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Let $f,g:A \to B$ be two ring homomorphisms of noetherian rings satisfying that for any prime ideal $\mathfrak{q} \subset B$, $f^{-1}(\mathfrak{q})=g^{-1}(\mathfrak{q})=:\mathfrak{p}$ and the induced morphisms from $A/\mathfrak{p} \to B/\mathfrak{q}$ are the same. Is it true that $f=g$? If not true in general, is there any known additional conditions on $A$ or $B$, for which we have a positive answer to the question?

Any reference/hint will be most welcome.

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    $\begingroup$ This is not true in general: take $A=B=k[[t]]$, where $k$ is a field, $f=\mathrm{Id}$ and $g$ the automorphism given by $t\mapsto t+t^2h(t)$, for any $h(t)\in k[[t]]$. I have no idea on the second question. $\endgroup$
    – abx
    Commented Apr 27, 2016 at 6:29
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    $\begingroup$ @abx: the condition in the question is not satisfied at $\mathfrak{q}=0$. $\endgroup$
    – Laurent Moret-Bailly
    Commented Apr 27, 2016 at 6:41
  • $\begingroup$ Oops! Right, of course. $\endgroup$
    – abx
    Commented Apr 27, 2016 at 6:59

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The condition just says that for each prime $\mathfrak{q}$ of $B$, both composite maps $A\rightrightarrows B\to B/\mathfrak{q}$ are equal. Hence for each $x\in A$, $f(x)-g(x)$ belongs to every prime of $B$, i.e. is nilpotent.

The converse is clear, so your condition is equivalent to the equality of both composites $A\rightrightarrows B\to B_\mathrm{red}$. No noetherian assumption is necessary.

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