# A question about the proof that for an integral domain $D \subset K$ there is a valuation ring

I have some questions about a proof in Atiyah-Macdonald. It starts on page 65 where they write "...We want to prove that $B$ is a valuation ring of $K$."

Let me modify the claim a tiny bit and present it as follows: If $A$ is an integral domain with field of fractions $K$ then there exists a valuation ring $B$ of $K$ such that $A \subset B \subset K$. Furthermore, if $f: A \to \Omega$ is a ring homomorphism into an algebraically closed field $\Omega$, $f$ extends to a ring homomorphism $g: B \to \Omega$.

The way to proceed is to consider $\Sigma$, the set of all pairs $(B,f)$ where $B$ is a ring such that $A \subset B \subset K$ and $f: B \to \Omega$. Then one can partially order $\Sigma$ and apply Zorn's lemma to get a maximal element $(B,g)$. For this $B$ and this $g$ we can then show that $B$ is local with maximal ideal $m = \ker(g)$ and that for any $x$ in $K$, we either have $m[x] \neq B[x]$ or $m[x^{-1}] \neq B[x^{-1}]$. Then one can use these two claims to show that $B$ is a valuation ring of $K$, the field of fractions of $A$.

Proof

Let $x$ be any element in $K$. We want to show that either $x \in B$ or $x^{-1} \in B$. By the previously proved claim we may assume that $m[x] \subsetneq B[x]$ so that there exists a maximal ideal $M$ such that $m[x] \subset M \subsetneq B[x]$. Now we consider $i: B \hookrightarrow B[x]$ and observe that this map induces a homomorphism $i_m : B/m \hookrightarrow B[x]/M$, which is well-defined since $M \cap B \supset m$ and hence $i(m) \subset M$.

Next AM write "...induces an embedding of the field $k = B/m$ in the field $k^\prime = B[x]/M$; also $k^\prime = k[\overline{x}]$ where $\overline{x}$ is the image of $x$ in $k^\prime$, hence $\overline{x}$ is algebraic over $k$, and therefore $k^\prime$ is a finite algebraic extension of $k$."

(i) I'm not clear about the implication "hence $\overline{x}$ is algebraic over $k$". Isn't the correct reasoning like this:

First we observe that $B[x]/M$ is a field. Then we observe that it is also the same as the extension $(B/m)[\overline{x}]$. Since it's a field, $-\overline{x}$ is also in it and hence $\overline{x}$ is a root of the polynomial $p(x) = x - \overline{x}$ so that $\overline{x}$ is algebraic over $B/m$.

(ii) I have a second question about the end of this proof (which I don't quite understand):

"Now the homomorphism $g$ induces an embedding $\overline{g}$ of $k = B/m$ in $\Omega$, since by (5.19) $m$ is the kernel of $g$. Since $\Omega$ is algebraically closed, $\overline{g}$ can be extended to an embedding $\overline{g}^\prime$ of $k^\prime$ into $\Omega$...."

Maybe I'm just tired but I don't understand the last part. How does it follow from $\Omega$ being algebraically closed that we can extend $\overline{g}$? Isn't that what we want to show?

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It seems that you assume $\overline x \in B/m$ to show that it is algebraic over $B/m,$ but I think this is not the case. – Andrew Jul 17 '12 at 17:01
But if $k'$ is algebraic over $k,$ then it should be possible to consider $k'$ as a subfield of $\Omega$ by mapping generators of $k'/k$ to elements of $\Omega$ with the same minimal polynomial over $k.$ – Andrew Jul 17 '12 at 17:07
@Andrew Thank you. But I don't understand where I assume that $\bar{x}$ is in $B/m$. I'll have to think about it. – Rudy the Reindeer Jul 18 '12 at 6:28
To show that $\overline x$ is algebraic over $k,$ the minimal polynomial $p$ must have coefficients in $k.$ The minimal polynomial of $\overline x$ is $p(x)=x-\overline x$ iff $\overline x \in k.$ – Andrew Jul 18 '12 at 14:07
@Andrew Thank you! I see my mistake now. – Rudy the Reindeer Jul 19 '12 at 17:16