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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.

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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 $B/\mathfrak m$.

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    Ay, there's the r$u$b. Th$a$nks!2012-07-19