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Let $R$ be a DVR with a uniformizer $\pi$ and let $M$ be a finitely generated flat $R$-algebra. Assuming that $M\otimes R/\pi R$ is a connected ring, is $M$ connected as well? My geometric intution tells me this should be true but I do not know how to actually prove it. An analogous statement with connectedness replaced by irreducibility is not true (consider $k[[t]][x]/(x^2-t^2)$ for an algebraically closed field of characteristic 0 $k$).

  • It seems to me that this does not need so many hypotheses: if $M$ is not connected, then $M=A\times B$, so $M\otimes_R R/I\simeq A\otimes_R R/I \times B\otimes_R R/I$. This does not need $M$ to be flat, or $R$ to be a DVR or $I$ to be maximal. Did I miss something? – Captain Lama Dec 11 '19 at 00:49
  • @CaptainLama consider $\mathbb{F}_p\times \mathbb{Q}_p$ as a $\mathbb{Z}_p$-algebra. –  Dec 11 '19 at 00:55
  • Ok, I see what you mean, one of the quotient factors might actually be $0$. – Captain Lama Dec 11 '19 at 01:02
  • Does Nakayama not help? If $M = A \times B$, then $A \otimes_{R} R/\pi R \cong A/(\pi R) A = 0$ means that $A = (\pi R) A$, so $A = 0$ by Nakayama. (In case I am making a stupid mistake somehow, another stray comment: since $R$ is a PID, the condition that $M$ be flat means that $M$ is free, in fact, hence faithfully flat as an $R$-module.) – Alex Wertheim Dec 11 '19 at 02:43
  • @AlexWertheim: $M$ is finitely generated as an algebra, not as a module. – Eric Wofsey Dec 11 '19 at 03:54
  • @EricWofsey: oh, thanks! I misunderstood. – Alex Wertheim Dec 11 '19 at 04:02

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No: for instance, if $M=R[\pi^{-1}]\times R$ then $M\otimes R/\pi R\cong R/\pi R$ is connected but $M$ is not connected.

Eric Wofsey
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