in A [x]$ , the division algorithm tells us that $f = 𝑞𝑔 + r$ with $q, r \in A [x]$ and either $r = 0$ or $\deg r < \deg g$ . It follows that $f (β) = r (β)$ , and therefore that $f (β)$ is in the $A$ -submodule generated by ${1, β, \dots, β^{\deg g - 1}}$ , so (ii) holds. Condition (ii) is clearly at least as strong as (iii). Suppose that (iii) holds. Then we may take the $A [β]$ -submodule $A [β]$ of $B$ , which being free over itself has trivial annihilator.

Finally, suppose that (iv) holds. Let

M = \sum_{i = 1}^{n} A γ_{i} \subseteq B

be a faithful $A [β]$ -module, and suppose without loss of generality that $β \neq 0$ . We have

β γ_{j} = \sum_{j = 1}^{n} c_{𝑖𝑗} γ_{i}

for some $c_{𝑖𝑗} \in A$ with $1 \leq i, j \leq n$ . The characteristic polynomial $c (x) \in A [x]$ of the matrix $C = (c_{𝑖𝑗})$ is monic, and as $β I_{n} - C$ takes $(γ_{1}, \dots, γ_{n}) \in M^{n}$ to zero, $c (β)$ acts as zero on $M$ . Since $M$ is a faithful $A [β]$ -module, we must have $c (β) = 0$ . Thus, $β$ is integral. □

Example 11.4.5. §

The element $\frac{1}{2} \in ℚ$ is not integral over $ℤ$ , as $ℤ [1, 2^{- 1}, \dots, 2^{- n}]$ for $n \geq 0$ is equal to $ℤ [2^{- n}]$ , which does not contain $2^{- (n + 1)}$ .

Definition 11.4.6. §

Let $B ∕ A$ be an extension of commutative rings. We say that $B$ is an integral extension of $A$ if every element of $B$ is integral over $A$ .

Example 11.4.7. §

The ring $ℤ [\sqrt{2}]$ is an integral extension of $ℤ$ . Given $α = a + b \sqrt{2}$ with $a, b \in ℤ$ , note that $α$ is a root of $x^{2} - 2 𝑎𝑥 + a^{2} - 2 b^{2}$ .

The integral extensions of a field are its algebraic field extensions.

Lemma 11.4.8. §

Let $B ∕ A$ be an integral extension of domains. Then $B$ is a field if and only if $A$ is a field.

Proof.

Suppose that $A$ is a field. Every $b \in B \subseteq Q (B)$ is the root of a monic polynomial with coefficients in $A$ , so $A (b) = A [b] \subseteq B$ . That is, $b \in B^{\times} \cup {0}$ , and thus $B$ is a field.

Now suppose that $B$ is a field. Then for any $a \in A - {0}$ , the element $a^{- 1}$ is integral over $A$ , so there is a monic polynomial $f \in A [x]$ with $f (a^{- 1}) = 0$ . Write $f (x) = x^{n} + g (x)$ with $\deg g < n$ for some $n \geq 1$ . Then $a^{- 1} = - a^{n - 1} g (a^{- 1}) \in A$ , so $a \in A^{\times}$ . □

Remark 11.4.9. §

An extension of fields is integral if and only if it is algebraic.

Lemma 11.4.10. §

Suppose that $B ∕ A$ is an extension of commutative rings such that $B$ is finitely generated as an $A$ -module, and let $M$ be a finitely generated $B$ -module. Then $M$ is a finitely generated $A$ -module.

Proof.

Let ${m_{1}, \dots, m_{n}}$ be a set of generators of $M$ as a $B$ -module, and let ${β_{1}, \dots, β_{k}}$ be a set of generators of $B$ as an $A$ -module. We claim that ${β_{i} m_{j} ∣ 1 \leq i \leq k, 1 \leq j \leq n}$ is a set of generators of $M$ as an $A$ -module. To see this, let $m \in M$ and write

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

with $b_{j} \in B$ for $1 \leq j \leq n$ . For $1 \leq j \leq n$ , we then write

b_{j} = \sum_{i = 1}^{k} a_{𝑖𝑗} β_{i}

with $a_{𝑖𝑗} \in A$ for $1 \leq i \leq k$ . We then have

m = \sum_{i = 1}^{k} \sum_{j = 1}^{n} a_{𝑖𝑗} β_{i} m_{j},

as desired. □

We now give a criterion for a finitely generated algebra over a ring to be finitely generated as a module.

Proposition 11.4.11. §

Let $B ∕ A$ be an extension of commutative rings and suppose that

B = A [β_{1}, β_{2}, \dots, β_{k}]

for some $k \geq 0$ and $β_{i} \in B$ with $1 \leq i \leq k$ . Then the following are equivalent:

i.: the ring $B$ is integral over $A$ ,
ii.: each $β_{i}$ with $1 \leq i \leq k$ is integral over $A$ , and
iii.: the ring $B$ is finitely generated as an $A$ -module.

Proof.

Clearly, (i) implies (ii), so suppose that (ii) holds. By definition, each $β_{i}$ is then integral over any commutative ring containing $A$ . By Proposition 11.4.4, each $A [β_{1}, \dots, β_{j}]$ with $1 \leq j \leq k$ is a finitely generated $A [β_{1}, \dots, β_{j - 1}]$ -module, generated by ${1, β_{j}, \dots, β_{j}^{n_{j}}}$ for some $n_{j} \geq 0$ . Assuming recursively that $A [β_{1}, \dots, β_{j - 1}]$ is finitely generated as an $A$ -module, Lemma 11.4.10 implies that $A [β_{1}, \dots, β_{j}] = A [β_{1}, \dots, β_{j - 1}] [β_{j}]$ is finitely generated as an $A$ -module as well. Therefore, (iii) holds. Finally, if (iii) holds and $β \in B$ , then since $B$ is a faithful $A [β]$ -module (as $1 \in B$ ), the element $β$ is integral over $a$ by Proposition 11.4.4. Thus (i) holds. □

We derive the following important consequence.

Proposition 11.4.12. §

Suppose that $C ∕ B$ and $B ∕ A$ are integral extensions of commutative rings. Then $C ∕ A$ is an integral extension as well.

Proof.

Let $γ \in C$ , and let $f \in B [x]$ be a monic polynomial which has $γ$ as a root. Let $B^{'}$ be the subring of $B$ generated over $A$ by the coefficients of $f$ , which is integral over $A$ as $B$ is. By Proposition 11.4.11, the ring $B^{'}$ is then finitely generated as an $A$ -module. As $B^{'} [γ]$ is a finitely generated $B^{'}$ -module as well, we have $B^{'} [γ]$ is finitely generated as an $A$ -module. Hence, $B [γ]$ is itself an integral extension of $A$ . By definition of an integral extension, the element $γ$ is integral over $A$ . Since $γ \in C$ was arbitrary, we conclude that $C$ is integral over $A$ . □

Definition 11.4.13. §

Let $B ∕ A$ be an extension of commutative rings. The integral closure of $A$ in $B$ is the set of elements of $B$ that are integral over $A$ .

Proposition 11.4.14. §

Let $B ∕ A$ be an extension of commutative rings. Then the integral closure of $A$ in $B$ is a subring of $B$ .

Proof.

If $α$ and $β$ are elements of $B$ that are integral over $A$ , then $A [α, β]$ is integral over $A$ by Proposition 11.4.11. Therefore, every element of $A [α, β]$ , including $- α$ , $α + β$ , and $α \cdot β$ , is integral over $A$ as well. That is, the integral closure of $A$ in $B$ is closed under addition, additive inverses, and multiplication, and it contains $1$ , so it is a ring. □

Example 11.4.15. §

The integral closure of $ℤ$ in $ℤ [x]$ is $ℤ$ , since if $f \in ℤ [x]$ is of degree at least $1$ and $g \in ℤ [x]$ is nonconstant, then $g (f (x))$ has degree $\deg g \cdot \deg f$ in $x$ , hence cannot be $0$ .

Definition 11.4.16. §

a.: The ring of algebraic integers is the integral closure $\bar{ℤ}$ of $ℤ$ inside $ℂ$ .
b.: An algebraic integer is an element of $\bar{ℤ}$ .

Definition 11.4.17. §

Let $B ∕ A$ be an extension of commutative rings. We say that $A$ is integrally closed in $B$ if $A$ is its own integral closure in $B$ .

Definition 11.4.18. §

We say that an integral domain $A$ is integrally closed if it is integrally closed in its quotient field.

More generally, we may make the following definition.

Definition 11.4.19. §

A commutative ring is normal if it is integrally closed in its total ring of fractions.

In particular, a domain is is integrally closed if and only if it is normal.

Example 11.4.20. §

Every field is integrally closed.

Proposition 11.4.21. §

Let $A$ be an integrally closed domain, let $K$ be the quotient field, and let $L$ be a field extension of $K$ . If $β \in L$ is integral over $A$ with minimal polynomial $f \in K [x]$ , then $f \in A [x]$ .

Proof.

Since $β \in L$ is integral, it is the root of some monic polynomial $g \in A [x]$ such that $f$ divides $g$ in $K [x]$ . As $g$ is monic, every root of $g$ in an algebraic closure $\bar{K}$ containing $K$ is integral over $A$ . As every root of $f$ is a root of $g$ , the same is true of the roots of $f$ . Write $f = \prod_{i = 1}^{n} (x - β_{i})$ for $β_{i} \in \bar{K}$ integral over $A$ . As the integral closure of $A$ in $\bar{K}$ is a ring, it follows that every coefficient of $f$ is integral over $A$ , being sums of products of the elements $β_{i}$ . Since $f \in K [x]$ and $A$ is integrally closed, we then have $f \in A [x]$ . □

In particular, we have the following result on the norm and trace on quotient fields on integral extensions of domains.

Corollary 11.4.22. §

Let $B ∕ A$ be an integral extension of domains, and suppose that $A$ is integrally closed in its quotient field $K$ . Let $L$ denote the quotient field of $B$ , and suppose that $L ∕ K$ is finite. Then $N_{L ∕ K} (β)$ and ${Tr}_{L ∕ K} (β)$ are elements of $A$ for every $β \in B$ .

Proof.

Since $β$ is integral over $A$ , so are all of its conjugates in an algebraic closure of $L$ , since they are also roots of the monic polynomial of which $β$ is a root. It follows from Proposition 10.1.4 that $N_{L ∕ K} (β)$ and ${Tr}_{L ∕ K} (β)$ are elements of $B \cap K$ , which is $A$ since $A$ is integrally closed. □

The following holds in the case of UFDs.

Proposition 11.4.23. §

Let $A$ be a UFD, let $K$ be the quotient field of $A$ , and let $L$ be a field extension of $K$ . If $β \in L$ is integral over $A$ with minimal polynomial $f \in K [x]$ , then $f \in A [x]$ .

Proof.

Let $β \in L$ be integral over $A$ , let $g \in A [x]$ be a monic polynomial of which it is a root, and let $f \in K [x]$ be the minimal polynomial of $β$ . Since $f$ divides $g$ in $K [x]$ and $A$ is a UFD with quotient field $K$ , there exists $d \in K$ such that $𝑑𝑓 \in A [x]$ and $𝑑𝑓$ divides $g$ in $A [x]$ . Since $f$ is monic, $d$ must be an element of $A$ (and in fact may be taken to be a least common denominator of the coefficients of $f$ ). The coefficient of the leading term of any multiple of $𝑑𝑓$ will be divisible by $d$ , so this forces $d$ to be a unit, in which case $f \in A [x]$ . □

Corollary 11.4.24. §

Every unique factorization domain is integrally closed.

Proof.

The minimal polynomial of an element $a$ of the quotient field $K$ of a UFD $A$ is $x - a$ . If $a \notin A$ , it follows from Proposition 11.4.23 that $a$ is not integral over $A$ . □

Examples 11.4.25. §

The ring $ℤ$ is integrally closed.

Example 11.4.26. §

The ring $ℤ [\sqrt{17}]$ is not integrally closed, since $α = \frac{1 + \sqrt{17}}{2}$ is a root of the monic polynomial $x^{2} - x - 4$ . In particular, $ℤ [\sqrt{17}]$ is not a UFD.

Proposition 11.4.27. §

Let $B ∕ A$ be an extension of commutative rings, and suppose that $B$ is an integrally closed domain. Then the integral closure of $A$ in $B$ is integrally closed.

Proof.

Let $\bar{A}$ denote the integral closure of $A$ in $B$ , and let $Q$ denote the quotient field of $\bar{A}$ . Let $α \in Q$ , and suppose that $α$ is integral over $\bar{A}$ . Then $\bar{A} [α]$ is integral over $\bar{A}$ , so $\bar{A} [α]$ is integral over $A$ , and therefore $α$ is integral over $A$ . That is, $α$ is an element of $\bar{A}$ , as desired. □

Example 11.4.28. §

The ring $\bar{ℤ}$ of algebraic integers is integrally closed.

Proposition 11.4.29. §

Let $A$ be an integral domain with quotient field $K$ , and let $L$ be an algebraic extension of $K$ . Then the integral closure $B$ of $A$ in $L$ has quotient field equal to $L$ inside $L$ . In fact, every element of $L$ may be written as $\frac{b}{d}$ for some $d \in A$ and $b \in B$ .

Proof.

Any $β \in L$ is the root of a nonconstant polynomial $f = \sum_{i = 0}^{n} a_{i} x^{i} \in K [x]$ , with $a_{n} = 1$ . Let $d \in A$ be such that $𝑑𝑓 \in A [x]$ . Then

d^{n} f (d^{- 1} x) = \sum_{i = 0}^{n} a_{i} d^{n - i} x^{i} \in A [x]

is both monic and has $𝑑𝛽$ as a root. In other words, $𝑑𝛽$ is contained in $B$ , as desired. □

Example 11.4.30. §

The quotient field of $\bar{ℤ}$ is $\bar{ℚ}$ .

Definition 11.4.31. §

A number field (or algebraic number field) is a finite field extension of $ℚ$ .

Definition 11.4.32. §

The ring of integers (or integer ring) of a number field $K$ is the integral closure of $ℤ$ in $K$ .

In other words, the ring of integers of a number field is the subring of algebraic integers it contains. The prototypical examples of rings of integers arise in the setting of quadratic fields.

