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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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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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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 the accepted 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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The question is right,isn't it? Actually,see an exercise (2.18b) of chap.2 of algebraic geometry by Hartshorne.Given a ring homomorphism $f:A \to B$,let $g:SpecB \to SpecA$ be the induced morphism.Then f is injective iff the map of sheaves$g^\sharp:O_{SpecA} \to g_*O_{SpecB}$ is injective.Note that $g^\sharp $ is injective iff for any $ a \in A$,$ g^ \sharp(D(a)):O_{SpecA}(D(a)) \to g_*O_{SpecB}(D(a))$ is injective i.e. for any $ a \in A$,$ g^ \sharp_a:A_a \to B_{f(a)}$ is injective .So the question becomes f is injective iff $g^\sharp_a$ is injective for any $a\in A$.The proof as follows: "if"part,assume $g^\sharp $ is injective,taking global section (note that taking global functor is left exact) we have $f:A \to B $ is injective."only if"part,if $f:A \to B $ is injective ,and $g^\sharp_a(c/a^n)=f(c)/f(a^n)=0 \in B_{f(a)}$,then there exists some intrger $m$ such that $ f(c)f(a)^m=0$ which implies $ f(ca^m)=0 $,since f is injective, $ ca^m=0$, so $c/a^n=0$.Q.E.D.By this conclusion, a ring homomorphism f is injective iff for any $p\in SpecB$,the induced map $g^\sharp_p:O_{SpecA,f^{-1}(p)} \to O_{SpecB,p}$ is injective.In the language of category,this fact states that since the category of affine schemes is equivalent to the opposite category of the category of commutative rings with identity,so injectivity of ring homomorphisms is equivalent to injectivity of morphisms of affine schemes,note that injectivity of morphisms of sheaves is equivalent to injectivity of morphisms of sheaves on stalks,hence the result is not strange.

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  • $\begingroup$ It is true that the map $g^\sharp\colon \mathcal O_{{\rm Spec}(A)} \rightarrow g_*\mathcal O_{{\rm Spec}(B)}$ is injective. But this does not mean that $g^\flat\colon g^{-1}\mathcal O_{{\rm Spec}(A)} \rightarrow \mathcal O_{{\rm Spec}(B)}$ is injective. Notice that one has $g^\flat_{\mathfrak p}\colon \mathcal O_{{\rm Spec}(A), f^{-1}(\mathfrak p)} \rightarrow \mathcal O_{{\rm Spec}(B), \mathfrak p}$ for all $\mathfrak p\in {\rm Spec}(B)$, but computing the map on stalks from $g^\sharp$ is much more difficult! $\endgroup$ – Claudius Aug 7 '17 at 14:58
  • $\begingroup$ @Claudius,Actually,I don't understand what's your meaning.First,the map $g^\sharp_p$ from $O_{SpecA,f^{-1}(p)}$ to $O_{SpecB,p}$ for any $p \in SpecB$,ringht?Second,what you mean states that even though $g_*$ and $g^{-1}$ are adjont,the injectivity is not equivalent,right?,Finally,I don't use the $g^b$ at all in my argument.Certainly,I have wrong understanding about this question,and could you explain the question explictly? I am eager to know the right understanding,thank you in advanced! $\endgroup$ – Jiabin Du Aug 8 '17 at 14:16
  • $\begingroup$ What I mean is, you have the map $g^\sharp\colon \mathcal O_{{\rm Spec}(A)} \rightarrow g_*\mathcal O_{{\rm Spec}(B)}$ but this does not give you a map $g_{\mathfrak p}^\sharp\colon \mathcal O_{{\rm Spec}(A), f^{-1}(\mathfrak p)} \rightarrow \mathcal O_{{\rm Spec}(B),\mathfrak p}$. In fact, $g^\sharp$ is a map of sheaves on ${\rm Spec}(A)$; given $\mathfrak p\in {\rm Spec}(B)$, then in general $(g_*\mathcal O_{{\rm Spec}(B)})_{f^{-1}(\mathfrak p)} \neq \mathcal O_{{\rm Spec}(B),\mathfrak p}$. $\endgroup$ – Claudius Aug 9 '17 at 3:58
  • $\begingroup$ To elaborate on my previous comment, let’s take the example in the accepted answer: There we have $\mathfrak p = xB$ and $f^{-1}(\mathfrak p) = XA$ (which is the origin of the line ${\rm Spec}(A)$). Then $\mathcal O_{{\rm Spec}(B), \mathfrak p}$ is isomorphic to $K[Y,Y^{-1}]$. On the other hand, $(g_*\mathcal O_{{\rm Spec}(B)})_{f^{-1}(\mathfrak p)}$ equals $S^{-1}B$, where $S = \{h(X)\in k[X] \mid h(X)\notin (X)\} \subseteq B$. Notice that $y$ is not invertible in $S^{-1}B$ (because $y$ does not divide any element in $S$). Therefore, $S^{-1}B$ and $K[Y,Y^{-1}]$ are really different rings. $\endgroup$ – Claudius Aug 9 '17 at 8:40
  • $\begingroup$ @Claudius First,the map $g^\sharp:O_{SpecA} \to g_*O_{SpecB}$ induces exactly a map on stalk,that is, $g^\sharp_p:O_{SpecA,g(p)} \to O_{SpecB,p}$,(note that $g(p)=f^{-1}(p)$),about the fact, you can refer to any algebraic geometry book,such as Hartshorne's (Page72)or Qing Liu's (page37) book.It's easy to check (the stalk of a sheaf at one point is just taking direct limit over neighborhoods of that point).Yeah,$g^\sharp$ is a map of sheaves on $ SpecA $,if you choose a point $ p\in SpecA $,note that $f(p)$ is not necessarily a point in $SpecB$,in fact after extension just get an ideal of B. $\endgroup$ – Jiabin Du Aug 9 '17 at 9:35

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