# The Stacks Project

## Tag 00IA

Lemma 10.49.2. Let $K$ be a field. Let $A \subset K$ be a local subring. Then there exists a valuation ring with fraction field $K$ dominating $A$.

Proof. We consider the collection of local subrings of $K$ as a partially ordered set using the relation of domination. Suppose that $\{A_i\}_{i \in I}$ is a totally ordered collection of local subrings of $K$. Then $B = \bigcup A_i$ is a local subring which dominates all of the $A_i$. Hence by Zorn's Lemma, it suffices to show that if $A \subset K$ is a local ring whose fraction field is not $K$, then there exists a local ring $B \subset K$, $B \not = A$ dominating $A$.

Pick $t \in K$ which is not in the fraction field of $A$. If $t$ is transcendental over $A$, then $A[t] \subset K$ and hence $A[t]_{(t, \mathfrak m)} \subset K$ is a local ring distinct from $A$ dominating $A$. Suppose $t$ is algebraic over $A$. Then for some $a \in A$ the element $at$ is integral over $A$. In this case the subring $A' \subset K$ generated by $A$ and $ta$ is finite over $A$. By Lemma 10.35.17 there exists a prime ideal $\mathfrak m' \subset A'$ lying over $\mathfrak m$. Then $A'_{\mathfrak m'}$ dominates $A$. If $A = A'_{\mathfrak m'}$, then $t$ is in the fraction field of $A$ which we assumed not to be the case. Thus $A \not = A'_{\mathfrak m'}$ as desired. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 11161–11166 (see updates for more information).

\begin{lemma}
\label{lemma-dominate}
Let $K$ be a field. Let $A \subset K$ be a local subring.
Then there exists a valuation ring with fraction field $K$
dominating $A$.
\end{lemma}

\begin{proof}
We consider the collection of local subrings
of $K$ as a partially ordered set using the relation of domination.
Suppose that $\{A_i\}_{i \in I}$ is a totally ordered
collection of local subrings of $K$. Then $B = \bigcup A_i$
is a local subring which dominates all of the $A_i$.
Hence by Zorn's Lemma, it suffices to show that if $A \subset K$
is a local ring whose fraction field is not $K$, then there
exists a local ring $B \subset K$, $B \not = A$ dominating $A$.

\medskip\noindent
Pick $t \in K$ which is not in the fraction field of $A$.
If $t$ is transcendental over $A$, then $A[t] \subset K$
and hence $A[t]_{(t, \mathfrak m)} \subset K$ is a local ring
distinct from $A$ dominating $A$. Suppose $t$ is algebraic over $A$.
Then for some $a \in A$ the element $at$ is integral over $A$.
In this case the subring $A' \subset K$ generated by $A$ and
$ta$ is finite over $A$.
By Lemma \ref{lemma-integral-overring-surjective} there exists
a prime ideal $\mathfrak m' \subset A'$ lying over
$\mathfrak m$. Then $A'_{\mathfrak m'}$ dominates
$A$. If $A = A'_{\mathfrak m'}$, then $t$
is in the fraction field of $A$ which we assumed not to be the case.
Thus $A \not = A'_{\mathfrak m'}$ as desired.
\end{proof}

Comment #1238 by Jonathan Wise on January 6, 2015 a 3:47 am UTC

This proof does not appear to be complete. In the last paragraph, the element $at$ may be in $A$, so $A' = A$ and does not contradict the maximality of $A$.

Comment #1251 by Johan (site) on January 6, 2015 a 11:52 pm UTC

If $A = A'_{\mathfrak m'}$ then $t$ is in the fraction field of $A$ which we assumed not to be the case. I have added this explanation to the proof, see here.

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