Theorem 11.4.33. §

Let $d \neq 1$ be a square-free integer. The ring $𝒪$ of integers in $ℚ (\sqrt{d})$ is

𝒪 = {\begin{matrix} ℤ [\frac{1 + \sqrt{d}}{2}] & if d \equiv 1 mod 4, \\ ℤ [\sqrt{d}] & if d \equiv 2, 3 mod 4 . \end{matrix}

Proof.

Suppose that $α = a + b \sqrt{d}$ is integral for $a, b \in ℚ$ . If $b = 0$ , then we must have $a \in ℤ$ . If $b \neq 0$ , then the minimal polynomial of $α$ is $f = x^{2} - 2 𝑎𝑥 + a^{?} ? 2 - b^{2} d$ . Since $α$ is integral, then we must have $f \in ℤ [x]$ , so $2 a \in ℤ$ . If $a \in ℤ$ , then since $a^{2} - b^{2} d \in ℤ$ and $d$ is square-free, we have $b \in ℤ$ as well. If $a \notin ℤ$ , then $2 a = a^{'}$ and $2 b = b^{'}$ for some odd $a^{'}, b^{'} \in ℤ$ , and ${(a^{'})}^{2} \equiv {(b^{'})}^{2} d mod 4$ . As ${(ℤ ∕ 4 ℤ)}^{2} = {0, 1}$ , this is impossible if $d ≢ 1 mod 4$ . If $d \equiv 1 mod 4$ , then clearly we can take $a^{'} = b^{'} = 1$ . □

Definition 11.4.34. §

Let $B ∕ A$ be an integral extension of domains such that $A$ is integrally closed, and suppose that $B$ is free of rank $n$ as an $A$ -module. Let $(β_{1}, \dots, β_{n})$ be an ordered basis of $B$ as a free $A$ -module. The discriminant $B$ over $A$ relative to the basis $(β_{1}, \dots, β_{n})$ is $D (β_{1}, \dots, β_{n})$ .

Lemma 11.4.35. §

Let $A$ be an integrally closed domain with quotient field $K$ . Let $L$ be a finite separable extension of $K$ , and let $B$ denote the integral closure of $A$ in $L$ . Let $(α_{1}, \dots, α_{n})$ be any ordered basis of $L$ as a $K$ -vector space that is contained in $B$ . Let $β \in L$ be such that ${Tr}_{L ∕ K} (𝛼𝛽) \in A$ for all $α \in B$ . Then

D (α_{1}, \dots, α_{n}) β \in \sum_{i = 1}^{n} A α_{i} .

Proof.

Since $β \in L$ , we may write

β = \sum_{i = 1}^{n} a_{i} α_{i}

for some $a_{i} \in K$ for $1 \leq i \leq n$ . For any $i$ , we have that

{Tr}_{L ∕ K} (α_{i} β) = \sum_{j = 1}^{n} a_{j} {Tr}_{L ∕ K} (α_{i} α_{j}) .

(11.4.1)

The right-hand side of (11.4.1) is the $i$ th term of the product of the matrix $Q = ({Tr}_{L ∕ K} (α_{i} α_{j}))$ times the column vector with $i$ th entry $a_{i}$ . Since the determinant of $Q$ is $d = D (α_{1}, \dots, α_{n})$ , letting $Q^{*} \in M_{n} (A)$ denote the adjoint matrix to $Q$ , we have $Q^{*} Q = d I_{n}$ . Thus, we have $d a_{i} \in A$ for each $i$ . In other words, $𝑑𝛽$ lies in the $A$ -module generated by the $α_{i}$ , so we are done. □

Proposition 11.4.36. §

Let $A$ be an integrally closed domain with quotient field $K$ . Let $L$ be a finite separable extension of $K$ , and let $B$ denote the integral closure of $A$ in $L$ . There exists an ordered basis $(α_{1}, \dots, α_{n})$ of $L$ as a $K$ -vector space contained in $B$ . Moreover, for any such basis, we have

\sum_{i = 1}^{n} A α_{i} \subseteq B \subseteq \sum_{i = 1}^{n} A d^{- 1} α_{i},

where $d = D (α_{1}, \dots, α_{n})$ .

Proof.

First, take any ordered basis $(β_{1}, \dots, β_{n})$ of $L ∕ K$ . By Proposition 11.4.29, there exists $a \in A - {0}$ such that $α_{i} = a β_{i} \in B$ for each $1 \leq i \leq n$ . Clearly, $(α_{1}, \dots, α_{n})$ is a basis of $L ∕ K$ , so in particular, the $A$ -module generated by the $α_{i}$ is free and contained in $B$ . The other containment is simply a corollary of Lemma 11.4.35 and the fact that ${Tr}_{L ∕ K} (B) \subseteq A$ . □

The following notion of rank is most interesting for finitely generated modules, though we shall have occasion to use it without this assumption.

Definition 11.4.37. §

The rank of a module $M$ over a domain $A$ is

{rank}_{A} (M) = \dim_{K} (K \otimes_{A} M) .

Corollary 11.4.38. §

Let $A$ be an integrally closed noetherian domain with quotient field $K$ . Let $L$ be a finite separable extension of $K$ , and let $B$ denote the integral closure of $A$ in $L$ . Then $B$ is a finitely generated, torsion-free $A$ -module of rank $[L : K]$ .

Proof.

By Proposition 11.4.36, we have free $A$ -modules $M$ and $M^{'}$ of rank $n = [L : K]$ such that $M \subseteq A \subseteq M^{'}$ . Since $M^{'}$ has no $A$ -torsion, neither does $B$ . We have

K \otimes_{A} M \subseteq K \otimes_{A} B \subseteq K \otimes_{A} M^{'} .

As $M$ and $M^{'}$ are both isomorphic to $A^{n}$ , their tensor products over $A$ with $K$ are $n$ -dimensional $K$ -vector spaces, which forces $K \otimes_{A} B$ to have $K$ -dimension $n$ as well. Moreover, $B$ is finitely generated being a submodule of a finitely generated module over $A$ , as $A$ is noetherian. □

Proposition 11.4.39. §

Let $A$ be an integrally closed noetherian domain with quotient field $K$ . Let $L$ be a finite separable extension of $K$ , and let $B$ denote the integral closure of $A$ in $L$ . Then any finitely generated, nonzero $B$ -submodule of $L$ is a torsion-free $A$ -module of rank $[L : K]$ .

Proof.

Let $M$ be a finitely generated, nonzero $B$ -submodule of $L$ . If $β \in L^{\times}$ , then the multiplication-by- $β$ map $B \to 𝐵𝛽$ is an isomorphism of $B$ -modules, so $𝐵𝛽$ has rank $[L : K]$ as an $A$ -module. In particular, ${rank}_{A} (M) \geq {rank}_{A} (B)$ , taking $β \in M$ . Since $M$ is $B$ -finitely generated and contained in the quotient field of $B$ , there exists $α \in B$ such that $𝛼𝑀 \subseteq B$ . Since multiplication by $α$ is an isomorphism, ${rank}_{A} (M) \leq {rank}_{A} (B)$ . The result now follows from Corollary 11.4.38. □

Corollary 11.4.40. §

Let $A$ be a PID with quotient field $K$ , let $L$ be a finite separable extension of $K$ , and let $B$ denote the integral closure of $K$ in $L$ . Then any finitely generated $B$ -submodule of $L$ is a free $A$ -module of rank $[L : K]$ .

Proof.

By the structure theorem for modules over a PID, any torsion-free rank $n$ module over $A$ is isomorphic to $A^{n}$ . The result is then immediate from Proposition 11.4.39. □

We have the following application to number fields.

Lemma 11.4.41. §

Let $K$ be a number field. Then the discriminant of $𝒪_{K}$ over $ℤ$ is independent of the choice of ordered basis of $𝒪_{K}$ as a free $ℤ$ -module.

Proof.

By Corollary 11.4.40, the ring $𝒪_{K}$ is free of rank $n = [K : ℚ]$ over $ℤ$ . If $β_{1}, \dots, β_{n}$ and $α_{1}, \dots, α_{n}$ are bases of $𝒪_{K}$ as a free $ℤ$ -module, then there exists a $ℚ$ -linear homomorphism $T : K \to K$ such that $T (α_{i}) = β_{i}$ for all $i$ . Then

D (β_{1}, \dots, β_{n}) = \det {(T)}^{2} D (α_{1}, \dots, α_{n}),

and $\det (T)$ is a unit in $ℤ$ , so in ${\pm 1}$ , which is to say that $\det {(T)}^{2} = 1$ . □

Definition 11.4.42. §

If $K$ is a number field, the discriminant $disc (K)$ of $K$ is the discriminant of $𝒪_{K}$ over $ℤ$ relative to any basis of $𝒪_{K}$ as a free $ℤ$ -module.

Noting Theorem 11.4.33, the case of quadratic fields is immediately calculated.

Proposition 11.4.43. §

Let $K = ℚ (\sqrt{d})$ , where $d \neq 1$ is a square-free integer. Then

disc (K) = {\begin{matrix} d & d \equiv 1 mod 4, \\ 4 d & d \equiv 2, 3 mod 4 . \end{matrix}

The following theorem will be useful to us later.

Theorem 11.4.44 (Noether’s normalization lemma). §

Let $F$ be a field, and let $A$ be a finitely generated commutative $F$ -algebra with generators $z_{1}, \dots, z_{r} \in A$ . Then there exists $s \leq r$ and $F$ -algebraically independent elements $t_{1}, \dots, t_{s} \in A$ such that $A$ is integral over $F [t_{1}, \dots, t_{s}]$ .

Proof.

The result is obvious for $r = 0$ , so suppose $r \geq 1$ . If the elements $z_{1}, \dots, z_{r}$ are algebraically independent over $F$ , then we may take $s = r$ and $t_{i} = z_{i}$ for all $i$ , so suppose not. In this case, there exists a nonzero polynomial $f \in F [x_{1}, \dots, x_{r}]$ such that $f (z_{1}, \dots, z_{r}) = 0$ . Let $d$ be the maximum of the degrees of $f$ viewed as a polynomial in each of the $x_{i}$ . Since $f$ is nonconstant, without loss of generality we may take it to be nonconstant as a polynomial in $x_{1}$ with coefficients in $F [x_{2}, \dots, x_{r}]$ .

Consider the polynomial

g (x_{1}, \dots, x_{r}) = f (x_{1}, x_{2} + x_{1}^{d + 1}, \dots, x_{r} + x_{1}^{{(d + 1)}^{r - 1}}) .

Each monomial $x_{1}^{k_{1}} \dots x_{r}^{k_{r}}$ in $f$ has $k_{i} \leq d$ for all $i$ and gives rise to a sum of monomials in $g$ , exactly one of which has the form a constant in $F$ times $x_{1}$ to the power $\sum_{i = 1}^{r} k_{i} {(1 + d)}^{i - 1}$ . Each of these powers for the different monomials in $f$ is distinct, so the highest degree term in $g$ viewed as a polynomial in $x_{1}$ has a nonzero coefficient $c$ that lies in $F$ . That is, $c^{- 1} g$ is monic as a polynomial in $x_{1}$ with coefficients in $F [x_{2}, \dots, x_{r}]$ .

Set

w_{i} = z_{i} - z_{1}^{{(d + 1)}^{i - 1}}

for $2 \leq i \leq r$ , and note that $g (z_{1}, w_{2}, \dots, w_{r}) = f (z_{1}, \dots, z_{r}) = 0$ . It follows that $z_{1}$ is integral over $B = F [w_{2}, \dots, w_{r}]$ . By induction, there exist $s \leq r$ and elements $t_{1}, \dots, t_{s} \in B$ that are algebraically independent over $F$ and for which $B$ is integral over $F [t_{1}, \dots, t_{s}]$ . Then $A = B [z_{1}]$ is integral over $F [t_{1}, \dots, t_{s}]$ by the transitivity of integral extensions, proving the theorem. □

Corollary 11.4.45. §

Let $K$ be an extension of a field $F$ that is finitely generated as an $F$ -algebra. Then $K$ is a finite extension of $F$ .

Proof.

By Noether’s normalization lemma, $K$ is an integral extension of $F [t_{1}, \dots, t_{s}]$ for some algebraically independent elements $t_{1}, \dots, t_{s} \in K$ . However, $K$ is a field so contains the quotient field $K (t_{1}, \dots, t_{s})$ . Since no $t_{i}^{- 1}$ is integral over $F [t_{1}, \dots, t_{s}]$ , we must have $s = 0$ . Thus $K$ is integral over $F$ , which is to say it is an algebraic extension of $F$ , but then it is clearly finite being that it is generated by finitely many elements. □

11.5. Going up and going down

We use $B ∕ A$ to denote an extension of commutative rings.

Definition 11.5.1. §

Let $B ∕ A$ be an extension of commutative rings. We say that an ideal $𝔟$ of $B$ lies over an ideal $𝔞$ of $A$ if $𝔟 \cap A = 𝔞$ .

We begin by noting the following simple lemma.

Lemma 11.5.2. §

Let $A$ be an integral domain, and let $B$ be a commutative ring extension of $A$ that is integral over $A$ . If $𝔟$ is an ideal of $B$ that contains a nonzero element which is not a zero divisor, then $𝔟$ lies over a nonzero ideal of $A$ .

Proof.

That $𝔟 \cap A$ is an ideal is clear, so it suffices to show that $𝔟 \cap A$ is nonzero. Let $β \in 𝔟$ be nonzero and not a zero divisor. The element $β$ is a root of some monic polynomial $g \in A [x]$ . Write $g = x^{n} f$ for some nonzero $f \in A [x]$ with nonzero constant term. Since $β \in 𝔟$ , we have $f (β) - f (0) \in 𝔟$ , and as $f (β) = 0$ given that $β$ is not a zero divisor, we have $f (0) \in 𝔟$ . But $f (0) \neq 0$ , so $𝔟$ has a nonzero element. □

The following are also easily verified.

Lemma 11.5.3. §

If $B ∕ A$ is integral and $𝔟$ is an ideal of $B$ that lies over $𝔞$ , then $B ∕ 𝔟$ is integral over $A ∕ 𝔞$ .

