ANT week 3.2

$\mathcal O_K$ is a free Z module and $\mathrm{rank}(\mathcal O_K)=\dim_\mathbb Q K$.Theorem Proposition. Let $K/\mathbb Q$ be a finite extension, then $\mathcal O_K$ is a free $\mathbb Z$ module and $\mathrm{rank}(\mathcal O_K)=\dim_{\mathbb Q} K$.

Here is the proof of the lemma in week 3.2, I will not use this lemma.

Lemma

Let $R$$R$ be a PID, and let

$$A\hookrightarrow B\hookrightarrow R^n$$

be inclusions of finitely generated, torsion-free $R$$R$–modules. If $\mathrm{rank}(A)=m$$\rank(A)=m$, then

$$m=\mathrm{rank}(A)\le \mathrm{rank}(B)\le n.$$

Proof. Over a PID, any finitely generated, torsion-free module is free, so the "dimension" of $A$$A$ and $B$$B$ refers to their ranks as free modules. Let $F=\mathrm{Frac}(R)$$F = \mathrm{Frac}(R)$ be the field of fractions of $R$$R$. Since $F$$F$ is flat over $R$$R$, the functor

$$-{\otimes }_{R}F:R\text{-Mod}⟶F\text{-Vect}$$

is exact and in particular preserves injections. Tensoring our chain of injections with $F$$F$ gives

$$F{\otimes }_{R}A\hookrightarrow F{\otimes }_{R}B\hookrightarrow F{\otimes }_R{R}^n.$$

But

$$F{\otimes }_{R}A\cong F^m,F{\otimes }_{R}B\cong F^{\mathrm{rank}(B)},F{\otimes }_R{R}^n\cong F^n,$$

so we obtain

$$F^m\hookrightarrow F^{\mathrm{rank}(B)}\hookrightarrow F^n⟹m\le \mathrm{rank}(B)\le n.$$

This completes the proof. $◻$$\square$

${\mathcal{O}}_K$$\mathcal O_K$ is a free Z module and $\mathrm{rank}({\mathcal{O}}_K)={\dim }_{\mathbb{Q}}K$$\mathrm{rank}(\mathcal O_K)=\dim_\mathbb Q K$.

Theorem

Let $A$$A$ be a (countable) Noetherian integral domain with fraction field $K$$K$, and let $L/K$$L/K$ be a finite separable extension of degree $n$$n$. Denote the integral closure

$$B=\{x\in L:x\text{ is integral over }A\}$$

of $A$$A$ in $L$$L$. Then $B$$B$ is finitely generated as an $A$$A$–module.

Proof. We proceed in six steps.

Step 1 (Choose a $K$$K$–basis and clear denominators). Since $[L:K]=n$$[L:K]=n$, pick any $K$$K$–basis

$$e_1,\dots ,e_n\subset L.$$

Each $e_i$$e_i$ is algebraic over the field $K$$K$, so its minimal polynomial has coefficients in $K$$K$. Clearing denominators in that polynomial shows there is a nonzero element $D_i\in A$$D_i\in A$ such that $D_i{e}_i$$D_i e_i$ is integral over $A$$A$. Taking

$$D=D_1{D}_2\cdots D_n\in A\setminus \{0\},$$

we then have

$${\omega }_i:=D{e}_i\in B,i=1,\dots ,n.$$

Since scaling by $D\ne 0$$D\neq0$ preserves linear independence over $K$$K$, $\{{\omega }_i\}$$\{\omega_i\}$ is again a $K$$K$–basis of $L$$L$.

Step 2 (Trace pairing and dual basis). Because $L/K$$L/K$ is separable, the trace form

$${\text{Tr}}_{L/K}:L\times L⟶K$$

defined by $\text{Tr}(x,y)={\text{Tr}}_{L/K}(xy)$$\text{Tr}(x,y)=\text{Tr}_{L/K}(xy)$ is a nondegenerate $K$$K$–bilinear form. Indeed, an inner product.

Hence there is a unique dual basis $\{{\omega }_1^{\ast },\dots ,{\omega }_n^{\ast }\}\subset L$$\{\omega_1^*,\dots,\omega_n^*\}\subset L$ characterized by

$${\text{Tr}}_{L/K}({\omega }_i^{\ast }{\omega }_j)={\delta }_{ij},1\le i,j\le n.$$

Step 3 (Build a free $A$$A$–module $M$$M$). Consider the $A$$A$–submodule of ${\text{Hom}}_K(L,K)$$\text{Hom}_K(L,K)$ given by

$$M=\underset{i=1}{\overset{n}{⨁}}A{\omega }_i^{\ast }.$$

Since $\{{\omega }_i^{\ast }\}$$\{\omega_i^*\}$ is a $K$$K$–basis of ${\text{Hom}}_K(L,K)$$\text{Hom}_K(L,K)$, we have $M\cong A^n$$M\cong A^n$. In particular, $M$$M$ is a free (hence Noetherian) $A$$A$–module of rank $n$$n$.

