Let $A$ be a finitely generated $k$-algebra ($k$ algebraically closed) of dimension one, integrally closed in its quotient field $K$. Let $R\subseteq K$ be a valuation ring. It's true that $A\subseteq R$ ?
Thank you!
Asked
Active
Viewed 356 times
1
-
@user26857 A must be a domain as it is contained in a field: its quotient field! – awllower Oct 05 '14 at 09:33
-
Right, maybe OP missed some words. – awllower Oct 05 '14 at 09:51
-
You're right, I forgot to write that A is a domain. – Andrea Oct 05 '14 at 19:14
2 Answers
3
By Krull-Akizuki theorem every valuation ring containing a Dedekind domain is a DVR. If take $R$ a valuation ring which is not a DVR you can conclude.
An explicit example: let $L$ be a field, $k=L(Y)$, and $A=k[X]$. Then $K=k(X)=L(X,Y)$ and now consider for $K$ a non-noetherian valuation ring.
1
No. Let $A=k[x]$ be the ring of univariate polynomials with coefficients in $k$. Then $K=k(x)$ is the field of rational functions, and $A$ is integrally closed in $K$. The valuation ring known as the "place at infinity" $$ R=\left\{\frac{p(x)}{q(x)}\in K\,\bigg\vert\,\deg p(x)\le \deg q(x)\right\} $$ is a DVR in $K$. Clearly it does not contain $A$.
Jyrki Lahtonen
- 140,891
-
3The definition Jyrki gives for $v_\infty$ requires $p\neq 0$. For $p=0$ the map $v_\infty$ is defined to be equal to $\infty$. Nothing to check here. Moreover: the valuation ring of $v_\infty$ does not contain $k[x]$ but $k[x^{-1}]$. The latter is a polynomial ring too, and $v_\infty$ ist just the $x^{-1}$-adic valuation from that point of view. – Hagen Knaf Oct 05 '14 at 14:57
-
@user26857: Sorry about creating confusion. I ignored the problem with $p=0$ as I was concentrating on the DVR as opposed to the valuation. Thanks to Hagen for clarifying. – Jyrki Lahtonen Oct 05 '14 at 15:53
-
1@user26857: Usually when extending the degree function to zero, you set $\deg 0 = -\infty$. – Oct 05 '14 at 19:53