Lemma 11.5.4. §

Let $S$ be a multiplicatively closed subset of $A$ . If $B ∕ A$ is integral, then so is $S^{- 1} B ∕ S^{- 1} A$ .

Proposition 11.5.5. §

Let $B ∕ A$ be an integral extension. If $𝔭$ is a prime ideal of $A$ , then there exists a prime ideal $𝔮$ of $B$ lying over $𝔭$ .

Proof.

The ring $B_{𝔭} = S_{𝔭}^{- 1} B$ is integral over $A_{𝔭}$ by Lemma 11.5.4. Let $𝔐$ be a maximal ideal of $B_{𝔭}$ . Set $𝔪 = 𝔐 \cap A_{𝔭}$ . Since the field $B_{𝔭} ∕ 𝔐$ is an integral extension of $A_{𝔭} ∕ 𝔪$ , Lemma 11.4.8 tells us that $A_{𝔭} ∕ 𝔪$ is a field as well, so $𝔪$ is a maximal ideal of $A_{𝔭}$ . Since $A_{𝔭}$ is local, we have $𝔪 = 𝔭 A_{𝔭}$ . Let $ι_{B} : B \to B_{𝔭}$ and $ι_{A} : A \to A_{𝔭}$ be the localization maps so that $𝔮 = ι_{B}^{- 1} (𝔐)$ is prime, and

𝔮 \cap A = ι_{B}^{- 1} (𝔐) \cap A = ι_{A}^{- 1} (𝔪) = ι_{A}^{- 1} (𝔭 A_{𝔭}) = 𝔭 .

□

Theorem 11.5.6 (Going up). §

Let $B ∕ A$ be an integral extension. Suppose that $𝔭_{1} \subseteq 𝔭_{2}$ are prime ideals of $A$ and $𝔮_{1}$ is a prime ideal of $B$ lying over $𝔭_{1}$ . Then there exists a prime ideal $𝔮_{2}$ of $B$ containing $𝔮_{1}$ and lying over $𝔭_{2}$ .

Proof.

Let $\bar{A} = A ∕ 𝔭_{1}$ and $\bar{B} = B ∕ 𝔭_{2}$ , and let $π : B \to \bar{B}$ be the quotient map. Let ${\bar{𝔭}}_{2}$ be the image of $𝔭_{2}$ in $\bar{A}$ . By Proposition 11.5.5, there exists a prime ideal ${\bar{𝔮}}_{2}$ of $\bar{B}$ lying over ${\bar{𝔭}}_{2}$ . Then $𝔮_{2} = π^{- 1} ({\bar{𝔮}}_{2})$ contains $𝔮_{1}$ and satisfies

𝔮_{2} \cap A = π^{- 1} ({\bar{𝔮}}_{2} \cap \bar{A}) = π^{- 1} ({\bar{𝔭}}_{2}) = 𝔭_{2},

since $𝔭_{2}$ contains $𝔭_{1}$ . □

Proposition 11.5.7. §

Let $B ∕ A$ be an integral extension of domains. Let $𝔞$ be an ideal of $A$ . An element $β \in B$ is a root of $f \in A [x]$ with $f = x^{n} + \sum_{i = 0}^{n - 1} a_{i} x^{i}$ with $a_{i} \in 𝔞$ for all $0 \leq i \leq n - 1$ if and only if $β \in \sqrt{𝔞𝐵}$ .

Proof.

First, suppose that $β \in B$ is a root of $f$ as in the statement of the proposition. Then $β^{n} = - \sum_{i = 0}^{n - 1} a_{i} β^{i} \in 𝔞𝐵$ , so $β \in \sqrt{𝔞𝐵}$ .

Conversely, suppose that $β^{n} \in 𝔞𝐵$ for some $n \geq 1$ . Let $a_{1}, \dots, a_{k} \in 𝔞$ and $b_{1}, \dots, b_{k} \in B$ for $k \geq 1$ be such that $β^{n} = \sum_{i = 1}^{k} a_{i} b_{i}$ . Set $M = A [b_{1}, \dots, b_{k}]$ , which is a finitely generated faithful $A$ -module as a finitely generated integral $A$ -algebra with the property that $β^{n} M \subseteq 𝔞𝑀$ . Given a generating set of $M$ with $n$ elements, as in the proof of Lemma 11.4.4, we may form a matrix in $M_{n} (A)$ given by the action of $β^{n}$ on the generating set by multiplication, and it has entries in $𝔞$ . Then its characteristic polynomial $c (x) = \sum_{i = 1}^{n} c_{i} x^{i} \in A [x]$ has $c_{i} \in 𝔞$ for $i < n$ . Since $c (β^{n})$ annihilates $M$ , we have $c (β^{n}) = 0$ as $M$ is faithful. We then have $β^{n} = - \sum_{i = 1}^{n} c_{i} β^{i} \in 𝔞𝐵$ . □

Proposition 11.5.8. §

Let $B ∕ A$ be an integral extension of domains such that $A$ is integrally closed, and suppose that $β \in B$ is the root of a monic polynomial in $A [x]$ with non-leading coefficients in a radical ideal $𝔞$ of $A$ . Then the minimal polynomial of $β$ is also such a polynomial.

Proof.

Let $K = Q (A)$ and $f \in K [x]$ be the minimal polynomial of $β$ . Note that $f \in A [x]$ by Proposition 11.4.21. Let $L$ be an extension of $K$ that contains all of the roots of $f$ , and let $C$ be the integral closure of $A$ in $L$ .

By Proposition 11.5.7, we have $β \in \sqrt{𝔞𝐶}$ , so there exists $g \in A [x]$ monic with $g (β) = 0$ and which has image $x^{n}$ in $A ∕ 𝔞 [x]$ for $n = \deg g$ . Then $f$ divides $g$ in $K [x]$ , so in particular $g (γ) = 0$ for every root $γ$ of $f$ , which again tells us that $γ \in \sqrt{𝔞𝐶}$ .

Since the non-leading coefficients of $f$ are symmetric polynomials in the roots of $f$ , we now have that they lie in $\sqrt{𝔞𝐶}$ . Once again, such a coefficient $a$ is a root of a monic $g \in A [x]$ which reduces to $x^{\deg g}$ modulo $𝔞$ , but then $a^{\deg g} \in 𝔞$ as $a \in A$ . As we have assumed that $𝔞$ is radical, we then have $a \in 𝔞$ , so $f$ has the desired form. □

Lemma 11.5.9. §

Let $𝔭$ be a prime ideal of $A$ . There exists a prime ideal $𝔮$ of $B$ lying over $𝔭$ if and only if $𝔭𝐵 \cap A = 𝔭$ .

Proof.

If $𝔮 \cap A = 𝔭$ , then $𝔮$ contains $𝔭𝐵$ , and then

𝔭 = 𝔮 \cap A \supseteq 𝔭𝐵 \cap A \supseteq 𝔭,

so $𝔭𝐵 \cap A = 𝔭$ .

Conversely, if $𝔭𝐵 \cap A = 𝔭$ , then $𝔭𝐵$ is disjoint from $S_{𝔭}$ , so there exists a maximal ideal $𝔐$ of $B_{𝔭}$ with $𝔭𝐵$ contained in $𝔐$ . Let $𝔮$ be the inverse image of $𝔐$ in $B$ . Then $𝔪 = 𝔮 \cap A$ is a prime ideal containing $𝔭$ with $𝔪 \cap S_{𝔭} = \emptyset$ , which forces $𝔪 = 𝔭$ . □

Theorem 11.5.10 (Going down). §

Let $B ∕ A$ be an integral extension of integral domains with $A$ integrally closed. Suppose that $𝔭_{2} \subseteq 𝔭_{1}$ are prime ideals of $A$ and $𝔮_{1}$ is a prime ideal of $B$ lying over $𝔭_{1}$ . Then there exists a prime ideal $𝔮_{2}$ of $B$ contained in $𝔮_{1}$ and lying over $𝔭_{2}$ .

Proof.

Note that the maps $B \to B_{𝔮_{1}}$ and $A \to A_{𝔭_{1}}$ are injective as $B$ is a domain. By Lemma 11.5.9, it is enough to show that $𝔭_{2} B_{𝔮_{1}} \cap A = 𝔭_{2}$ . That is, in this case there exists a prime ideal $𝔔_{2}$ of $B_{𝔮_{1}}$ lying over $𝔭_{2} A_{𝔭_{1}}$ , and then we can take $𝔮_{2} = 𝔔_{2} \cap B$ .

If $β = \frac{b}{s} \in 𝔭_{2} B_{𝔮_{1}}$ with $b \in 𝔭_{2} B$ and $s \in B - 𝔮_{1}$ , then by Proposition 11.5.8, the minimal polynomial $f = x^{n} + \sum_{i = 0}^{n - 1} a_{i} x^{i}$ of $b$ has non-leading coefficients in $𝔭_{2}$ . If $β$ is also in $A$ , then $s = β^{- 1} b$ has minimal polynomial $β^{- n} f (𝛽𝑥) \in Q (A) [x]$ . Since $s$ is integral over $A$ , this polynomial lies in $A [x]$ , and therefore $β^{i - n} a_{i} \in A$ for all $i$ . If $β \notin 𝔭_{2}$ , then for all $i$ we have $β^{i - n} a_{i} \in 𝔭_{2}$ since $a_{i} \in 𝔭_{2}$ . But then $s^{n} \in 𝔭_{2} B$ , so $s \in 𝔮_{1}$ , a contradiction. Thus, $β \in 𝔭_{2}$ , as required. □

11.6. Hilbert’s Nullstellensatz

We use $K$ to denote a fixed algebraically closed field in this section. Much but certainly not all of what is done here can be generalized to fields which are not algebraically closed as well, but for this brief introduction, we feel it suffices to focus on the more specific setting. This section assumes some basic knowledge of topological spaces.

Fix a nonnegative integer $n$ .

Definition 11.6.1. §

Let $S$ be a subset of $K [x_{1}, \dots, x_{n}]$ . The zero set, or vanishing locus, of $S$ is

V (S) = {(a_{1}, \dots, a_{n}) \in K^{n} ∣ f (a_{1}, \dots, a_{n}) = 0 for all f \in S} .

An algebraic set in $K^{n}$ is any subset of $K^{n}$ that is a zero set of some set of polynomials in $K [x_{1}, \dots, x_{n}]$ .

From now on, let us set $R = K [x_{1}, \dots, x_{n}]$ for brevity.

Notation 11.6.2. §

If $S = {f_{1}, \dots, f_{n}} \subset R$ , we also write $V (f_{1}, \dots, f_{n})$ for $V (S)$ . At times, for $a = (a_{1}, \dots, a_{n}) \in K^{n}$ , we write $f (a)$ for $f (a_{1}, \dots, a_{n})$ .

Example 11.6.3. §

We have $V (\emptyset) = K^{n}$ and $V (R) = V (1) = \emptyset$ .

Example 11.6.4. §

Consider $f (x, y) = x - y$ and $g (x, y) = x^{2} + y^{2} - 2$ in $ℂ [x, y]$ . Then $V (f, g) = V (f) \cap V (g) = {(- 1, - 1), (1, 1)}$ .

Remark 11.6.5. §

For any subset $S$ of $R$ , the zero set $V (S)$ equals the zero set of the ideal $(S)$ generated by $S$ .

Proposition 11.6.6. §

a.: The intersection of any collection of algebraic sets in $K^{n}$ is also an algebraic set.
b.: The union of any finite collection of algebraic sets in $K^{n}$ is also an algebraic set.

Proof.

Let ${S_{i} ∣ i \in I}$ be a collection of subsets of $R$ . Then $⋂_{i \in I} V (S_{i}) = V (S)$ is algebraic, so we have part a. If $S$ and $T$ are subsets of $R$ , set $I = (S)$ and $J = (T)$ . We clearly have

V (S) \cup V (T) = V (I) \cup V (J) \subseteq V (I \cap J) .

If $a \in V (I \cap J)$ and $a \notin V (I)$ , then there exists $f \in I$ with $f (a) \neq 0$ . If $g \in J$ , then $𝑓𝑔 \in I \cap J$ , so $f (a) g (a) = 0$ , so $g (a) = 0$ . Thus $a \in V (J)$ , and we have part b. □

It follows from the proposition that the following definition does in fact yield a topology.

Definition 11.6.7. §

The Zariski topology on $K^{n}$ is the topology ${K^{n} - V (S) ∣ S \subseteq R}$ with closed sets the algebraic sets in $K^{n}$ .

Definition 11.6.8. §

For $n \geq 0$ , the affine $n$ -space over $K$ is the set $𝔸_{K}^{n} = K^{n}$ endowed with the Zariski topology.

Remark 11.6.9. §

For any $(a_{1}, \dots, a_{n}) \in 𝔸_{K}^{n}$ , we have $V (x_{1} - a_{1}, \dots, x_{n} - a_{n}) = {(a_{1}, \dots, a_{n})}$ , so points in $𝔸_{K}^{n}$ are closed. However, it is not a Hausdoff topology: for instance, for $n = 1$ , the only closed sets other than $X$ are finite, so any two nonempty open sets will intersect as $K$ is infinite.

Notation 11.6.10. §

Let $Z \subseteq 𝔸_{K}^{n}$ . Then

I (Z) = {f \in K [x_{1}, \dots, x_{n}] ∣ f (a) = 0 for all a \in Z} .

The set $I (Z)$ is clearly an ideal: it is the ideal of $R$ of elements that vanish on all of $Z$ .

Remark 11.6.11. §

Note that if $f \in R$ satisfies $f^{k} \in I (Z)$ for some subset $Z$ of $𝔸_{K}^{n}$ and $k \geq 1$ , then $f {(a)}^{k} = 0$ for all $a \in Z$ , so $f$ vanishes on $Z$ , which is to say that $f \in I (Z)$ . Hence, $I (Z)$ is a radical ideal.

Example 11.6.12. §

For $a = (a_{1}, \dots, a_{n}) \in 𝔸_{K}^{n}$ , we have $I ({a}) = (x_{1} - a_{1}, \dots, x_{n} - a_{n})$ .

