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Let $\alpha:A\to B$ be a ring homomorphism, $Q\subset B$ a prime ideal, $P=\alpha^{-1}(Q)\subset A$ a prime ideal. Consider the natural map $\alpha_Q:A_P\to B_Q$ defined by $\alpha_Q(a/b)=\alpha(a)/\alpha(b)$. Suppose that $\alpha$ is injective. Then is $\alpha_Q$ always injective?

I think so, but I'm clearly being too dense to prove it! My argument goes as follows.

Let $\alpha(a)/\alpha(b)=0$. Then $\exists c \in B\setminus Q$ s.t. $c\alpha(a)=0$. If $B$ is a domain we are done. If not we must exhibit some $d\in A\setminus P$ s.t. $da=0$. Obviously this is true if $c =\alpha(d)$. But I don't see how I have any information to prove this!

Am I wrong and this is actually false? If so could someone show me the trivial counterexample I must be missing?

Many thanks!

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I wish someone would provide us a counterexample. I tried to find it, but failed. – Makoto Kato Oct 30 '12 at 0:29
up vote 11 down vote accepted

Take $A=K[X]$, $B=K[X,Y]/(XY)$ and $\alpha$ the following application $$A=K[X]\subset K[X,Y]\rightarrow K[X,Y]/(XY)=B.$$ Obviously $\alpha$ is injective. Write $B=K[x,y]$ with $xy=0$. Let $Q=xB$. It is obvious that $Q$ is prime ($B/Q\cong K[Y]$) and $P=\alpha^{-1}(Q)=XA$. Now choose $\frac{X}{1}\in A_P$ and observe that $\alpha(\frac{X}{1})=\frac{x}{1}$. But $\frac{x}{1}=\frac{0}{1}$ in $B_Q$ because $yx=0$ and $y\in B-Q$.

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That's a great counterexample - thanks! – Edward Hughes Oct 30 '12 at 8:42

Since algebraic geometry is one of the tags, let me give a geometric account of the problem: one is given Spec $B \to $ Spec $A$ dominant, and one wants to show that this isn't necessarily dominant in a n.h. of some point $Q$ of Spec $B$. The basic way to achieve this is to arrange Spec $B$ to be the union of two components, one which maps dominantly to Spec $A$ and one which doesn't, and take $Q$ to be a point lying (only) on the component that doesn't map dominantly. This is what happens in YACP's answer: Spec $B$ is two lines crossing, Spec $A$ is a single line, and the map is the identity on the first line and constant on the second line. The point $Q$ is then taken to be the generic point of the second line.

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