Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

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 $S^{-1}A$.

share|improve this answer
    
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.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.