In particular $I$ and $V$ provide bijections between the points of $𝔸_{K}^{n}$ and a subset of the maximal ideals of $R$ , i.e., those of the form $(x_{1} - a_{1}, \dots, x_{n} - a_{n})$ for some $a \in 𝔸_{K}^{n}$ . The statement the latter maximal ideals are all of the maximal ideals of $R$ is known as the weak form of Hilbert’s Nullstellensatz.

Theorem 11.6.13. §

Every maximal ideal of $K [x_{1}, \dots, x_{n}]$ has the form $(x_{1} - a_{1}, \dots, x_{n} - a_{n})$ for some $(a_{1}, \dots, a_{n}) \in 𝔸_{K}^{n}$ .

Proof.

Let $𝔪$ be a maximal ideal of $R = K [x_{1}, \dots, x_{n}]$ , and consider $L = R ∕ 𝔪$ , which is a field containing $K$ that is finitely generated over $K$ . By Corollary 11.4.45, the field $L$ is an algebraic extension of the algebraically closed field $K$ , so it is equal to $K$ . Under the quotient map $R \to L = K$ , each $x_{i}$ is sent to some $a_{i} \in L$ , so $x_{i} - a_{i} \in 𝔪$ . Since $(x_{1} - a_{1}, \dots, x_{n} - a_{n})$ is maximal, it equals $𝔪$ . □

In other words, $V$ and $I$ give inverse bijections between the maximal ideals of $R$ and the singleton subsets of $𝔸_{K}^{n}$ . We now prove the stronger form of this statement, one which boils down to the statement that $I (V (𝔞)) = \sqrt{𝔞}$ for ideals $𝔞$ of $R$ .

Theorem 11.6.14 (Hilbert’s Nullstellensatz). §

The maps $I$ and $V$ provide mutually inverse, inclusion-reversing bijections

Hilbert's Nullstellensatz correspondence. A full diagram description follows.

Diagram description: Hilbert's Nullstellensatz correspondence

The two displayed maps are mutually inverse, inclusion-reversing bijections between the specified collections.

Objects, listed by row and column:

Row 1, from left to right: column 1: left brace radical ideals of K[x subscript (1), ellipsis , x subscript (n)] right brace; column 2: left brace algebraic sets in blackboard A subscript (K) superscript (n) right brace.

Arrows and lines:

An arrow from left brace radical ideals of K[x subscript (1), ellipsis , x subscript (n)] right brace to left brace algebraic sets in blackboard A subscript (K) superscript (n) right brace, labelled V.
An arrow from left brace algebraic sets in blackboard A subscript (K) superscript (n) right brace to left brace radical ideals of K[x subscript (1), ellipsis , x subscript (n)] right brace, labelled I.

Proof.

The operation $I$ is by definition inclusion-reversing on subsets of $𝔸_{K}^{n}$ , and the operation $V$ is inclusion-reversing on subsets of $R = K [x_{1}, \dots, x_{n}]$ . It is immediate from the definitions and Remark 11.6.11 that if $𝔞$ is an ideal of $R$ , then $I (V (𝔞))$ contains $\sqrt{𝔞}$ , and if $Z$ is a subset of $R$ , then $V (I (Z))$ contains $Z$ . If $Z = V (𝔞)$ for some ideal $𝔞$ , then

V (I (Z)) = V (I (V (𝔞))) \subseteq V (𝔞) = Z,

since $V$ is inclusion-reversing. Thus, on algebraic sets $Z$ , we have $V (I (Z)) = Z$ .

It remains to show that $I (V (𝔞)) \subseteq \sqrt{𝔞}$ for any ideal $𝔞$ of $R$ . Let $f \in I (V (𝔞))$ . For an indeterminate $y$ , let $J$ be the ideal of $R [y]$ generated by $𝔞$ and $1 - 𝑓𝑦$ . We view $R [y]$ as $K [x_{1}, \dots, x_{n}, y]$ and consider the vanishing set of $J$ in $𝔸_{K}^{n + 1}$ . If $a = (a_{1}, \dots, a_{n + 1}) \in V (J)$ , then $(a_{1}, \dots, a_{n}) \in V (𝔞)$ , in which case we have

(1 - 𝑓𝑦) (a_{1}, \dots, a_{n + 1}) = 1 - f (a_{1}, \dots, a_{n}) a_{n + 1} = 1 \neq 0 .

Thus $V (J) = \emptyset$ . By the weak form of the Nullstellensatz, if $J$ were a proper ideal, then its vanishing locus would contain the point in the vanishing locus of a maximal ideal containing it, so $J = R [y]$ .

In the quotient $R [y] ∕ (1 - 𝑓𝑦)$ , which we identify with $R [f^{- 1}]$ via $y \mapsto f^{- 1}$ , it follows that $1 \in 𝔞𝑅 [f^{- 1}]$ . In other words, $f^{N} \in 𝔞$ for some $N \geq 0$ , so $f \in \sqrt{𝔞}$ , as was desired. □

Remark 11.6.15. §

We record the following simple consequences of the Nullstellensatz.

a.: For any $S \subseteq R$ , we have $V (S) = V (\sqrt{(S)})$ , and $I (V (S)) = \sqrt{(S)}$ .
b.: For any $Z \subseteq 𝔸_{K}^{n}$ , we have $I (Z) = I (\bar{Z})$ , where $\bar{Z}$ is the closure of $Z$ in the Zariski topology (i.e., the smallest algebraic set containing $Z$ ), and $V (I (Z)) = \bar{Z}$ .

Definition 11.6.16. §

We say that an algebraic set is irreducible if it is not a union of two proper algebraic subsets.

We have seen that maximal ideals correspond to singleton sets under $V$ and $I$ . Hilbert’s Nullstellensatz tells us that prime ideals correspond to irreducible algebraic sets.

Corollary 11.6.17. §

The maps $I$ and $V$ restrict to mutually inverse, inclusion-reversing bijections

Prime ideals and irreducible algebraic sets. A full diagram description follows.

Diagram description: Prime ideals and irreducible algebraic sets

The two displayed maps are mutually inverse, inclusion-reversing bijections between the specified collections.

Objects, listed by row and column:

Row 1, from left to right: column 1: left brace prime ideals of K[x subscript (1), ellipsis , x subscript (n)] right brace; column 2: left brace irreducible algebraic sets in blackboard A subscript (K) superscript (n) right brace.

Arrows and lines:

An arrow from left brace prime ideals of K[x subscript (1), ellipsis , x subscript (n)] right brace to left brace irreducible algebraic sets in blackboard A subscript (K) superscript (n) right brace, labelled V.
An arrow from left brace irreducible algebraic sets in blackboard A subscript (K) superscript (n) right brace to left brace prime ideals of K[x subscript (1), ellipsis , x subscript (n)] right brace, labelled I.

Proof.

An algebraic set $Z$ is by definition the vanishing locus of some radical ideal $I$ of $R$ . By the Nullstellensatz, such a set $Z$ is irreducible if and only if $I = I (Z)$ cannot be written as an intersection of two radical ideals properly containing $I$ . Note that if $I$ were the intersection of two arbitrary ideals, then it would also be the intersection of their radicals. Conversely, if $I$ is an irreducible ideal, then so is its radical, and then its vanishing locus is irreducible as well. Since the irreducible ideals in $R$ are exactly the primary ideals, and those which are radical are the prime ideals, irreducible algebraic sets correspond exactly to the prime ideals of $R$ . □

Remark 11.6.18. §

By the primary decomposition theorem and Corollary 11.6.17, every algebraic set is a finite union of irreducible algebraic sets.

Remark 11.6.19. §

An irreducible algebraic set $Z \subseteq 𝔸_{K}^{n}$ together with its subspace topology is also what is called an (affine) algebraic variety. It has an associated coordinate ring $K [Z] = R ∕ I (Z)$ . Note that the ring $R ∕ I (Z)$ is a domain, since $I (Z)$ is prime. The radical ideals of $K [Z]$ correspond to algebraic subsets of $Z$ , and via this bijection the maximal ideals of $K [Z]$ correspond to the points (or more precisely, singleton subsets) of $Z$ .

11.7. Spectra of rings

In the previous section, we saw that the points of $𝔸_{K}^{n}$ for an algebraically closed field $K$ correspond to the maximal ideals of $K [x_{1}, \dots, x_{n}]$ . Making this identification, we may think of the Zariski topology as endowing the set of maximal ideals of $K [x_{1}, \dots, x_{n}]$ with a topology. We now aim to mimic this for the larger set of prime ideals, in an arbitrary commutative ring $R$ .

Definition 11.7.1. §

The spectrum $Spec R$ of a commutative ring $R$ is the set of prime ideals of $R$ .

Example 11.7.2. §

For a PID $R$ , we have $Spec R = {(0)} \cup {(f) ∣ f irreducible}$ .

Notation 11.7.3. §

For any subset $T$ of $R$ , we set

V (T) = {𝔭 \in Spec R ∣ T \subseteq 𝔭} .

For any subset $Y$ of $Spec R$ , we set

I (Y) = ⋂_{𝔭 \in Y} 𝔭 .

Remark 11.7.4. §

We have $V (T) = V ((T))$ for the ideal $(T)$ generated by $T$ . In fact, for any ideal $I$ of $R$ , we have $V (I) = V (\sqrt{I})$ since if $I \subseteq 𝔭$ , then $\sqrt{I} \subseteq \sqrt{𝔭} = 𝔭$ .

The following lemma is easily verified.

Lemma 11.7.5. §

a.: We have $V ((0)) = Spec R$ and $V (R) = \emptyset$ .
b.: If $𝔞$ and $𝔟$ are ideals of $R$ , then $V (𝔞) \cup V (𝔟) = V (𝔞 \cap 𝔟) = V (𝔞𝔟)$ .
c.: If ${𝔞_{j} ∣ j \in X}$ is a collection of ideals of $R$ , then
$⋂_{j \in X} V (𝔞_{j}) = V (⋃_{j \in X} 𝔞_{j}) = V (\sum_{j \in X} 𝔞_{j}) .$

In particular, the sets $V (I)$ for $I$ an ideal of $R$ form a topology on $Spec R$ .

Definition 11.7.6. §

The Zariski topology on $Spec R$ is the unique topology with closed sets the $V (I)$ with $I$ an ideal of $R$ .

Remark 11.7.7. §

In $Spec R$ , the singleton sets ${𝔪}$ with $𝔪$ maximal are closed, since $V (𝔪) = {𝔪}$ . However, points in general need not be closed. The closure of ${𝔭}$ with $𝔭$ prime is the smallest closed subset containing $𝔭$ , which is exactly $V (𝔭)$ , the set of prime ideals containing $𝔭$ . So, ${𝔭}$ is closed if and only if $𝔭$ is maximal. E.g., in an integral domain, the closure of $(0)$ is $Spec R$ !

Definition 11.7.8. §

The closed points of $Spec R$ are the maximal ideals of $R$ .

Definition 11.7.9. §

The closure of a subset $Y$ of $Spec R$ in the Zariski topology on $R$ is known as the Zariski closure of $Y$ .

The analogue of the Nullstellensatz for $Spec R$ is considerably less difficult.

Proposition 11.7.10. §

The maps $I$ and $V$ provide mutually inverse, inclusion-reversing bijections

Radical ideals and closed subsets of a spectrum. A full diagram description follows.

Diagram description: Radical ideals and closed subsets of a spectrum

The two displayed maps are mutually inverse, inclusion-reversing bijections between the specified collections.

Objects, listed by row and column:

Row 1, from left to right: column 1: left brace radical ideals of R right brace; column 2: left brace closed subsets of Spec R right brace.

Arrows and lines:

An arrow from left brace radical ideals of R right brace to left brace closed subsets of Spec R right brace, labelled V.
An arrow from left brace closed subsets of Spec R right brace to left brace radical ideals of R right brace, labelled I.

In fact, for any ideal $𝔞$ of $R$ , we have $I (V (𝔞)) = \sqrt{𝔞}$ , and for any subset $Y$ of $Spec R$ , the set $V (I (Y))$ is the Zariski closure of $Y$ .

Proof.

That $V$ and $I$ are inclusion-reversing is clear. Let $𝔞$ be an ideal of $R$ . Then

I (V (𝔞)) = ⋂_{𝔞 \subseteq 𝔭} 𝔭 = \sqrt{𝔞}

by Proposition 11.2.10. Conversely, if $Y$ is a subset of $Spec R$ , then its closure $\bar{Y}$ is $V (𝔞)$ for some ideal $𝔞$ of $R$ , and

V (I (Y)) \subseteq V (I (\bar{Y})) = V (I (V (𝔞))) = V (\sqrt{𝔞}) = V (𝔞) = \bar{Y},

but $V (I (Y))$ is closed an contains $Y$ , so $V (I (Y)) = \bar{Y}$ . □

Corollary 11.7.11. §

The Zariski closure of a subset $Y$ of $Spec R$ is the set of all prime ideals containing some element of $Y$ .

Proof.

The set $V (I (Y))$ consists of the prime ideals $𝔭$ containing the intersection of all prime ideals containing $Y$ . By Lemma 3.10.19b, these are exactly the ideals that contain some element of $Y$ . □

Let us compare $V$ and $I$ of Definition 11.7.3 with our prior maps with these notations for the polynomial ring $R = K [x_{1}, \dots, x_{n}]$ over an algebraically closed field $K$ .

Proposition 11.7.12. §

Let $R = K [x_{1}, \dots, x_{n}]$ for some $n \geq 0$ and algebraically closed field $K$ . Let us use $V^{'}$ and $I^{'}$ to denote the maps which take vanishing loci of algebraic sets in $𝔸_{K}^{n}$ and the ideal of vanishing of subsets of $R$ , respectively.

a.: The injective map $ι : 𝔸_{K}^{n} \to Spec R$ given by taking a point to its corresponding maximal ideal is a homeomorphism onto its image, which we use to identify $𝔸_{K}^{n}$ with a subspace of $Spec R$ .
b.: For any ideal $𝔞$ of $R$ , we have $V^{'} (𝔞) = V (𝔞) \cap 𝔸_{K}^{n}$ .
c.: For any Zariski closed subset $Y$ of $Spec R$ , we have $I (Y) = I^{'} (Y \cap 𝔸_{K}^{n})$ .

Proof.

