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Let $f:X\rightarrow Y$ be a morphism of algebraic schemes (over an algebraically closed field with char 0). If $f$ is proper and the fibres are connected, then $X$ is connected whenever $Y$ is (see https://stacks.math.columbia.edu/tag/0377). Is the same result true if we replace "connected" by irreducible? The proof cannot be adapted to the irreducible case, I think.

Note that I do not assume that all the fibres have the same dimension.

Vector
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  • The answer at the linked duplicate exactly addresses your question - a proper map is closed, so the proof there using assumption #1 (and not necessarily assumption #2 about equidimensionality) applies. – KReiser Jul 29 '20 at 02:33
  • @KReiser but I think the proof uses that all fibres have the same dimension? And also is for varieties, not schemes – Vector Jul 29 '20 at 10:16
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    Ah, my apologies - I misread your edit. I've posted an answer. – KReiser Jul 29 '20 at 16:45

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This is not true. Consider $V(XY)\subset \Bbb P^2$ mapping to $\Bbb P^1$ by projection on to the $X$-axis. Then the target is irreducible and the fibers are all irreducible, but $V(XY)$ is clearly not irreducible.

If one assumes that the fibers are all of the same dimension, you'll have a chance: see this question.

KReiser
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  • It's been a long time, so I don't know if anyone will read this. In trying to disprove the statement, I thought exactly of the example you mentioned (or rather of the affine analogue), but I'm stuck trying to prove that the morphism is proper. Is there a simple way to see it? – abho Mar 27 '23 at 19:30
  • @abho Yes, see Stacks 01W6. – KReiser Mar 27 '23 at 19:55