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?

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    $\begingroup$ Does it say generically reduced? Correct assumption is the generic fiber is reduced. $\endgroup$
    – Mohan
    Commented Sep 19, 2023 at 21:51
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    $\begingroup$ 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. $\endgroup$ Commented Sep 20, 2023 at 13:02


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