If $Y = V (𝔞)$ for some ideal $𝔞$ of $Spec R$ , then $Y \cap 𝔸_{K}^{n}$ is the set of maximal ideals of $R$ containing $𝔞$ , which equals $V^{'} (𝔞)$ , proving part a. This implies that the intersection of $Y$ with $𝔸_{K}^{n}$ is closed and that the image of a closed set $Z = V^{'} (𝔞)$ in the Zariski topology on $𝔸_{K}^{n}$ is closed under the subspace topology on $𝔄_{K}^{n}$ from the Zariski topology on $Spec R$ . Thus, $ι$ is a homeomorphism onto its image, proving part b.

Finally, if $Y = V (𝔞)$ is a closed subset of $Spec R$ with $𝔞$ radical, and if $Z = Y \cap 𝔸_{K}^{n} = V^{'} (𝔞)$ , then $𝔞 = I^{'} (Z)$ is the intersection of all prime ideals containing $𝔞$ . That is, $𝔞$ is the intersection of all prime ideals in $V (𝔞) = Y$ , so $𝔞 = I (Y)$ as well, and we have part c. □

Definition 11.7.13. §

If $a \in R$ , then $U_{a} = Spec R - V ((a))$ is called a principal open set of $R$ .

Proposition 11.7.14. §

The sets $U_{a}$ for $a \in R$ form a basis for the Zariski topology on $Spec R$ .

Proof.

Let $U$ be an open set in $Spec R$ . Then $U = Spec R - V (I)$ for some ideal $I$ , and $V (I) = ⋂_{a \in I} V ((a))$ , so $U = ⋃_{a \in I} U_{a}$ . □

We have the following simple lemma.

Lemma 11.7.15. §

Let $φ : R \to S$ be a ring homomorphism.

a.: If $𝔮$ is a prime ideal of $S$ , then $φ^{- 1} (𝔮)$ is a prime ideal of $R$ .
b.: Suppose that $φ$ is surjective and $𝔭$ is a prime ideal of $R$ containing the kernel of $φ$ . Then $φ (𝔭)$ is a prime ideal of $S$ .

Proof.

Set $𝔭 = φ^{- 1} (𝔮)$ for $𝔮$ a prime ideal of $S$ . Then $a, b \in R$ satisfy $𝑎𝑏 \in 𝔭$ if and only if $φ (𝑎𝑏) \in 𝔮$ , so if and only if either $φ (a) \in 𝔮$ or $φ (b) \in 𝔮$ , i.e., $a \in 𝔭$ or $b \in 𝔭$ .

If $𝔭$ is a prime ideal of $R$ containing $\ker φ$ , then $𝔮 = φ (𝔭)$ is an ideal of $S$ by the surjectivity of $φ$ , For $a, b \in S$ with $𝑎𝑏 \in 𝔮$ , write $a = φ (c)$ and $b = φ (d)$ for some $c, d \in R$ . Then $𝑐𝑑 \in φ^{- 1} (𝔮) = 𝔭$ since $\ker φ \subseteq 𝔭$ , so $c \in 𝔭$ or $d \in 𝔭$ , and therefore $a \in 𝔮$ or $b \in 𝔮$ . □

By Lemma 11.7.15, the following definition makes sense.

Definition 11.7.16. §

Let $φ : R \to S$ be a ring homomorphism. The pullback map

φ^{*} : Spec S \to Spec R

is the function given by $φ^{*} (𝔮) = φ^{- 1} (𝔮)$ for $𝔮 \in Spec S$ .

The following lemma is simple.

Lemma 11.7.17. §

If $φ$ is a ring homomorphism, then the pullback map $φ^{*}$ is continuous with respect to the Zariski topologies.

Remark 11.7.18. §

The map that takes a ring to its spectrum and ring homomorphism to the corresponding pullback map is a contravariant functor from $𝐑𝐢𝐧𝐠$ to $𝐓𝐨𝐩$ .

Example 11.7.19. §

Let $π : R \to F$ be a surjective ring homomorphism, where $F$ is a field. Then $π^{*} ((0)) = \ker π$ is the maximal ideal that is the kernel of $f$ .

Example 11.7.20. §

Let $𝔭$ be a prime ideal. The localization map $φ : R \to R_{𝔭}$ has pullback $φ^{*} (𝔮 R_{𝔭}) = 𝔮$ for prime ideals $𝔮$ of $R$ contained in $𝔭$ .

Example 11.7.21. §

Consider the map $φ : ℂ [x] \to ℂ [x]$ given by $φ (f) (x) = f (x^{2})$ . Then $φ^{*} ((0)) = (0)$ , and for $a \in ℂ$ irreducible,

φ^{*} ((x - a)) = {g \in ℂ [x] ∣ g (a^{2}) = 0} = (x - a^{2}) .

In particular, $φ^{*}$ is $2$ -to- $1$ on closed points, taking both $(x - a)$ and $(x + a)$ to $(x - a^{2})$ , except for $a = 0$ , in which case only $(x)$ is carried to $(x)$ .

11.8. Krull dimension

We continue to use $R$ to denote a commutative ring.

Definition 11.8.1. §

The length of an ascending chain ${(𝔭_{i})}_{i = 0}^{n}$ of distinct prime ideals is $n$ . We often refer to such a finite strictly ascending chain more simply as a chain of prime ideals, where minimal confusion can arise.

Example 11.8.2. §

If $R$ is an integral domain and $n \geq 0$ , then the ring $R [x_{1}, \dots, x_{n}]$ contains a chain of primes of length $n$ :

(0) \subset (x_{1}) \subset (x_{1}, x_{2}) \subset \dots \subset (x_{1}, \dots, x_{n}) .

Definition 11.8.3. §

The Krull dimension, or dimension, $\dim R$ of a commutative ring $R$ is the length of the longest ascending chain of distinct prime ideals in $R$ , if it exists, and is otherwise said to be infinite.

Remark 11.8.4. §

In set-theoretic terms, if finite, $\dim R$ is one less than the maximum of the cardinalities of all chains in $Spec R$ .

Examples 11.8.5. §

Let $F$ be a field.

a.

The Krull dimension of $F$ is $0$ : its only prime ideal is $(0)$ . In fact, the Krull dimension of $F [x] ∕ (x^{n})$ for $n \geq 0$ is $1$ , since its unique prime ideal is $(x)$ .

b.

The Krull dimension of $ℤ$ is $1$ : the longest chains are all of the form $(0) \subset (p)$ for some prime number $p$ . In fact, $\dim R = 1$ for every PID $R$ that is not a field.

c.

The Krull dimension of $F [{(x_{i})}_{i \geq 1}]$ is infinite, since

(0) \subset (x_{1}) \subset (x_{1}, x_{2}) \subset \dots

is an ascending chain of prime ideals that is not eventually constant.

Lemma 11.8.6. §

Let $π : R \to S$ be a surjective map of rings. Then $\dim R \geq \dim S$ .

Proof.

Let ${(𝔮_{i})}_{i = 0}^{n}$ be a chain of primes in $S$ of length $n$ , and set $𝔭_{i} = φ^{- 1} (𝔮_{i})$ for each $i$ . Then $𝔮_{i} = φ (𝔭_{i})$ for each $i$ , so each $𝔭_{i}$ is distinct. □

Lemma 11.8.7. §

Let $𝔭$ be a non-minimal prime of $R$ . Then $\dim R \geq \dim R ∕ 𝔭 + 1$ .

Proof.

Any chain in $R ∕ 𝔭$ of maximal length has inverse image in $R$ of the same length, and such a chain can be extended by adding in a minimal prime properly contained in $𝔭$ . □

Proposition 11.8.8. §

If $B ∕ A$ is an integral extension of domains, then $A$ has finite Krull dimension if and only if $B$ does, in which case $\dim B = \dim A$ .

Proof.

The the going up theorem tells us that $\dim B \geq \dim A$ . Suppose that $\dim B = n$ , let ${(𝔮_{i})}_{i = 0}^{n}$ be a maximal ascending chain of prime ideals of $B$ , and set $𝔭_{i} = 𝔮_{i} \cap A$ for all $i$ . We have $𝔭_{1} \cap A \neq (0)$ by Lemma 11.5.2, so $\dim B ∕ 𝔮_{1} = \dim B - 1$ , and $\dim A ∕ 𝔭_{1} \leq \dim A - 1$ by Lemma 11.8.7. By induction on $\dim B$ , we then have the remaining inequality $\dim B \leq \dim A$ . □

Proposition 11.8.9. §

Let $F$ be a field and $n$ be a nonnegative integer. The ring $F [x_{1}, \dots, x_{n}]$ has Krull dimension $n$ .

Proof.

Example 11.8.2 tells us that $R = F [x_{1}, \dots, x_{n}]$ has dimension at least $n$ . We may suppose that $n \geq 1$ . Let ${(𝔭_{i})}_{i = 0}^{m}$ be an ascending chain of prime ideals in $R$ . We may suppose that $𝔭_{0} = (0)$ and that $𝔭_{1}$ is minimal, generated by an irreducible element $g \in R$ , as otherwise we may extend the chain to contain such primes.

Consider the quotient $R = S ∕ 𝔭_{1} = S ∕ (g)$ . Since the images of the $x_{i}$ in $S$ satisfy an equation of algebraic dependence over $F$ , and these images generate $S$ as an $F$ -algebra, the quotient field of $S$ has transcendence degree at most $n - 1$ over $F$ . Thus, no set of more than $n - 1$ elements of $S$ can be algebraically independent.

By Noether’s normalization lemma, there exist algebraically independent elements $t_{1}, \dots, t_{s} \in R$ such that $S$ is integral over $F [t_{1}, \dots, t_{s}]$ . From what we have already shown, we must have $s \leq n - 1$ . Then $\dim S = s \leq n - 1$ by Proposition 11.8.8 and induction. On the other hand, the images ${\bar{𝔭}}_{i}$ in $S$ of the ideals $𝔭_{i}$ with $1 \leq i \leq m$ remain prime in $S$ by Lemma 11.7.15, and they are distinct, so $m - 1 \leq s \leq n - 1$ . Therefore, $m = n$ . □

In fact, the following result, for which we omit the proof, holds more generally.

Theorem 11.8.10. §

Let $R$ be a noetherian domain of finite Krull dimension. Then

\dim R [x] = \dim R + 1 .

Definition 11.8.11. §

The height $ht (𝔭)$ of a prime ideal $𝔭$ of $R$ is the length of the longest chain of primes of $R$ contained in $𝔭$ . A prime of height $0$ is called a minimal prime of $R$ .

Note that a minimal prime of $R$ is just an isolated prime of $(0)$ .

Examples 11.8.12. §

a.: In $F [x_{1}, \dots, x_{n}]$ for a field $F$ , the height of $(x_{1}, \dots, x_{k})$ for $k \leq n$ is $k$ .
b.: In a UFD, the primes of height one are principal, generated by the irreducible elements.
c.: In a product of fields $R = \prod_{i = 1}^{n} F_{i}$ , the minimal primes are the maximal ideals, the kernels of projection maps $R \to F_{k}$ for some $1 \leq k \leq n$ .

Remark 11.8.13. §

Suppose $R = K [x_{1}, \dots, x_{n}]$ with $K$ algebraically closed. The prime ideals $𝔭$ of $R$ correspond to algebraic sets in $𝔸_{K}^{n}$ . The dimension of the algebraic set $V$ that is the vanishing locus of $𝔭$ is defined to be $n - ht (𝔭)$ . In particular, $𝔸_{K}^{n}$ has dimension $n$ , as one would expect. We often refer to $ht (𝔭)$ as the codimension of $V$ in $𝔸_{K}^{n}$ . In particular, the vanishing locus of a single nonconstant polynomial in $R$ has codimension $1$ .

11.9. Dedekind domains

Definition 11.9.1. §

A Dedekind domain is a noetherian, integrally closed domain of Krull dimension at most $1$ .

The condition of having Krull dimension at most $1$ is the same as every nonzero prime ideal being maximal. We have the following class of examples.

Lemma 11.9.2. §

Every PID is a Dedekind domain.

Proof.

A PID is noetherian, and it is a UFD, so it is integrally closed. Its nonzero prime ideals are maximal, generated by its irreducible elements. □

Proposition 11.9.3. §

Let $A$ be a Dedekind domain, and let $B$ be the integral closure of $A$ in a finite, separable extension of the quotient field of $A$ . Then $B$ is a Dedekind domain.

Proof.

Note that $B$ is a finitely generated $A$ -module by Corollary 11.4.38. If $𝔟$ is an ideal of $B$ , then $𝔟$ is an $A$ -submodule of $B$ , and as $A$ is noetherian, it is therefore finitely generated. Thus, $B$ is noetherian. That $B$ is integrally closed is just Proposition 11.4.27. That every nonzero prime ideal in $A$ is maximal follows from Lemma 11.8.8. □

We have the following immediate corollary.

Corollary 11.9.4. §

The ring of integers of any number field is a Dedekind domain.

More examples of Dedekind domains can be produced as follows.

Proposition 11.9.5. §

Let $A$ be a Dedekind domain, and let $S$ be a multiplicatively closed subset of $A$ . Then $S^{- 1} A$ is also a Dedekind domain.

Proof.

Given an ideal $𝔟$ of $S^{- 1} A$ , set $𝔞 = A \cap 𝔟$ . Then $𝔞$ is an ideal of $A$ , and $𝔟 = S^{- 1} 𝔞$ . It follows that any set of generators of $𝔞$ as an ideal of $A$ generates $S^{- 1} 𝔞$ as an ideal of $S^{- 1} A$ . Hence $S^{- 1} A$ is noetherian. If, moreover, $𝔟$ is a nonzero prime, then clearly $𝔞$ is as well, and $𝔞$ is maximal since $A$ is a Dedekind domain. Then $S^{- 1} A ∕ 𝔟≅𝐴 ∕ 𝔞$ is a field, so $𝔟$ is maximal as well.

