1
$\begingroup$

Let $(K,|\cdot|)$ be a complete valued field and let $L$ be a field extension with $[L:K]<\infty$. Let $\mathcal{O}_K$ be the valuation ring in $K$ and let $\mathcal{O}_L$ be the integral closure of $\mathcal{O}_K$ in $L$.

I am reading a theorem in which it is proved that $|\cdot|_K$ extends uniquely to $L$ and that $\mathcal{O}_L$ is the valuation ring in $L$ corresponding to this valuation. In the proof it is implicitely assumed that the fraction field of $\mathcal{O}_L$ is equal to $L$, but I have difficulties deducing this. Could someone lend me a hand with a tip?

Thank you in advance.

$\endgroup$
1
  • $\begingroup$ For any integral domain $A$, $K$ its field of fractions, and $L$ an algebraic extension of $K$ we have that $L$ is the field of fractions of the integral closure of $A$ in $L$. $\endgroup$
    – user26857
    Commented Jan 3, 2015 at 14:19

1 Answer 1

3
$\begingroup$

The proof that the field of fractions of $\mathcal{O}_L$ is $L$ is just a matter of clearing denominators. Let $x\in L$. Then $x$ satisfies a monic irreducible polynomial with coefficients in $K$:

$$x^n+c_{n-1}x^{n-1} + \cdots +c_0=0, c_i \in K$$

Now in a valuation field $K$, the fraction field of the valuation ring $\mathcal{O}_K$ is the field $K$. Thus, each $c_i = \frac{a_i}{b_i}$ with $a_i, b_i \in \mathcal{O}_K$. We may find a common denominator $e \in \mathcal{O}_K$ for all these coefficients, so that for some $d_i \in \mathcal{O}_K$ the coefficients are now $\frac{d_i}{e}$. Now multiplying our polynomial by $e^n$ gives

$$(ex)^n +d_{n-1}(ex)^{n-1} + \cdots +e^{n-2}d_1(ex) + e^{n-1}d_0=0.$$

Thus, $ex$ satisfies a monic polynomial in $\mathcal{O}_K$, so $ex \in \mathcal{O}_L$. This shows that $x \in \text{Frac}(\mathcal{O}_L)$, since $x=\frac{ex}{e}$ with $ex, e \in \mathcal{O}_L$.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .