2
$\begingroup$

Let $A$, $B$ rings with a morphism $f : A \to B$ and suppose that $B$ is integral over $A$. Let $\mathfrak{n} \subseteq B$ a maximal ideal, and $\mathfrak{m}$ its preimage under $f$ (so $\mathfrak{m}$ is maximal in $A$). The question is:

Is the induced ring homomorphism $A_{\mathfrak{m}} \to B_{\mathfrak{n}}$ always integral?

The answer will be "No", but of course my aim it to find a counterexample. Luckily, I have an hint:

Consider $A = k[X^2 - 1] \subseteq B = k[X]$

I see that in that case we have an integral A-algebra, but unfortunately I am finding some difficulties in proceeding. I need a maximal ideal of $B$ that does the job, so I tried with $(x)$. After having done a lot of computation, I am feeling a bit lost in my results...please, could you help me in any way? Also just saying "Yes, $(x)$ works, try to think better." or "Instead of $(x)$, try to consider the ideal $\mathfrak{b}$" will be appreciated comments.

Any suggestion is welcome :)

Cheers

$\endgroup$
0
0
$\begingroup$

For obtaining a simpler contraction I'd choose $\mathfrak n=(X-1)$. Then $\mathfrak m=\ ?$, and $\dfrac{1}{X+1}\in B_{\mathfrak n}$ is not integral over $A_{\mathfrak m}$. (Suppose it is and find that $X+1$ is invertible in $S^{-1}B$, where $S=A-\mathfrak m$, a contradiction.)

For a bigger picture, let's suppose we have a ring extension $A\subset B$ and $\mathfrak p$ a prime ideal of $A$. Set $S=A-\mathfrak p$. If $A\subset B$ is integral, then $S^{-1}A=A_{\mathfrak p}\subset S^{-1}B$ is also integral. (This last ring is usually denoted by $B_{\mathfrak p}$.) Now note that if $P$ is a prime ideal of $B$ lying over $\mathfrak p$ we have $S\cap P=\emptyset$, and moreover $(S^{-1}B)_{S^{-1}P}=B_P$. If $A_{\mathfrak p}\subset B_P$ is integral, then $B_{\mathfrak p}\subset B_P$ is also integral. But a ring of fractions is integral over the base ring iff the multiplicative set is made of units (why?). Thus $B_{\mathfrak p}\subset B_P$ is integral iff $S^{-1}B-S^{-1}P$ is made of units.

$\endgroup$
6
  • $\begingroup$ Thanks really a lot for the fantastic answer!!! I want to be sure to understand it entirely. Please, could you help me in some points? For example, the first is: why exactly is $\frac{1}{X+1} \in S^{-1}B$ not integral over $A_{\mathfrak{m}}$? $\endgroup$
    – user233650
    May 14 '15 at 19:28
  • $\begingroup$ @user233650 I simply can't, because as I said the ring extension $S^{-1}A\subset S^{-1}B$ is integral. But it seems you missed that $X+1$ is not in $A$, hence not in $S$. Instead it is in $B-\mathfrak n$. (I suggest you to write down an integral dependence relation for $\frac{1}{X+1}$ and then multiply it by $X+1$ to the right power.) $\endgroup$
    – user26857
    May 14 '15 at 20:23
  • $\begingroup$ sorry I am not sure to have understood, but I probably explained myself not so clearly. When you say "Suppose it is and find that $X+1$ is invertible in $S^{−1}B$, where $S= A− \mathfrak{m}$, a contradiction" - may you write down the details, please? $\endgroup$
    – user233650
    May 14 '15 at 20:52
  • $\begingroup$ @user233650 Please see the linked post. $\endgroup$
    – user26857
    May 14 '15 at 20:53
  • 1
    $\begingroup$ @user233650 Ok. I've also forgotten. But I won't delete this answer because I've added some general considerations which can show that almost never $A_n\subset B_m$ is integral, no matter if $A\subset B$ is integral. $\endgroup$
    – user26857
    May 14 '15 at 20:57