Let $K$ be the quotient field of $A$ . Any $α \in K$ that is integral over $S^{- 1} A$ satisfies a monic polynomial $f$ with coefficients in $S^{- 1} A$ . Set $n = \deg f$ . If $d \in S$ is the product of the denominators of these coefficients, then $d^{n} f (d^{- 1} x) \in A [x]$ is monic with $𝑑𝛼 \in K$ as a root. Since $A$ is integrally closed, we have $𝑑𝛼 \in A$ , so $α \in S^{- 1} A$ . That is, $S^{- 1} A$ is integrally closed. □

Lemma 11.9.6. §

Let $A$ be a noetherian domain, and let $𝔞$ be a nonzero ideal of $A$ .

a.: There exist $k \geq 0$ and nonzero prime ideals $𝔭_{1}, \dots, 𝔭_{k}$ of $A$ such that $𝔭_{1} \dots 𝔭_{k} \subseteq 𝔞$ .
b.: Suppose that $\dim A \leq 1$ . If $𝔭_{1}, \dots, 𝔭_{k}$ are as in part a and $𝔭$ is a prime ideal of $A$ containing $𝔞$ , then $𝔭 = 𝔭_{i}$ for some positive $i \leq k$ .

Proof.

Consider the set $X$ of nonzero ideals of $A$ for which the statement of the first part of the lemma fails, and order $X$ by inclusion. Suppose by way of contradiction that $X$ is nonempty. As $A$ is noetherian, $X$ must contain a maximal element $𝔞$ by Proposition 5.1.15. Now $𝔞$ is not prime since it lies in $X$ , so let $a, b \in A - 𝔞$ with $𝑎𝑏 \in 𝔞$ . Then $𝔞 + (a)$ and $𝔞 + (b)$ both properly contain $𝔞$ , so by maximality of $𝔞$ , there exist prime ideals $𝔭_{1}, \dots, 𝔭_{k}$ and $𝔮_{1}, \dots, 𝔮_{l}$ of $A$ for some $k, l \geq 0$ such that $𝔭_{1} \dots 𝔭_{k} \subseteq 𝔞 + (a)$ and $𝔮_{1} \dots 𝔮_{l} \subseteq 𝔞 + (b)$ . We then have

𝔭_{1} \dots 𝔭_{k} 𝔮_{1} \dots 𝔮_{l} \subseteq (𝔞 + (a)) (𝔞 + (b)) \subseteq 𝔞,

a contradiction of $𝔞 \in X$ . This proves part a.

Now, suppose that $𝔞$ is proper, and let $𝔭$ be a prime ideal containing $𝔞$ . Assume that $\dim A \leq 1$ . If no $𝔭_{i}$ equals $𝔭$ , then since $𝔭_{i}$ is maximal, there exist $b_{i} \in 𝔭_{i} - 𝔭$ for each $1 \leq i \leq k$ . We then have $b_{1} \dots b_{k} \notin 𝔭$ as $𝔭$ is prime, so $b_{1} \dots b_{k} \notin 𝔞$ , a contradiction. Hence we have part b. □

Definition 11.9.7. §

A fractional ideal of a domain $A$ is a nonzero $A$ -submodule $𝔞$ of the quotient field of $A$ for which there exists a nonzero $d \in A$ such that $𝑑𝔞 \subseteq A$ .

Remark 11.9.8. §

Every nonzero ideal in a domain $A$ is a fractional ideal, which is sometimes referred to as an integral ideal. Every fractional ideal of $A$ that is contained in $A$ is an integral ideal.

Example 11.9.9. §

The fractional ideals of $ℤ$ are exactly the $ℤ$ -submodules of $ℚ$ generated by a nonzero rational number.

Lemma 11.9.10. §

Let $A$ be a noetherian domain. An $A$ -submodule of the quotient field of $A$ is a fractional ideal if and only if it is finitely generated.

Proof.

If $𝔞$ is a finitely generated $A$ -submodule of the quotient field of $A$ , then let $d \in A$ denote the product of the denominators of a set of generators. Then $𝑑𝔞 \subseteq A$ . Conversely, suppose that $𝔞$ is a fractional ideal and $d \in A$ is nonzero and satisfies $𝑑𝔞 \subseteq A$ . Then $𝑑𝔞$ is an ideal of $A$ , hence finitely generated. Moreover, the multiplication-by- $d$ map carries $𝔞$ isomorphically onto $𝑑𝔞$ . □

Definition 11.9.11. §

Let $A$ be a domain with quotient field $K$ , and let $𝔞$ and $𝔟$ be fractional ideals of $A$ .

a.: The inverse of $𝔞$ is $𝔞^{- 1} = {b \in K ∣ 𝑏𝔞 \subseteq A}$ .
b.: The product of $𝔞$ and $𝔟$ is the $A$ -submodule of $K$ generated by the set ${𝑎𝑏 ∣ a \in 𝔞, b \in 𝔟}$ .

Lemma 11.9.12. §

Let $A$ be a domain, and let $𝔞$ and $𝔟$ be fractional ideals of $A$ . Then $𝔞^{- 1}$ , $𝔞 + 𝔟$ , $𝔞𝔟$ , $𝔞 \cap 𝔟$ are fractional ideals of $A$ as well.

Proof.

Let $K$ denote the quotient field of $A$ . Let $c, d \in A$ be nonzero such that $𝑐𝔞 \subseteq A$ and $𝑑𝔟 \subseteq A$ . Then $c (𝔞 \cap 𝔟) \subseteq A$ , $𝑐𝑑 (𝔞 + 𝔟) \subseteq A$ , and $𝑐𝑑𝔞𝔟 \subseteq A$ .

Note that $𝔞^{- 1}$ is an $A$ -submodule of $K$ which is nonzero since there exists $d \in A$ with $𝑑𝔞 \subseteq A$ in that $𝔞$ is a fractional ideal. Let $a \in 𝔞$ be nonzero, and let $e \in A$ be its numerator in a representation of $a$ as a fraction, so $e \in 𝔞$ as well. For any $t \in 𝔞^{- 1}$ , we have $𝑡𝑒 \in A$ by definition, so $e 𝔞^{- 1} \subseteq A$ , and therefore $𝔞^{- 1}$ is a fractional ideal. □

Remark 11.9.13. §

By definition, multiplication of fractional ideals is an associative (and commutative) operation, so the set $I (A)$ of fractional ideals in $A$ is a monoid.

Definition 11.9.14. §

We say that a fractional ideal $𝔞$ of a domain $A$ is invertible if there exists a fractional ideal $𝔟$ of $A$ such that $𝔞𝔟 = A$ .

Lemma 11.9.15. §

A fractional ideal $𝔞$ of a domain $A$ is invertible if and only if $𝔞^{- 1} 𝔞 = A$ .

Proof.

For the nonobvious direction, suppose that $𝔞$ is invertible. Then we must have $𝔟 \subseteq 𝔞^{- 1}$ by definition of $𝔞^{- 1}$ . On the other hand,

A = 𝔟𝔞 \subseteq 𝔞^{- 1} 𝔞 \subseteq A,

so we must have $𝔞^{- 1} 𝔞 = A$ . □

Example 11.9.16. §

Consider the maximal ideal $(x, y)$ of $ℚ [x, y]$ . If $f \in ℚ {(x, y)}^{\times}$ is such that $𝑓𝑥 \in ℚ [x, y]$ (resp., $𝑓𝑦 \in ℚ [x, y]$ ) then its denominator is a divisor of $x$ (resp., $y$ ). Therefore ${(x, y)}^{- 1} = ℚ [x, y]$ , and we have

(x, y) \cdot {(x, y)}^{- 1} = (x, y) \neq ℚ [x, y] .

Thus, $(x, y)$ is not invertible as a fractional ideal.

Definition 11.9.17. §

A principal fractional ideal of $A$ is an $A$ -submodule $(a)$ generated by a nonzero element $a$ of the quotient field of $A$ .

Lemma 11.9.18. §

Let $𝔞$ be a fractional ideal of a PID. Then $𝔞$ is principal.

Proof.

There exists $d \in A$ such that $𝑑𝔞 = (b)$ for some $b \in A$ . Then $\frac{b}{d} \in 𝔞$ and given any $c \in 𝔞$ , we have $𝑑𝑐 = 𝑏𝑎$ for some $a \in A$ , so $c = a \frac{b}{d}$ . That is, $𝔞 = (\frac{b}{d})$ . □

Lemma 11.9.19. §

Let $A$ be a domain, and let $a$ be a nonzero element of its quotient field. Then $(a)$ is invertible, and ${(a)}^{- 1} = (a^{- 1})$ .

Proof.

If $x \in {(a)}^{- 1}$ , then $𝑥𝑎 = b$ for some $b \in A$ , so $x = b a^{- 1} \in (a^{- 1})$ . If $x \in (a^{- 1})$ , then $x = a^{- 1} b$ for some $b \in A$ . On other hand, any $z \in (a)$ has the form $z = 𝑦𝑎$ for some $y \in A$ , and we have $𝑥𝑧 = a^{- 1} 𝑏𝑦𝑎 = 𝑏𝑦 \in A$ , so $x \in {(a)}^{- 1}$ . We then have

(a) {(a)}^{- 1} = (a) (a^{- 1}) = (a a^{- 1}) = A,

completing the proof. □

Lemma 11.9.20. §

Let $A$ be a Dedekind domain, and let $𝔭$ be a nonzero prime ideal of $A$ . Then $𝔭 𝔭^{- 1} = A$ .

Proof.

Let $a \in 𝔭$ be nonzero. Noting Lemma 11.9.6a, we let $k \geq 1$ be minimal such that there exist nonzero prime ideals $𝔭_{1}, \dots, 𝔭_{k}$ of $A$ with $𝔭_{1} \dots 𝔭_{k} \subseteq (a)$ . By Lemma 11.9.6b, we may without loss of generality suppose that $𝔭_{k} = 𝔭$ . By the minimality of $k$ , we may choose $b \in 𝔭_{1} \dots 𝔭_{k - 1}$ be such that $b \notin (a)$ . Then $a^{- 1} b \notin A$ , but we have

a^{- 1} 𝑏𝔭 \subseteq a^{- 1} 𝔭_{1} \dots 𝔭_{k} \subseteq A,

which implies that $a^{- 1} b \in 𝔭^{- 1}$ . Moreover, if $𝔭^{- 1} 𝔭 = 𝔭$ , then $a^{- 1} 𝑏𝔭 \subseteq 𝔭$ . Since $𝔭$ is finitely generated, Proposition 11.4.4 tells us that $a^{- 1} b$ is integral over $A$ . But $A$ is integrally closed, so we have a contradiction. That is, we must have $𝔭 ⊊ 𝔭^{- 1} 𝔭 \subseteq A$ , from which it follows that $𝔭^{- 1} 𝔭 = A$ by maximality of $𝔭$ . □

Theorem 11.9.21. §

Let $A$ be a Dedekind domain, and let $𝔞$ be a fractional ideal of $A$ . Then there exist $k \geq 0$ and distinct nonzero prime ideals $𝔭_{1}, \dots, 𝔭_{k}$ and $r_{1}, \dots, r_{k} \in ℤ - {0}$ such that $𝔞 = 𝔭_{1}^{r_{1}} \dots 𝔭_{k}^{r_{k}}$ , and this decomposition is unique up to ordering. Moreover, $𝔞$ is an ideal of $A$ if and only if every $r_{i}$ is positive.

Proof.

First suppose that $𝔞$ is a nonzero ideal of $A$ . We work by induction on a minimal nonnegative integer $m$ such that there are nonzero prime ideals $𝔮_{1}, \dots, 𝔮_{m}$ of $𝔞$ (not necessarily distinct) with $𝔮_{1} \dots 𝔮_{m} \subseteq 𝔞$ , which exists by Lemma 11.9.6a. If $m = 0$ , then $A \subseteq 𝔞$ , so $𝔞 = A$ . In general, for $m \geq 1$ , we know that $𝔞$ is proper, so there exists a nonzero prime ideal $𝔭$ that contains $𝔞$ and $𝔭 = 𝔮_{i}$ for some $i \leq m$ . Without loss of generality, we take $i = m$ . Then

𝔮_{1} \dots 𝔮_{m - 1} \subseteq 𝔮_{1} \dots 𝔮_{m} 𝔭^{- 1} \subseteq 𝔞 𝔭^{- 1} \subseteq A .

By induction, there exist nonzero prime ideals $𝔮_{1}^{'}, \dots, 𝔮_{ℓ}^{'}$ of $A$ for some $ℓ < m$ such that $𝔞 𝔭^{- 1} = 𝔮_{1}^{'} \dots 𝔮_{ℓ}^{'}$ . The desired factorization is given by multiplying by $𝔭$ , applying Lemma 11.9.20, and gathering together nondistinct primes.

In general, for a fractional ideal $𝔞$ , we let $d \in A$ be such that $𝑑𝔞 \subseteq A$ . We write $𝑑𝔞 = 𝔮_{1} \dots 𝔮_{m}$ for some $m \geq 0$ and prime ideals $𝔮_{i}$ for $1 \leq i \leq m$ . We also write $(d) = 𝔩_{1} \dots 𝔩_{n}$ for some $n \geq 0$ and prime ideals $𝔩_{i}$ for $1 \leq i \leq n$ . By Lemma 11.9.20, we then have

𝔞 = {(d)}^{- 1} (𝑑𝔞) = 𝔩_{1}^{- 1} \dots 𝔩_{n}^{- 1} 𝔮_{1} \dots 𝔮_{m} .

If $𝔮_{i} = 𝔩_{j}$ for some $i$ and $j$ , then we may use Lemma 11.9.20 to remove $𝔮_{i} 𝔩_{j}^{- 1}$ from the product. Hence we have the desired factorization.

Now suppose that

𝔞 = 𝔭_{1}^{r_{1}} \dots 𝔭_{k}^{r_{k}}

for some $k \geq 0$ , distinct primes $𝔭_{1}, \dots, 𝔭_{k}$ and nonzero $r_{1}, \dots, r_{k}$ . For each prime $𝔭$ of $A$ , consider the localization $A_{𝔭}$ , which is a Dedekind domain with unique nonzero prime ideal $𝔭 A_{𝔭}$ . Note that $𝔮 A_{𝔭} = A_{𝔭}$ if $𝔮$ is a nonzero prime of $A$ other than $𝔭$ . We therefore have

𝔞 A_{𝔭} = 𝔭_{1}^{r_{1}} \dots, 𝔭_{k}^{r_{k}} A_{𝔭} = 𝔭^{r} A_{𝔭},

where $r = r_{i}$ if $𝔭 = 𝔭_{i}$ for some $i$ , and $r = 0$ otherwise. Moreover, if $𝔭^{k} A_{𝔭} = 𝔭^{l} A_{𝔭}$ for some integers $k \leq l$ , then $𝔭^{l - k} A_{𝔭} = A_{𝔭}$ , which since $𝔭$ is nonzero, can only happen if $k = l$ . Therefore, the primes $𝔭_{i}$ and corresponding integers $r_{i}$ are uniquely determined by $𝔞$ . □

We have the following immediate corollary of Theorem 11.9.21.

