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I'm trying to understand a proof that $f:A\to B$ is integral implies $S^{-1}f:S^{-1}A\to S^{-1}B$ is integral. Here $S$ is a multiplicative subset of $A$.

Take $\alpha\in B$, since $B$ is integral over $A$, then there is a relation $$ \alpha^n+a_{n-1}\alpha^{n-1}+\cdots+a_1\alpha+a_0=0 $$ for $a_i\in A$, by writing $a\beta$ for $f(a)\beta$ for $a\in A$ and $\beta\in B$.

Canonically projecting this into $S^{-1}A$ and $S^{-1}B$ yields $$ (\alpha/1)^n+[a_{n-1}/1](\alpha/1)^{n-1}+\cdots+[a_1/1](\alpha/1)+[a_0/1]=0/1 $$ so $\alpha/1\in S^{-1}B$ is integral over $S^{-1}A$. I don't see how this proves all of $S^{-1}B$ is integral over $S^{-1}A$, it seems to only prove that the canonical image of $B$ in $S^{-1}B$ is integral.

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up vote 1 down vote accepted

$\bf Hint:$ Divide the equation $\alpha^n+a_{n-1}\alpha^{n-1}+...+a_1\alpha^1+a_0$ by $s^n$, where $s$ is any element of your multiplicative set, and rearrange to produce an expression for $\frac{\alpha^n}{s^n}$ with coefficients in $S^{-1}A$.

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Thanks azarel, I see how this works. – Waldott Jan 22 '12 at 0:52

Azarel's answer does it. However, you could note more generally that $B$ generates $S^{-1}B$ as an algebra over $S^{-1} A$, and then use the general fact (applied with $S^{-1} A$ in place of $A$ and $S^{-1} B$ and in place of $B$) that if $B$ is an $A$-algebra which is generated (as an $A$-algebra) by a set of elements that are integral over $B$, then every element of $B$ is integral over $A$.

Can you prove this general fact? (Azarel's answer describes how to prove the special case you need for your particular question, which is slightly easier than the general fact.)

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