Let $A,B$ be commutative rings with identity and $f:A\to B$ a homomorphism of rings. For any prime ideal $\mathfrak q$ of $B$, denote $f^{-1}(\mathfrak q)$ by $\mathfrak p$, then $f$ induces the canonical homomorphism $g: A_{\mathfrak p}\to B_{\mathfrak q}$, do we have $(\ker f)_{\mathfrak p}\simeq \ker g$?
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By Eric Wofsey's answer, it is not true. https://math.stackexchange.com/questions/2997819/how-to-show-the-canonical-homomorphism-a-mathfrak-p-to-b-mathfrak-q-is-i – Born to be proud Nov 14 '18 at 07:12
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I wonder if jgon's answer is correct, can anybody help me check it? 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 07:15