Corollary 11.9.22. §

The set of fractional ideals $I (A)$ of a Dedekind domain $A$ is a group under multiplication of fractional ideals with identity $A$ , the inverse of $𝔞 \in I (A)$ being $𝔞^{- 1}$ .

Definition 11.9.23. §

Let $A$ be a Dedekind domain. The group $I (A)$ of fractional ideals of $A$ is called the ideal group of $A$ .

Definition 11.9.24. §

Let $A$ be a Dedekind domain. Then we let $P (A)$ denote the set of its principal fractional ideals. We refer to this as the principal ideal group.

Corollary 11.9.25. §

Let $A$ be a Dedekind domain. The group $P (A)$ is a subgroup of $I (A)$ .

Definition 11.9.26. §

The class group (or ideal class group) of a Dedekind domain $A$ is $Cl (A) = I (A) ∕ P (A)$ , the quotient of the ideal group by the principal ideal group.

Lemma 11.9.27. §

A Dedekind domain $A$ is a PID if and only if $Cl (A)$ is trivial.

Proof.

Every element of $I (A)$ has the form $𝔞 𝔟^{- 1}$ where $𝔞$ and $𝔟$ are nonzero ideals of $A$ . If $A$ is a PID, then both $𝔞$ and $𝔟$ are principal and, therefore, so is $𝔞 𝔟^{- 1}$ . On the other hand, if $𝔞$ is a nonzero ideal of $A$ with $𝔞 = (a)$ for some $a \in K$ , then clearly $a \in A$ , so $Cl (A)$ being trivial implies that $A$ is a PID. □

Notation 11.9.28. §

Let $K$ be a number field. We let $I_{K}$ , $P_{K}$ , and ${Cl}_{K}$ denote the ideal group, principal ideal group, and class group of $𝒪_{K}$ , respectively. We refer to these as the ideal group of $K$ , the principal ideal group of $K$ , and the class group of $K$ , respectively.

Example 11.9.29. §

Let $K = ℚ (\sqrt{- 5})$ . Then $𝒪_{K} = ℤ [\sqrt{- 5}]$ . The ideal $𝔞 = (2, 1 + \sqrt{- 5})$ is non-principal. To see this, note that $N_{K ∕ ℚ} (2) = 4$ and $N_{K ∕ ℚ} (1 + \sqrt{- 5}) = 6$ , so any generator $x$ of $𝔞$ must satisfy $N_{K ∕ ℚ} (x) \in {\pm 1, \pm 2}$ . But

N_{K ∕ ℚ} (a + b \sqrt{- 5}) = a^{2} + 5 b^{2}

for $a, b \in ℤ$ , which forces $x = \pm 1$ . This would mean that $𝔞 = ℤ [\sqrt{- 5}]$ . To see that this cannot happen, define $ϕ : ℤ [\sqrt{- 5}] \to ℤ ∕ 6 ℤ$ by $ϕ (a + b \sqrt{- 5}) = a - b$ for $a, b \in ℤ$ . This is a ring homomorphism as

\begin{matrix} ϕ ((a + b \sqrt{- 5}) (c + d \sqrt{- 5})) = ϕ (𝑎𝑐 - 5 𝑏𝑑 + (𝑎𝑑 + 𝑏𝑐) \sqrt{- 5}) = 𝑎𝑐 - 5 𝑏𝑑 - 𝑎𝑑 - 𝑏𝑐 \\ = 𝑎𝑐 + 𝑏𝑑 - 𝑎𝑑 - 𝑏𝑐 = (a - b) (c - d) . \end{matrix}

Moreover, $ϕ (1 + \sqrt{- 5}) = 0$ , so the kernel of $ϕ$ contains (and is in fact equal to) $(1 + \sqrt{- 5})$ . Therefore, $ϕ$ induces a surjection (in fact, isomorphism),

ℤ [\sqrt{- 5}] ∕ 𝔞 \to ℤ ∕ 6 ℤ ∕ (2) \to ℤ ∕ 2 ℤ,

so $𝔞 \neq ℤ [\sqrt{- 5}]$ , and $x$ does not exist. Therefore, ${Cl}_{ℚ (\sqrt{- 5})}$ is nontrivial.

We end with the following important theorem.

Theorem 11.9.30. §

A Dedekind domain is a UFD if and only if it is a PID.

Proof.

We need only show that a Dedekind domain that is a UFD is a PID. Let $A$ be such a Dedekind domain. By Theorem 11.9.21, it suffices to show that each nonzero prime ideal $𝔭$ of $A$ is principal. Since $𝔭$ is prime and $A$ is a UFD, any nonzero element of $𝔭$ is divisible by an irreducible element in $𝔭$ . If $π$ is such an element, then $(π)$ is maximal and contained in $𝔭$ , so $𝔭 = (π)$ . □

11.10. Discrete valuation rings

Definition 11.10.1. §

A discrete valuation ring, or DVR, is a principal ideal domain that has exactly one nonzero prime ideal.

Lemma 11.10.2. §

The following are equivalent conditions on a principal ideal domain $A$ .

i.: $A$ is a DVR,
ii.: $A$ has a unique nonzero maximal ideal,
iii.: $A$ has a unique nonzero irreducible element up to associates.

Proof.

This is a simple consequence of the fact that in a PID, every nonzero prime ideal is maximal generated by any irreducible element it contains. □

Definition 11.10.3. §

A uniformizer of a DVR is a generator of its maximal ideal.

Moreover, we have the following a priori weaker but in fact equivalent condition for a domain to be a DVR.

Proposition 11.10.4. §

A domain $A$ is a DVR if and only if it is a local Dedekind domain that is not a field.

Proof.

A DVR is a PID, hence a Dedekind domain, and it is local by definition. Conversely, suppose that $A$ is noetherian, integrally closed, and has a unique nonzero prime ideal $𝔭$ . We must show that $A$ is a PID. Since nonzero ideals factor uniquely as products of primes in $A$ , every ideal of $A$ has the form $𝔭^{n}$ for some $n$ . In particular, $𝔭 = (π)$ for any $π \in 𝔭 - 𝔭^{2}$ , and then $𝔭^{n} = (π^{n})$ for all $n$ . Therefore, $A$ is a PID and hence a DVR. □

Theorem 11.10.5. §

A noetherian domain $A$ is a Dedekind domain if and only if its localization at every nonzero prime ideal is a DVR.

Proof.

We have seen in Proposition 11.9.5 that $A_{𝔭}$ is a Dedekind domain for all nonzero prime ideals $𝔭$ . By Proposition 11.10.4, each such localization is therefore a DVR.

Conversely, if $A$ is a noetherian integral domain such that $A_{𝔭}$ is a DVR for every nonzero prime ideal $𝔭$ , we consider the intersection $B = ⋂_{𝔭} A_{𝔭}$ over all nonzero prime ideals $𝔭$ of $A$ , taken inside the quotient field $K$ of $A$ . Clearly, $B$ contains $A$ , and if $\frac{c}{d} \in B$ for some $c, d \in A$ , then we set

𝔞 = {a \in A ∣ 𝑎𝑐 \in (d)} .

By definition of $B$ , we may write $\frac{c}{d} = \frac{r}{s}$ with $r \in A$ and $s \in A - 𝔭$ , and we see that $𝑠𝑐 = 𝑟𝑑$ , so $s \in 𝔞$ . In other words, we have $𝔞 ⊈ 𝔭$ for all prime ideals $𝔭$ of $A$ , which forces $𝔞 = A$ . This implies that $c \in (d)$ , so $\frac{c}{d} \in A$ .

Next, suppose that $𝔮$ is a nonzero prime ideal of $A$ , and let $𝔪$ be a maximal ideal containing it. Then $𝔮 A_{𝔪}$ is a nonzero prime ideal of $A_{𝔪}$ , which is a DVR, so $𝔮 A_{𝔪} = 𝔪 A_{𝔪}$ . Since $𝔮$ and $𝔪$ are prime ideals contained in $𝔪$ , we therefore have

𝔮 = A \cap 𝔮 A_{𝔪} = A \cap 𝔪 A_{𝔪} = 𝔪 .

Thus, $A$ has Krull dimension at most $1$ .

Finally, each $A_{𝔭}$ is integrally closed in $K$ by Corollary 11.4.24, and then the intersection $A$ is as well, since any element of $K$ that is integral over $A$ is integral over each $A_{𝔭}$ , hence contained in each $A_{𝔭}$ . That is, $A$ satisfies the conditions in the definition of a Dedekind domain. □

To make some sense of the name “discrete valuation ring”, we define the notion of a discrete valuation. For this purpose, we adjoin an element $\infty$ to $ℤ$ which is considered larger than any element of $ℤ$ , and we set $x + y = \infty$ if $x, y \in ℤ \cup {\infty}$ and either $x$ or $y$ equals $\infty$ .

Definition 11.10.6. §

Let $K$ be a field. A discrete valuation on $K$ is a surjective map $v : K \to ℤ \cup {\infty}$ such that

i.: $v (a) = \infty$ if and only $a = 0$ ,
ii.: $v (𝑎𝑏) = v (a) + v (b)$ , and
iii.: $v (a + b) \geq \min (v (a), v (b))$

for all $a, b \in K$ .

Definition 11.10.7. §

If $v$ is a discrete valuation on a field $K$ , then the quantity $v (a)$ for $a \in K$ is said to be the valuation of $a$ with respect to $v$ .

The following are standard examples of discrete valuations.

Example 11.10.8. §

Let $p$ be a prime number. Then the $p$ -adic valuation $v_{p}$ on $ℚ$ is defined by $v_{p} (0) = \infty$ and $v_{p} (a) = r$ for $a \in ℚ^{\times}$ if $a = p^{r} a^{'}$ for some $r \in ℤ$ and $a^{'} \in ℚ^{\times}$ such that $p$ divides neither the numerator nor denominator of $a^{'}$ in reduced form.

Example 11.10.9. §

Let $F$ be a field, and consider the function field $F (t)$ . The valuation at $\infty$ on $F (t)$ is defined by $v_{\infty} (\frac{g}{h}) = \deg h - \deg g$ for $g, h \in F [t]$ with $h \neq 0$ , taking $\deg 0 = \infty$ .

More generally, we have the following.

Definition 11.10.10. §

Let $A$ be a Dedekind domain with quotient field $K$ , and let $𝔭$ be a nonzero prime ideal of $A$ . The $𝔭$ -adic valuation $v_{𝔭}$ on $K$ is defined on $a \in K^{\times}$ as the unique integer such that $(a) = 𝔭^{v_{p} (a)} 𝔟 𝔠^{- 1}$ for some nonzero ideals $𝔟$ and $𝔠$ of $A$ that are not divisible by $𝔭$ .

Example 11.10.11. §

For the valuation at $\infty$ on $F (t)$ , where $F$ is a field, we may take $A = K [t^{- 1}]$ and $𝔭 = (t^{- 1})$ . Then the valuation $v_{\infty}$ on $F (t)$ is the $(t^{- 1})$ -adic valuation. To see this, note that for nonzero $g, h \in F [t]$ , one has

\frac{g (t)}{h (t)} = {(t^{- 1})}^{\deg h - \deg g} \frac{G (t^{- 1})}{H (t^{- 1})},

where $G (t^{- 1}) = t^{- \deg g} g (t)$ and $H (t^{- 1}) = t^{- \deg h} h (t)$ are polynomials in $t^{- 1}$ which have nonzero constant term.

Lemma 11.10.12. §

Let $A$ be a Dedekind domain with quotient field $K$ , and let $𝔭$ be a prime ideal of $A$ . The $𝔭$ -adic valuation on $K$ is a discrete valuation.

Proof.

Let $a, b \in K$ be nonzero (without loss of generality). Write $(a) = 𝔭^{r} 𝔞$ and $(b) = 𝔭^{s} 𝔟$ for $r = v_{𝔭} (a)$ and $s = v_{𝔭} (b)$ and fractional ideals $𝔞$ and $𝔟$ of $A$ . Note that $(𝑎𝑏) = 𝔭^{r + s} 𝔞𝔟$ , so $v_{p} (𝑎𝑏) = r + s$ . We have

(a + b) = 𝔭^{r} 𝔞 + 𝔭^{s} 𝔟 = 𝔭^{\min (r, s)} (𝔭^{r - \min (r, s)} 𝔞 + 𝔭^{s - \min (r, s)} 𝔟),

v_{𝔭} (a + b) = \min (r, s) + v_{𝔭} (𝔭^{r - \min (r, s)} 𝔞 + 𝔭^{s - \min (r, s)} 𝔟) \geq \min (r, s) .

□

Lemma 11.10.13. §

Let $v$ be a discrete valuation on a field $K$ . Then we have $v (- a) = v (a)$ for all $a \in K$ .

Proof.

Note that $2 v (- 1) = v (1) = 0$ , so we have $v (- a) = v (- 1) + v (a) = v (a)$ . □

Lemma 11.10.14. §

Let $v$ be a discrete valuation on a field $K$ . Then we have

v (a + b) = \min (v (a), v (b))

for all $a, b \in K$ with $v (a) \neq v (b)$ .

Proof.

If $v (a) < v (b)$ , then

v (a) = v ((a + b) - b) \geq \min (v (a + b), v (b)) \geq \min (v (a), v (b)) = v (a),

so we have $v (a) = \min (v (a + b), v (b))$ , which forces $v (a + b) = v (a)$ . □

Definition 11.10.15. §

Let $K$ be a field, and let $v$ be a discrete valuation on $K$ . Then

𝒪_{v} = {a \in K ∣ v (a) \geq 0}

is called the valuation ring of $v$ .

