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Let $B$ be a subring of some field $K$, $x$ some element in $K$, $m$ a maximal ideal in $B$ and $m[x]$ the extension of $m$ in $B[x]$ and $M$ a maximal ideal in $B[x]$ such that $m[x] \subset M $ and $M \cap B = m$.

Why is $B[x]/M$ algebraic over $B/m$? Thank you.

Rudy the Reindeer
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1 Answers1

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The $(B/\mathfrak m)$- algebra $B[x]/M$ is finitely generated (by $\bar x$) and is a field.
Hence by Zariski's lemma it is finite-dimensional and a fortiori algebraic over $B/\mathfrak m$.

Community
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Georges Elencwajg
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  • I'm now doubtful of my answer, because it doesn't seem like I proved anything nearly as strong. Was my reasoning incorrect? – Zev Chonoles Jul 19 '12 at 18:16
  • Dear @Zev, I see a difficulty in your proof: it is not clear why the polynomial $f$ you obtain that kills $x+M$ is non-zero: for example its leading coefficient $a_n+\frak m$ would be zero if $a_n\in \mathfrak m$ . In other words: if (for example) $M=\mathfrak m[x]$ we are in trouble! – Georges Elencwajg Jul 19 '12 at 18:41
  • Ay, there's the rub. Thanks! – Zev Chonoles Jul 19 '12 at 18:46