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Let $A$ be a commutative ring, consider the map $Spec(A[[t]])\rightarrow Spec(A)$, does it have geometrically connected fibers?

If $A$ is noetherian, it is clear because one has for $k$ a residue field of $A$:

$A[[t]]\otimes_{A}k=k[[t]]$

But such equality does not hold true in general. So what can we expect?

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  • $\begingroup$ I don't understand your proof in the Noetherian case. Can you please write down in commutative algebra language what you want to prove? $\endgroup$ Dec 10, 2022 at 12:19
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    $\begingroup$ $\mathbb{Z}[[t]]\otimes _{\mathbb{Z}}\mathbb{Q}$ is a strict subring of $\mathbb{Q}[[t]]$. $\endgroup$
    – abx
    Dec 10, 2022 at 14:25
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    $\begingroup$ If $A$ is noetherian and $\mathfrak{p}$ is a prime ideal, first using the fact that taking tensor product commutes with formation of quotient, we are reduced to study the generic fiber for $A$ noetherian integral, then if $K=Frac(A)$, $K\otimes_{A}A[[t]]$ is a subring of $K[[t]]$, thus integral. $\endgroup$
    – prochet
    Dec 11, 2022 at 10:31

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