Algebraic Number Theory

Note that we have a canonical isomorphism $K[x]{\otimes }_{K}M\underset{}{\overset{\sim }{\to }}M[x]$ that gives rise to the first map in the composition

the second isomorphism being the Chinese remainder theorem. The composition is $\kappa$. □

Write $L=K(𝜃)$, and let $f\in K[x]$ be the minimal polynomial of $𝜃$. Then we define $\kappa$ as the composition

where the first isomorphism is that of Proposition 1.1.1 and the second takes $x$ to $\sigma (𝜃)$ in the coordinate corresponding to $\sigma$. Any $\beta \in L$ has the form $g(𝜃)$ for some $g\in K[x]$, and since any $\sigma :L\hookrightarrow M$ fixing $K$ fixes the coefficients of $g$, we have $\kappa (\beta \otimes 1)$ is as stated. □

If $x\in L\cap M$ with $x\notin K$, then $x$ and $1$ are elements of $L$ that are $K$-linearly independent but not $M$-linearly independent, so $L$ and $M$ are not linearly disjoint over $K$. □

Suppose that ${\gamma }_1,\dots ,{\gamma }_s\in M$ are $L$-linearly dependent, and write ${\sum }_{i=1}^s{\beta }_i{\gamma }_i=0$ for some ${\beta }_i\in L$. If $\phi$ is injective, then we must have ${\sum }_{i=1}^s{\beta }_i\otimes {\gamma }_i=0$, which means that the ${\gamma }_i$ are $K$-linearly dependent.

Conversely, let $L$ and $M$ be linearly disjoint over $K$. Suppose that we have a nonzero

for some ${\beta }_i\in L$ and ${\gamma }_i\in M$, with $s$ taken to be minimal. If $x\ne 0$, then the ${\gamma }_i$ are $L$-linearly dependent, so they are $K$-linearly dependent. In this case, without loss of generality, we may suppose that

for some ${\alpha }_i$ in $K$. Then

contradicting minimality. Thus $\ker \phi =0$. □

Note that $\mathit{𝐿𝑀}$ is a union of subfields of the form $K(\alpha ,\beta )$ with $\alpha \in L$ and $\beta \in M$. Since $\alpha$ and $\beta$ are algebraic over $K$, we have $K(\alpha ,\beta )=K[\alpha ,\beta ]$, and every element of the latter ring is a $K$-linear combination of monomials in $\alpha$ and $\beta$. Thus $\phi$ of Proposition 1.1.6 is surjective, and the result follows from the latter proposition. □

Again, we have the surjection $\phi :L{\otimes }_{K}M\to \mathit{𝐿𝑀}$ given by multiplication which is an injection if and only if $L$ and $M$ are linearly disjoint by Proposition 1.1.6. As $L{\otimes }_{K}M$ has dimension $[L:K][M:K]$ over $K$, the result follows. □

We write $L=K(𝜃)$ for some $𝜃\in L$, and let $f\in K[x]$ be the minimal polynomial of $𝜃$. As $\mathrm{Gal}(\mathit{𝐿𝑀}∕M)\cong \mathrm{Gal}(L∕(L\cap M))$ by restriction, we have $L\cap M=K$ if and only if $[\mathit{𝐿𝑀}:M]=[L:K]$. Since $\mathit{𝐿𝑀}=M(𝜃)$, this occurs if and only if $f$ is irreducible in $M[x]$. The result then follows from Remark 1.1.9. □

Suppose that (i) holds. Then $\beta$ is a root of a monic polynomial $g\in A[x]$. Given any $f\in A[x]$, the division algorithm tells us that $f=\mathit{𝑞𝑔}+r$ with $q,r\in A[x]$ and either $r=0$ or $\mathrm{deg}r<\mathrm{deg}g$. It follows that $f(\beta )=r(\beta )$, and therefore that $f(\beta )$ is in the $A$-submodule generated by $\{1,\beta ,\dots ,{\beta }^{\mathrm{deg}g-1}\}$, so (ii) holds. Since this set is independent of $f$, it generates $A[\beta ]$ as an $A$-module, so (iii) holds. Suppose that (iii) holds. Then we may take $M=A[\beta ]$, which being free over itself has trivial annihilator.