Lemma 11.10.16. §

Let $K$ be a field, and let $v$ be a discrete valuation on $K$ . Then $𝒪_{v}$ is a DVR with maximal ideal

𝔪_{v} = {a \in K ∣ v (a) \geq 1} .

Proof.

That $𝒪_{v}$ is a ring follows from the fact that if $a, b \in 𝒪_{v}$ , then $v (𝑎𝑏) = v (a) + v (b) \geq 0$ , $v (- a) = v (a) \geq 0$ , and $v (a + b) \geq \min (v (a), v (b)) \geq 0$ . For $a \in 𝒪_{v}$ and $x, y \in 𝔪_{v}$ , we have $v (x + y) \geq \min (v (x), v (y)) \geq 1$ and $v (𝑎𝑥) = v (a) + v (x) \geq 1$ , so $𝔪_{v}$ is an ideal. It is also the unique maximal ideal: given $a \in 𝒪_{v} - 𝔪_{v}$ , we have $v (a^{- 1}) = v (a) + v (a^{- 1}) = v (1) = 0$ , so $a \in 𝒪_{v}^{\times}$ . Given an ideal $𝔞$ of $𝒪_{v}$ , let $a \in 𝔞$ be an element of minimal valuation $n$ . Let $π \in 𝒪_{v}$ with $v (π) = 1$ , and write $a = π^{n} u$ for some $u \in 𝒪_{v}^{\times}$ . Then $v (u) = 0$ , so $u \in 𝒪_{v}^{\times}$ . Therefore, $(π^{n}) \subseteq 𝔞$ . On the other hand, since $n$ is minimal, we have $𝔞 \subseteq (π^{n})$ , and therefore $𝔞$ is a principal. By Lemma 11.10.2, we conclude that $𝒪_{v}$ is a DVR. □

Example 11.10.17. §

In $ℚ$ , we have

𝒪_{v_{p}} = ℤ_{(p)} = {\frac{a}{b} ∣ a, b \in ℤ such that p ∤ b} .

11.11. Ramification of primes

The integral closure $B$ of a Dedekind domain $A$ in a finite extension $L$ of its quotient field $K$ is also a Dedekind domain. If $𝔭$ is a nonzero prime ideal of $A$ , then we can consider the ideal $𝔭𝐵$ of $B$ . This ideal may no longer be prime. Instead, it has a factorization

𝔭𝐵 = 𝔓_{1}^{e_{1}} \dots 𝔓_{g}^{e_{g}}

(11.11.1)

for some distinct nonzero prime ideals $𝔓_{i}$ of $B$ and positive integers $e_{i}$ , for $1 \leq i \leq g$ for some $g \geq 1$ . We make the following definitions.

Definition 11.11.1. §

Let $B ∕ A$ be an extension of commutative rings. We say that a prime ideal $𝔓$ of $B$ lies over (or above) a prime ideal $𝔭$ of $A$ if $𝔭 = 𝔓 \cap A$ . We then say that $𝔭$ lies under (or below) $𝔓$ .

In (11.11.1), the prime ideals of $B$ lying over $𝔭$ are exactly the $𝔓_{i}$ for $1 \leq i \leq g$ .

Definition 11.11.2. §

Let $A$ be a Dedekind domain, and let $B$ be the integral closure of $A$ in a finite extension $L$ of the quotient field $K$ of $A$ . Let $𝔭$ be a nonzero prime ideal of $A$ .

a.: We say that $𝔭$ ramifies (or is ramified) in $L ∕ K$ if $𝔭𝐵$ is divisible by the square of a prime ideal of $B$ . Otherwise, it is said to be unramified.
b.: We say that $𝔭$ is inert in $L ∕ K$ if $𝔭𝐵$ is a prime ideal.
c.: We say that $𝔭$ is split in $L ∕ K$ if there exist two distinct prime ideals of $B$ lying over $𝔭$ . Otherwise, $𝔭$ is non-split.

It follows directly that $𝔭$ is ramified in $L ∕ K$ if some $e_{i}$ in (11.11.1) is at least $2$ . On the other hand, $𝔭$ is inert in $L ∕ K$ if there is exactly one prime ideal of $B$ lying over $𝔭$ and its ramification index is $1$ , which is to say that $g = 1$ and $e_{1} = 1$ in (11.11.1). Finally, $𝔭$ is split in $L ∕ K$ if $g > 1$ .

Example 11.11.3. §

Let $A = ℤ$ and $L = ℚ (\sqrt{2})$ . The integral closure of $A$ in $L$ is $B = 𝒪_{L} = ℤ [\sqrt{2}]$ . The prime $𝔭 = (2)$ ramifies in $ℚ (\sqrt{2}) ∕ ℚ$ , since

2 ℤ [\sqrt{2}] = {(\sqrt{2})}^{2} .

Moreover, $𝔓 = (\sqrt{2})$ is a prime ideal of $ℤ [\sqrt{2}]$ , since $ℤ [\sqrt{2}] ∕ (\sqrt{2}) ≅ℤ ∕ 2 ℤ$ via the map that takes $a + b \sqrt{2}$ to $a mod 2$ . Therefore, $𝔭$ is ramified and non-split.

Next, consider the prime ideal $(3)$ of $ℤ$ . We have $ℤ [\sqrt{2}] ∕ (3) ≅ 𝔽_{3} [\sqrt{2}] ≅ 𝔽_{9}$ , so $(3)$ is inert in $ℚ (\sqrt{2}) ∕ ℚ$ . On the other hand, the prime factorization of $7 ℤ [\sqrt{2}]$ is exactly

7 ℤ [\sqrt{2}] = (3 + \sqrt{2}) (3 - \sqrt{2}),

since $ℤ [\sqrt{2}] ∕ (3 \pm \sqrt{2})$ is isomorphic to $ℤ ∕ 7 ℤ$ via the map that takes $a + b \sqrt{2}$ to $a \mp 3 b$ . That is, $(7)$ splits in $ℚ (\sqrt{2}) ∕ ℚ$ .

Definition 11.11.4. §

Let $A$ be a Dedekind domain, and let $𝔭$ be a nonzero prime ideal of $A$ . The residue field of $𝔭$ is $A ∕ 𝔭$ .

Remark 11.11.5. §

Let $A$ be a Dedekind domain, and let $B$ be the integral closure of $A$ in a finite extension $L$ the quotient field $K$ of $A$ . Let $𝔭$ be a nonzero prime ideal of $A$ , and let $𝔓$ be a prime ideal of $L$ lying over $K$ . Then $B ∕ 𝔓$ is a field extension of $A ∕ 𝔭$ via the natural map induced on quotients by the inclusion $A ↪ B$ .

Definition 11.11.6. §

Let $A$ be a Dedekind domain, and let $B$ be the integral closure of $A$ in a finite extension $L$ of the quotient field $K$ of $A$ . Let $𝔭$ be a nonzero prime ideal of $A$ , and let $𝔓$ be a prime ideal of $B$ lying over $𝔭$ .

a.: The ramification index $e_{𝔓 ∕ 𝔭}$ of $𝔓$ over $𝔭$ is the largest $e \geq 1$ such that $𝔓^{e}$ divides $𝔭𝐵$ .
b.: The residue degree $f_{𝔓 ∕ 𝔭}$ of a prime ideal of $𝔓$ lying over $𝔭$ is $[B ∕ 𝔓 : A ∕ 𝔭]$ .

Remark 11.11.7. §

It follows quickly from the definitions that ramification indices and residue degrees are multiplicative in extensions. That is, if $A \subseteq B \subseteq C$ are Dedekind domains with the quotient field of $C$ a finite extension of that of $A$ and $𝔓$ is a prime ideal of $C$ lying over $P$ of $B$ and $𝔭$ of $A$ , then

\begin{array}{rcl} e_{𝔓 ∕ 𝔭} = e_{𝔓 ∕ P} e_{P ∕ 𝔭} & and & f_{𝔓 ∕ 𝔭} = f_{𝔓 ∕ P} f_{P ∕ 𝔭} . \end{array}

Example 11.11.8. §

In Example 11.11.3, the residue degree of $(\sqrt{2})$ over $2 ℤ$ is $1$ , the residue degree of $3 ℤ [\sqrt{2}]$ over $3 ℤ$ is $2$ , and the residue degrees of $(3 \pm \sqrt{2})$ over $7 ℤ$ are each $1$ . The ramification indices are $2$ , $1$ , and $1$ , repsectively.

We shall require the following lemmas.

Lemma 11.11.9. §

Let $𝔭$ be a nonzero prime ideal in a Dedekind domain $A$ . For each $i \geq 0$ , the $A ∕ 𝔭$ -vector space $𝔭^{i} ∕ 𝔭^{i + 1}$ is one-dimensional.

Proof.

Let $x \in 𝔭^{i} - 𝔭^{i + 1}$ for some $i \geq 0$ . (Such an element exists by unique factorization of ideals.) We need only show that the image of $x$ spans $𝔭^{i} ∕ 𝔭^{i + 1}$ . For this, note that $(x) = 𝔭^{i} 𝔞$ for some nonzero ideal of $A$ not divisible by $𝔭$ . Then

(x) + 𝔭^{i + 1} = 𝔭^{i} (𝔞 + 𝔭) = 𝔭^{i},

the last step by the Chinese remainder theorem. □

Lemma 11.11.10. §

Let $A$ be a Dedekind domain and $P$ be a set of nonzero prime ideals of $A$ . Let $S$ a multiplicatively closed subset of $A$ such that $S \cap 𝔭 = \emptyset$ for all $𝔭 \in P$ . Let $𝔞$ be a nonzero ideal of $A$ that is divisible only by prime ideals in $P$ . Then the natural map

A ∕ 𝔞 \to S^{- 1} A ∕ S^{- 1} 𝔞

is an isomorphism.

Proof.

Suppose that $b \in S^{- 1} 𝔞 \cap A$ , and write $b = \frac{a}{s}$ for some $a \in 𝔞$ and $s \in S$ . Then $a = 𝑏𝑠$ , and since $𝔞$ divides $(a)$ while $(s)$ is relatively prime to $𝔞$ , we must have that $𝔞$ divides $(b)$ . In other words, $b \in 𝔞$ , and therefore the map is injective. Given $c \in A$ and $t \in S$ , the ideals $(t)$ and $𝔞$ have no common prime factor, so in that $A$ is a Dedekind domain, satisfy $(t) + 𝔞 = A$ . Thus, there exists $u \in A$ such that $𝑢𝑡 - 1 \in 𝔞$ . Then $𝑐𝑢 + 𝔞$ maps to $\frac{c}{t} + S^{- 1} 𝔞$ , so the map is surjective. □

The ramification indices and residue degrees of the primes over $𝔭$ satisfy the following degree formula.

Theorem 11.11.11. §

Let $A$ be a Dedekind domain, and let $B$ be the integral closure of $A$ in a finite separable extension $L$ the quotient field $K$ of $A$ . Let $𝔭$ be a nonzero prime ideal of $A$ , and write

𝔭𝐵 = 𝔓_{1}^{e_{1}} \dots 𝔓_{g}^{e_{g}}

for some distinct nonzero prime ideals $𝔓_{i}$ of $B$ and positive integers $e_{i}$ , for $1 \leq i \leq g$ and some $g \geq 1$ . For each $i$ , let $f_{i} = f_{𝔓_{i} ∕ 𝔭}$ . Then

\sum_{i = 1}^{g} e_{i} f_{i} = [L : K] .

Proof.

We prove that $\dim_{A ∕ 𝔭} B ∕ 𝔭𝐵$ equals both quantities in the desired equality. By the Chinese remainder theorem, we have a canonical isomorphism

B ∕ 𝔭𝐵≅ \prod_{i = 1}^{g} B ∕ 𝔓_{i}^{e_{i}},

of $A ∕ 𝔭$ -vector spaces, so

\dim_{A ∕ 𝔭} B ∕ 𝔭𝐵 = \sum_{i = 1}^{g} \dim_{A ∕ 𝔭} B ∕ 𝔓_{i}^{e_{i}} = \sum_{i = 1}^{g} \sum_{j = 0}^{e_{i} - 1} \dim_{A ∕ 𝔭} 𝔓_{i}^{j} ∕ 𝔓_{i}^{j + 1} .

By Lemma 11.11.9, each $𝔓_{i}^{j} ∕ 𝔓_{i}^{j + 1}$ is a $1$ -dimensional $B ∕ 𝔓_{i}$ -vector space, and we therefore have

\dim_{A ∕ 𝔭} B ∕ 𝔭𝐵 = \sum_{i = 1}^{g} e_{i} \dim_{A ∕ 𝔭} B ∕ 𝔓_{i} = \sum_{i = 1}^{g} e_{i} f_{i} .

Let $S$ denote the complement of $𝔭$ in $A$ . Then $S^{- 1} A = A_{𝔭}$ and $S^{- 1} B$ are Dedekind domains, and $A_{𝔭}$ is a DVR, hence a PID. Moreover, $S^{- 1} B$ is the integral closure of $A_{𝔭}$ in $L$ , being both integrally closed and contained in said integral closure. Thus, Corollary 11.4.40 tells us that $S^{- 1} B$ is free of rank $[L : K]$ over $A_{𝔭}$ . In particular, $S^{- 1} B ∕ 𝔭 S^{- 1} B$ is an $[L : K]$ -dimensional $A_{𝔭} ∕ 𝔭 A_{𝔭}$ -vector space. On the other hand, note that

S \cap 𝔓_{i} = S \cap A \cap 𝔓_{i} = S \cap 𝔭 = \emptyset

for each $1 \leq i \leq g$ . Therefore, Lemma 11.11.10 tells us that

S^{- 1} B ∕ 𝔭 S^{- 1} 𝐵≅𝐵 ∕ 𝔭𝐵

and $A_{𝔭} ∕ 𝔭 A_{𝔭} ≅𝐴 ∕ 𝔭$ . We thus have that $\dim_{A ∕ 𝔭} B ∕ 𝔭𝐵 = [L : K]$ , as required. □

In other words, Theorem 11.11.11 tells us that the sum over all primes lying over $𝔭$ of the products of their ramification indices with their residue degrees equals the degree of the field extension $L ∕ K$ .