5
$\begingroup$

Let $R\subset T$ be two commutative rings, and $T$ is integral over $R$. Let $\mathfrak m\in \operatorname{Max} R,\mathfrak n\in\operatorname{Max}T$ such that $\mathfrak m=\mathfrak n\cap R$. Show that $R_\mathfrak m\subset T_\mathfrak n$, i.e. the canonical map $\{\frac ru\mid r\in R,u\in R\setminus\mathfrak m\}\to\{\frac tv\mid t\in T,v\in T\setminus\mathfrak n\}$ must be an injection.

I feel confused because the exercise 4 from Chapter 5 in the book of Atiyah and Macdonald just assume that the localization is a ring extension. And I don't know if the condition "integral over" is necessary.

$\endgroup$
  • $\begingroup$ In exercise 4 from Chapter 5 the authors don't assume that $A_\mathfrak m\subset B_\mathfrak n$. They ask if $B_\mathfrak n$ is integral over $A_\mathfrak m$, that is, if the homomorphism $A_\mathfrak m\to B_\mathfrak n$ is integral. $\endgroup$ – user26857 May 15 '16 at 23:15
  • $\begingroup$ But your question makes sense independent of that exercise. $\endgroup$ – user26857 May 15 '16 at 23:16
  • $\begingroup$ @user26857 But the definition say that a ring "integral over" imply the ring extension $\endgroup$ – yaoliding May 17 '16 at 12:45
  • $\begingroup$ Which definition? See commalg.subwiki.org/wiki/Integral_morphism $\endgroup$ – user26857 May 17 '16 at 12:51
  • $\begingroup$ It seems that you are referring to the definition of integral dependence in Atiyah and Macdonald. Well, maybe they missed the general case or not been interested in it. However, for integral domains (as it is the case of their hint) we have $R_{\mathfrak m}\subset T_{\mathfrak n}$. $\endgroup$ – user26857 May 17 '16 at 14:16
15
$\begingroup$

Not necessarily!

Let $R=\mathbb Z_4$ be the ring of residue classes modulo $4$, and $T=\mathbb Z_4[X]/(X^2+X,2X)$. Denote by $x$ the residue class of $X$ modulo the ideal $(X^2+X,2X)$. Then $T=R[x]$.

We have that $R\subset T$ is an integral ring extension

since $x$ is integral over $R$.

The ideal $\mathfrak n=(2,x+1)$ is maximal in $T$ since $T/\mathfrak n\simeq\mathbb Z/2\mathbb Z$. Moreover, $\mathfrak n\cap R=\mathfrak m$, where $\mathfrak m=2\mathbb Z_4$.

Now we show that the map $R_{\mathfrak m}\to T_{\mathfrak n}$ is not injective.

In order to see this note that $\frac 21\ne\frac 01$ in $R_{\mathfrak m}$, but $\frac 21=\frac 01$ in $T_{\mathfrak n}$ since $2x=0$, and $x\in T\setminus\mathfrak n$.

$\endgroup$

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.