Finally, suppose that (iv) holds. Let

be such that $\mathit{𝛽𝑀}\subseteq M$, and suppose without loss of generality that $\beta \ne 0$. We have

for some $a_{\mathit{𝑖𝑗}}\in A$ with $1\le i\le n$ and $1\le j\le n$. Consider $A$-module homomorphism $T:B^n\to B^n$ represented by $(a_{\mathit{𝑖𝑗}})$. The characteristic polynomial $f(x)\in A[x]$ of $T$ is monic, and $f(\beta )$ acts as zero on $M$. Since $M$ is a faithful $A[\beta ]$-module, we must have $f(\beta )=0$. Thus, $\beta$ is integral. □

Let $\{m_1,\dots ,m_n\}$ be a set of generators of $M$ as a $B$-module, and let $\{{\beta }_1,\dots ,{\beta }_k\}$ be a set of generators of $B$ as an $A$-module. We claim that $\{{\beta }_i{m}_j\mid 1\le i\le k,1\le j\le n\}$ is a set of generators of $M$ as an $A$-module. To see this, let $m\in M$ and write

with $b_j\in B$ for $1\le j\le n$. For $1\le j\le n$, we then write

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

as desired. □

Clearly, (i) implies (ii), so suppose that (ii) holds. By definition, each ${\beta }_i$ is then integral over any commutative ring containing $A$. By Proposition 1.2.4, each $A[{\beta }_1,\dots ,{\beta }_j]$ with $1\le j\le k$ is a finitely generated $A[{\beta }_1,\dots ,{\beta }_{j-1}]$-module, generated by $\{1,{\beta }_j,\dots ,{\beta }_j^{n_j}\}$ for some $n_j\ge 0$. Assuming recursively that $A[{\beta }_1,\dots ,{\beta }_{j-1}]$ is finitely generated as an $A$-module, Lemma 1.2.8 implies that $A[{\beta }_1,\dots ,{\beta }_j]=A[{\beta }_1,\dots ,{\beta }_{j-1}][{\beta }_j]$ is finitely generated as an $A$-module as well. Therefore, (iii) holds. Finally, if (iii) holds and $\beta \in B$, then since $\mathit{𝛽𝐵}\subseteq B$, the element $\beta$ is integral over $a$ by Proposition 1.2.4. Thus (i) holds. □

Let $\gamma \in C$, and let $f\in B[x]$ be a monic polynomial which has $\gamma$ as a root. Let $B^{\prime }$ be the subring of $B$ generated over $A$ by the coefficients of $f$, which is integral over $A$ as $B$ is. By Proposition 1.2.9, the ring $B^{\prime }$ is then finitely generated over $A$. As $B^{\prime }[\gamma ]$ is finitely generated over $B^{\prime }$ as well, we have $B^{\prime }[\gamma ]$ is finitely generated over $A$. Hence, $B[\gamma ]$ is itself an integral extension of $A$. By definition of an integral extension, the element $\gamma$ is integral over $A$. Since $\gamma \in C$ was arbitrary, we conclude that $C$ is integral over $A$. □

If $\alpha$ and $\beta$ are elements of $B$ that are integral over $A$, then $A[\alpha ,\beta ]$ is integral over $A$ by Proposition 1.2.9. Therefore, every element of $A[\alpha ,\beta ]$, including $\alpha +\beta$ and $\alpha \cdot \beta$, 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. □

Since $\beta \in L$ 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 $\overline{K}$ containing $K$ is integral over $K$. As every root of $f$ is a root of $g$, the same is true of the roots of $f$. Write $f={\prod }_{i=1}^n(x-{\beta }_i)$ for ${\beta }_i\in \overline{K}$ integral over $A$. As the integral closure of $A$ in $\overline{K}$ is a ring, it follows that every coefficient of $f$ is integral over $A$, being sums of products of the elements ${\beta }_i$. Since $f\in K[x]$ and $A$ is integrally closed, we then have $f\in A[x]$. □

