# Is this property of flat morphisms accurate?

In Lazarsfeld's Positivity 1, pg. 246, the following is stated:

If $$f: X \to Y$$ is a flat mapping of schemes, with $$Y$$ integral and $$X$$ generically reduced, then $$X$$ must be everywhere reduced.

This seems wrong to me. For example, take $$Y = \operatorname{Spec} k$$ and $$X$$ any generically reduced, but non-reduced, $$k$$-scheme. (For example, introduce nilpotents to a closed point on a positive dimensional integral $$k$$ scheme. )

However, this is true if the map is assumed to be finite, and $$X$$ is assumed to be irreducible. A justification is recorded below.

Let's look locally so that $$f$$ is induced by $$A \hookrightarrow B$$, a flat finite extension of rings. We claim there is some $$\delta \in A$$ so that $$B_\delta$$ is reduced. To see this, we just need to show that if $$V \subset \operatorname{Spec} B$$ is the reduced locus, there is some nonempty open subset $$U$$ of $$\operatorname{Spec} A$$ so that $$f^{-1}(U) \subset V$$. To this end, let $$K = \operatorname{Spec} B - V$$. Since $$\operatorname{Spec} B$$ is irreducible, this is strictly of lower dimension than $$B$$. Hence, $$f(K) \subset \operatorname{Spec} A$$ is a strict closed subset, so its complement is the set $$U$$ we're looking for.

Once we have this $$\delta \in A$$ so that $$B_{\delta}$$ is reduced, an argument in the book (pg. 246) shows that $$B$$ is reduced.

Is the linked claim as false as I suspect it is? Also, is there a way to prove this when $$X$$ is not irreducible, or are there counterexamples in this case too?

• Does it say generically reduced? Correct assumption is the generic fiber is reduced. Commented Sep 19, 2023 at 21:51
• I agree with Mohan. See Prop. 4.3.8, p. 137 in Liu's book "Algebraic Geometry and Arithmetic Curves" for a proof of Lazarsfeld's statement. Commented Sep 20, 2023 at 13:02