Step 4 (Define the trace‐embedding $\iota$$\iota$). Define

$$\iota :L⟶{\text{Hom}}_K(L,K),\iota (x)(y)={\text{Tr}}_{L/K}(xy).$$

Nondegeneracy of the trace form implies $\iota$$\iota$ is injective.

Step 5 (Show $\iota (B)\subseteq M$$\iota(B)\subseteq M$). Take any integral element $x\in B$$x\in B$. Then for each basis vector ${\omega }_j\in B$$\omega_j\in B$, the product $x{\omega }_j$$x\,\omega_j$ is again integral, so

$$\iota (x)({\omega }_j)={\text{Tr}}_{L/K}(x{\omega }_j)\in A.$$

But writing $\iota (x)$$\iota(x)$ in the dual‐basis expansion

$$\iota (x)=\sum _{i=1}^n(\iota (x)({\omega }_i)){\omega }_i^{\ast }$$

we see that all the coefficients $\iota (x)({\omega }_i)$$\iota(x)(\omega_i)$ lie in $A$$A$. By definition of $M$$M$, this exactly says

$$\iota (x)\in M.$$

Since this holds for every $x\in B$$x\in B$, we conclude

$$\iota (B)\subseteq M,$$

so $B$$B$ is isomorphic (via $\iota$$\iota$) to an $A$$A$–submodule of the Noetherian module $M$$M$. Hence $B$$B$ is a Notherian Module.

Step 6 (Exhaustion by a chain and apply ACC). Because $B\subset L$$B\subset L$ is at most countable, enumerate its elements $\{b_1,b_2,b_3,\dots \}$$\{b_1,b_2,b_3,\dots\}$. For each $k\in \mathbb{N}$$k\in\mathbb{N}$, let

$$M_k=A[b_1,b_2,\dots ,b_k]\subseteq B\subseteq M.$$

Each $M_k$$M_k$ is finitely generated over $A$$A$ (adjoin one integral element at a time). We obtain an ascending chain of $A$$A$–submodules of the Noetherian module $M$$M$:

$$M_1\subseteq M_2\subseteq \cdots \subseteq \bigcup _{k=1}^{\mathrm{\infty }}{M}_k=B.$$

By the ACC property on $M$$M$, there is some $N$$N$ so that

$$M_N=M_{N+1}=\cdots .$$

Hence

$$B=\bigcup _{k=1}^{\mathrm{\infty }}{M}_k=M_N$$

is finitely generated as an $A$$A$–module. This completes the proof.

Proposition. Let $K/\mathbb{Q}$$K/\mathbb Q$ be a finite extension, then ${\mathcal{O}}_K$$\mathcal O_K$ is a free $\mathbb{Z}$$\mathbb Z$ module and $\mathrm{rank}({\mathcal{O}}_K)={\dim }_{\mathbb{Q}}K$$\mathrm{rank}(\mathcal O_K)=\dim_{\mathbb Q} K$.

Proof. Since $\mathbb{Z}$$\mathbb Z$ is a PID, hence a Noetherian ring, hence ${\mathcal{O}}_K$$\mathcal O_K$ is finite generated by lemma 1. Also, ${\mathcal{O}}_K$$\mathcal O_K$ is torsion free, hence it is free module by the structure theorem. Hence ${\mathcal{O}}_K\cong {\mathbb{Z}}^{\mathrm{rank}({\mathcal{O}}_K)}$$\mathcal O_K\cong\mathbb Z^{\mathrm{rank}(\mathcal O_K)}$. Now consider ${\mathcal{O}}_K{\otimes }_{\mathbb{Z}}\mathbb{Q}\cong K\cong {\mathbb{Q}}^{\mathrm{rank}({\mathcal{O}}_K)}$$\mathcal{O}_K\otimes_{\mathbb Z}\mathbb Q\cong K\cong\mathbb Q^{\mathrm{rank}(\mathcal O_K)}$ as $\mathbb{Q}$$\mathbb Q$ vector space.

Thus we have ${\dim }_{\mathbb{Q}}K=\mathrm{rank}({\mathcal{O}}_K)$$\dim_\mathbb Q K=\mathrm{rank}(\mathcal O_K)$. $◻$$\square$