Let $\beta \in L$ 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 $\beta$. Since $f$ divides $g$ in $K[x]$ and $A$ is a UFD with quotient field $K$, there exists $d\in K$ such that $\mathit{𝑑𝑓}\in A[x]$ and $\mathit{𝑑𝑓}$ 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 $\mathit{𝑑𝑓}$ will be divisible by $d$, so this forces $d$ to be a unit, in which case $f\in A[x]$. □

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 1.2.19 that $a$ is not integral over $A$. □

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

Any $\beta \in L$ is the root of a monic polynomial $f={\sum }_{i=0}^n{a}_i{x}^i\in K[x]$. Let $d\in A$ be such that $\mathit{𝑑𝑓}\in A[x]$. Then

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

Suppose that $\alpha =a+b\sqrt{d}$ is integral for $a,b\in \mathbb{Q}$. If $b=0$, then we must have $a\in \mathbb{Z}$. If $b\ne 0$, then the minimal polynomial of $\alpha$ is $f=x^2-2\mathit{𝑎𝑥}+a^2-b^{2}d$. Since $\alpha$ is integral, we must have $f\in \mathbb{Z}[x]$, so $2a\in \mathbb{Z}$. If $a\in \mathbb{Z}$, then since $a^2-b^{2}d\in \mathbb{Z}$ and $d$ is square-free, we have $b\in \mathbb{Z}$ as well. If $a\notin \mathbb{Z}$, then $2a=a^{\prime }$ and $2b=b^{\prime }$ for some odd $a^{\prime },b^{\prime }\in \mathbb{Z}$, and ${(a^{\prime })}^2\equiv {(b^{\prime })}^{2}dmod4$. As ${(\mathbb{Z}∕4\mathbb{Z})}^2=\{0,1\}$, this is impossible if $d\not\equiv 1mod4$. If $d\equiv 1mod4$, then $a+b\sqrt{d}$ lies in the claimed ring, since it contains $\sqrt{d}$, and clearly $(1+\sqrt{d})∕2$ is integral. □

We claim that the characteristic polynomial of the $K$-linear transformation $m_{\alpha }$ is $f^s$. First suppose that $L=K(\alpha )$. Note that $\{1,\alpha ,\dots ,{\alpha }^{d-1}\}$ forms a $K$-basis of $K(\alpha )$, and with respect to this basis, $m_{\alpha }$ is given by the matrix

where $a_i\in K$ for $1\le i\le d$ are such that

Expanding the determinant of $\mathit{𝑥𝐼}-A$ using its first row, we see that

where $A^{\prime }$is the $(1,1)$-minor of $A$. By induction on the dimension of $A$, we may assume that

so $\mathrm{char}{m}_{\alpha }=f$. Since

we have by expanding out the factorization of $f$ in $\overline{K}[x]$ that $N_{L∕K}\alpha$ and ${\mathrm{Tr}}_{L∕K}\alpha$ are as stated in this case.

In general, if $\{{\beta }_1,\dots ,{\beta }_s\}$ is a basis for $L∕K(\alpha )$, then $\{{\beta }_i{\alpha }^j\mid 1\le i\le s,0\le j\le d-1\}$is a basis for $L∕K$. The matrix of $m_{\alpha }$ with respect to this basis (with the lexicographical ordering on the pairs $(i,j)$) is the block diagonal matrix consisting of $s$ copies of $A$. In other words, $\mathrm{char}{m}_{\alpha }$ is the $f^s$, from which the result now follows easily. □

