Re: Irreducible constant dimensional fibres --> irreducibility?

markzorromz@xxxxxxxxx in litteris
<b54924bf-832c-483d-ac93-6dc7d8f691a8@xxxxxxxxxxxxxxxxxxxxxxxxxxx>
scripsit:
Let X, Y be affine varieties over an algebraically closed field of
characteristic zero. Assume Y is irreducible. Let f: X -->Y be a
surjective regular morphism with the property that the fibre over each
point of Y is irreducible, and of the same dimension. I was wondering
if:

1) This implies that X is irreducible.

No. Let Y be the affine line, let X be the disjoint union of {0} in Y
and its (open) complement. Then the obvious map f from X to Y is a
monomorphism whose fiber over every point y of Y (closed or not) is a
single point (in fact, f is an isomorphism over each point), yet Y is
irreducible and X is not.

--
David A. Madore
(david.madore@xxxxxx,
http://www.madore.org/~david/ )

.

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