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Let $A$ and $B$ be commutative rings with identity. Let $$\varphi:A\rightarrow B$$ be a unit preserving homomorphism.

Prove or give a counterexample that if $\varphi$ is injective then $$\varphi:A_{\varphi^{-1}(\mathfrak{p})}\rightarrow B_{\mathfrak{p}},$$ the induced map in localization, is injective for all prime $\mathfrak{p}\subseteq B$.

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    Notice that localization is exact, so it preserves injectivity. However, here you localize w.r.t to different multiplicative subsets. – 2017-01-27
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    @Watson You got a reference or a proof for that, as you are just restating my question. – 2017-01-27
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    Atiyah and Macdonald, chap. 4, Prop. 33, page 39. – 2017-01-27
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    @Watson He is localising modules over a ring $A$ by $S\subseteq A$. Thus $S$ is fixed throughout his argument. The problem I am having here is one is localizing by $B-\mathfrak{p}$ and by $A-\varphi^{-1}(\mathfrak{p})$ and they may be quite different. – 2017-01-27
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    Yes, this is why I said "However" is my first comment. At least, if $A$ is a domain then it is true. – 2017-01-27
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    Ok sorry, why is it true if $A$ is a domain ? the problem is there may be $a\in A$ such that $b\varphi(a)=0$ but $b$ is not in the image of $\varphi$. – 2017-01-27
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    Sorry, I wanted to say "if $B$ is a domain". Here is why. Assume that $$\phi(a/s) = 0 \in S^{-1}B,$$ where $a \in A, s \in T := \phi^{-1}(S)$. Then $\dfrac{\phi(a)}{s} \sim \dfrac 0 1$, that is: $$\exists s' \in S, \; s'(\phi(a) \cdot 1 - s \cdot 0) = 0,$$ which is equivalent to $s' \phi(a)=0 \in S$. If $B$ is a domain then either $s' = 0 \in S$ (which not the case if $S = B \setminus \mathfrak p$, or $\phi(a)=0$ which implies $a=0$. – 2017-01-27
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    I think you mean $B$ is a domain. Yeah this other question gives a nice counterexample. So I think this question should be marked duplicate, or I could just delete it. – 2017-01-27

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