# Proposition 5.21 in Atiyah-MacDonald

There's just one step in this proof I can't see for the life of me.

Set up: We have a field K and an algebraically closed field $\Omega$. $(B, g)$ is maximal in the set $\Sigma$ of ordered pairs $(A, f)$ where $A$ is a subring of K and $f$ a homomorphism into $\Omega$, where $\Sigma$ has the partial order $(A, f) \leq (A', f')$ if $A$ is a subring of $A'$ and $f'|_{A} = f$. The overall claim is that $(B, g)$ is a valuation of $K$. We let $M$ be the unique maximal ideal of $B$ (which exists). We take $x \in K$ with $x \neq 0$ and may assume that $M[x]$ is not the unit ideal of $B' = B[x]$(by a lemma) and so is contained in some maximal ideal $M'$. Let $k = B/M$ and $k' = B'/M'$.

The claim I don't understand: Since $k' = k[\bar{x}]$ for $\bar{x}$ the image of x in k' (which I see), $\bar{x}$ is algebraic over k.

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This is the 1000th commutatve-algebra post! Woot! – mixedmath Jun 11 '12 at 20:37
Nevermind, $\bar{x}$ transcendental would give us an indeterminate in $k'$, which is a field since $M'$ isn't just $M[x]$ but is a maximal ideal containing it. Sorry. – Thank you Jun 11 '12 at 23:17

## 2 Answers

If your doubt is what I think it is, then remember: if $\,k/F\,$ is a fields extension and $\,w\in k\,$, then $\,F(w)=F[w]\Longleftrightarrow w\,$ is algebraic over $\,F\,$.

In you case, since $\,k'\,$ is a field, it then equals the polynomial ring in $\,k[\overline{x}]\,$ iff this element is algebraic over $\,k\,$

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Yes, it is what you think:+1 – Georges Elencwajg Jun 11 '12 at 20:52

Suppose $\bar x$ is not algebraic over $k$. Then $k[\bar x]$ is isomorphic to the polynomial ring $k[X]$. However $k[X]$ is not a field(any polynomial of degree $\geq 1$ in $k[X]$ is not invertible in $k[X]$). Hence $k[\bar x]$ is not a field. This is a contradiction. Hence $\bar x$ is algebraic over $k$.

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