I was having some trouble with the following exercise from Atiyah-Macdonald.

Let $A$ be a subring of $B$ such that $B$ is integral over $A$. Let $\mathfrak{n}$ be a maximal ideal of $B$ and let $\mathfrak{m}=\mathfrak{n} \cap A$ be the corresponding maximal ideal of $A$. Is $B_{\mathfrak{n}}$ integral over $A_{\mathfrak{m}}$?

The book gives a hint which serves as a counter-example. Consider the subring $k[x^{2}-1]$ of $k[x]$ where $k$ is a field, and let $\mathfrak{n}=(x-1)$. I am trying to show that $1/(x+1)$ could not be integral over $k[x^{2}-1]_{\mathfrak{n}^{c}}$.

I have understood why this situation serves as a counterexample. But I am essentially stuck at trying to draw a contradiction. A hint or any help would be great.


Maybe you already noticed that $\mathfrak n^c=(x^2-1)$. Now apply the definition of integrality and after clearing the denominators you get $\sum_{i=0}^n a_is_i(x+1)^{n-i}=0$ with $a_i\in A$, $a_n=1$, and $s_i\in A-\mathfrak n^c$. Then $x+1\mid s_n$ (in $B$), so $s_n\in (x+1)B\cap A=(x^2-1)$, a contradiction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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