Let $A,B$ be commutative rings with identity and $f:A\to B$ an injective homomorphism of rings. For any prime ideal $\mathfrak q$ of $B$, denote $f^{-1}(\mathfrak q)$ by $\mathfrak p$, how to show the canonical homomorphism $A_{\mathfrak p}\to B_{\mathfrak q}$ is injective?
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This seems like the kind of thing that might follow from the universal property. Did you try using that? – A. Thomas Yerger Nov 14 '18 at 05:50
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@AlfredYerger Yes, I factor it as $A_{\mathfrak p}\to B_{\mathfrak p}\to B_{\mathfrak q}$, the first is injective, but I wonder if the second is injective. – Born to be proud Nov 14 '18 at 05:57
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@AlfredYerger I wonder if jgon's answer is correct. https://math.stackexchange.com/questions/2997043/do-we-also-have-to-use-the-condition-that-f-is-an-immersion – Born to be proud Nov 14 '18 at 06:02
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Related – Viktor Vaughn Nov 14 '18 at 07:41
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This is false. For instance, let $k$ be a field, let $A=k[x]$ and $B=k[x,y]/(xy)$, and let $f:A\to B$ be the obvious inclusion. Then the ideal $\mathfrak{q}=xB\subset B$ is prime, and the induced map $A_{f^{-1}(\mathfrak{q})}\to B_\mathfrak{q}$ is not injective: $x$ is nonzero in $A_{f^{-1}(\mathfrak{q})}$ but is $0$ in $B_\mathfrak{q}$ since $xy=0$ and $y\not\in\mathfrak{q}$.
Eric Wofsey
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1I wonder if jgon's answer is correct, please help me check it, thank you. https://math.stackexchange.com/questions/2997043/do-we-also-have-to-use-the-condition-that-f-is-an-immersion – Born to be proud Nov 14 '18 at 06:15