To show a certain integral closure is an order.

Question

I have a Dedekind domain $R$ with field of fractions $K$ and a finite separable field extension $L$ of $K$. Let $S$ be the closure of $R$ in $L$.

Is there a quick way to show that $S$ is finitely generated over $R$ and that $K.S=L$?

For the second part do I need to show that $S$ is torsion free over $R$?

I'll really appreciate either a proof or a reference to a proof in the literature as I have so far been unable to find one.

http://math.stackexchange.com/questions/50332/the-integral-closure-of-a-finite-separable-field-extension-of-the-fraction-field — user26857, Jun 11 '14 at 20:19

score 2 · Accepted Answer · answered Jun 11 '14 at 20:18

It's almost immediate that $KS=L$. Let $x\in L$. Then $x$ is algebraic over $K$, and therefore there exists $a\in R-\{0\}$ (for instance, a common denominator of the coefficients of the minimal polynomial of $x$ over $K$), such that $ax$ is integral over $R$, so $ax\in S$, and thus $x\in KS$.

To show a certain integral closure is an order.

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