# Ring of integers of a finite separable extension of a complete discrete valued field is a free module.

I'm trying to prove that

Suppose $K$ is complete with respect to a discrete valuation, and $L/K$ is finite and separable. Then $$\mathfrak{o}_L \cong \mathfrak{o}_K^{[L:K]}$$ as $\mathfrak{o}_K$-modules.

We know that $\mathcal{o}_L$ is finitely generated as an $\mathfrak{o}_K$ module. Then in my notes it says

As $\mathfrak{o}_K$ is a PID and $\mathfrak{o}_L \subset L$ is torsion free, $R\mathfrak{o}_L$ is free, necessarily of rank $n$.

Now, $R$ is a PID because it is a DVR. I don't however see why $L$ is torsion free? Wouldn't torsion free mean char$(K)=0$? I don't see why this is true. Nor do I see how you can conclude from these two facts that $\mathfrak{o}_L$ is free.

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## 1 Answer

Over a PID, a module is torsion-free if and only if it is flat. Now $\mathfrak o_L$ is flat over $\mathfrak o_K$ just because $\mathfrak o_L$ is an integral domain containing the Dedekind ring $\mathfrak o_K$. Now, over a PID, finitely generated $+$ torsion-free imply free.

I hope the following will be clear. I divide the argument in two steps (At the end, you just have to set $A=\mathfrak o_K$ and $B=\mathfrak o_L$).

Let $A$ be a noetherian domain with $K=\textrm{Frac}\,A$.

Step 1. If $L/K$ is finite separable and $A$ is integrally closed in $K$, then its integral closure $B$ in $L$ is finitely generated over $A$.

You already know this step! but let me explain briefly the strategy.

Let $n=[L:K]$. If $\omega_1,\dots,\omega_n$ is a $K$-basis of $L$ and $\omega'_1,\dots,\omega'_n$ is the dual basis, then there exists a nonzero $c\in A$ such that $c\omega'_i\in B$ for all $i=1,\dots,n$ (such a $c$ exists because $L/K$ is algebraic). So this says that

$$A(c\omega'_1)\oplus\dots\oplus A(c\omega'_n)\subseteq B.$$

Moreover, one has, after a small calculation, that $$B\subseteq A(c^{-1}\omega'_1)\oplus\dots\oplus A(c^{-1}\omega'_n).\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\star)$$

Now, the latter inclusion plus the noetherianity of $A$ tell us that $B$ is finitely generated over $A$.

Step 2. If, in addition, $A$ is a principal ring, then $B$ is free of rank $n$.

Indeed, we "surrounded" $B$ by two free $A$-modules of rank $n$. So, if we prove it's free we are done for its rank. The unique argument I know to say why $B$ is free is the one above: over a PID, finitely generated plus torsion-free together imply free. And the "torsion-free" part follows from the displayed equation $(\star)$.

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Can you do this without flatness? It is not a concept with which I am familar. – anne123 Oct 28 '12 at 17:30
Dear anne123, I modified the answer as you requested; I hope it will be clear. – Brenin Oct 28 '12 at 19:34