The distinct conjugates of $\alpha$ in a fixed algebraic closure $\overline{K}$ of $K$ are exactly the $\mathit{𝜏𝛼}$ for $\tau$ in the set $𝔗$ of distinct embeddings of $K(\alpha )$ in $\overline{K}$. These $\mathit{𝜏𝛼}$ are the distinct roots of the minimal polynomial of $\alpha$ over $K$, each occuring with multiplicity the degree ${[K(\alpha ):K]}_i$ of inseparability of $K(\alpha )∕K$. Now, each of these embeddings extends to ${[L:K(\alpha )]}_s$ distinct embeddings of $L$ into $\overline{K}$, and each extension $\sigma \in 𝔖$ of $\tau$ sends $\alpha$ to $\tau (\alpha )$. By Proposition 1.3.3, we have

and similarly for the trace. □

We prove this for norm maps. Let $𝔖$ denote the set of embeddings of $L$ into $\overline{K}$ that fix $K$, let $𝔗$ denote the set of embeddings of $M$ into $\overline{K}$ that fix $L$, and let $𝔘$ denote the set of embeddings of $M$ into $\overline{K}$ that fix $K$. Since ${[M:K]}_i={[M:L]}_i\cdot {[L:K]}_i$, it suffices by Proposition 1.3.4 to show that

We extend each $\sigma$ to an automorphism $\tilde{\sigma }$ of $\overline{K}$ fixing $K$. We then have

For the trace map, we simply replace the products by sums.

We claim that the subset $X=\{\tilde{\sigma }\circ \tau \mid \sigma \in 𝔖,\tau \in 𝔗\}$ of $𝔘$ is exactly $𝔘$, which will finish the proof. Let $\sigma ,{\sigma }^{\prime }\in 𝔖$ and $\tau ,{\tau }^{\prime }\in 𝔗$, and suppose that

Since $\tilde{\sigma }\circ \tau {|}_L=\sigma {|}_L$, we have that $\sigma ={\sigma }^{\prime }$. Since $\tilde{\sigma }$ is an automorphism, we then apply its inverse to (1.3.1) to obtain $\tau ={\tau }^{\prime }$. As there are then $|𝔖||𝔗|=|𝔘|$ elements of $X$, we have the result. □

Let $X$ be a set of linear characters $G\to K^{\times }$. Suppose by way of contradiction that $m\ge 2$ is minimal such that there $m$ distinct, linearly dependent elements of $G$. Choose $a_i\in K$ and ${\chi }_i\in X$ with $1\le i\le m$ for which $a_1\ne 0$ and

Also, let $h\in G$ be such that ${\chi }_1(h)\ne {\chi }_m(h)$. Set $b_i=a_i({\chi }_i(h)-{\chi }_m(h))$ for $1\le i\le m-1$. For any $g\in G$, we then have

Since $b_1\ne 0$ and ${\sum }_{i=1}^{m-1}{b}_i{\chi }_i$ has only $m-1$ terms, this contradicts the existence of $m$. □

Set $n=[L:K]$. Let $\beta \in L$, and note that

Next, suppose that $\alpha \in \ker N_{L∕K}$, and set

for $\gamma \in L$. The elements of $\mathrm{Gal}(L∕K)$, which is to say the powers of $\sigma$, are distinct $L$-valued characters on $L^{\times }$, and therefore they are $L$-linearly independent. Thus, there exists $\gamma \in L^{\times }$ such that $x_{\gamma }\ne 0$. We then note that

so ${\alpha }^{-1}=\sigma (x_{\gamma })x_{\gamma }^{-1}$, finishing the proof. □

Since $\beta$ is integral over $A$, so are all of its conjugates in an algebraic closure $\overline{L}$ of $L$, since they are also roots of the monic polynomial of which $\beta$ is a root. It follows from Proposition 1.3.4 and the fact that the integral closure of $A$ in $\overline{L}$ is a ring that $N_{L∕K}(\beta )$ and ${\mathrm{Tr}}_{L∕K}(\beta )$ are elements of $K$ integral over $A$, so $A$ is integrally closed. □

Suppose first that the $v_i$ form a basis of $V$. Let $A=(a_{\mathit{𝑖𝑗}})$ denote the matrix of $T$ with respect to the ordered basis $(v_1,\dots ,v_n)$. for each $i$. We may then write

As matrices, we then have

and the result follows as $\det T=\det A=\det A^T$.

If the $v_i$ do not form a basis of $V$, then there is an ordered basis $(e_1,\dots ,e_n)$ of $V$ and a linear transformation $U:V\to V$ of with $U(e_i)=v_i$ for all $i$. As $\det U=0$, we have