Is the preimage of a nonreduced subscheme via a proper map nonreduced?

Question

Let $f\colon X \to Y$ a surjective proper map between smooth varieties over an algebraically closed field $k$ of characteristic zero. Let $Z\subset Y$ be a closed non-reduced subscheme. Is the preimage $f^{-1}(Z)$ nonreduced?

(The situation I am interested in is the resolution of an ideal sheaf, but I do not know if this hypothesis helps.)

Notes:

1- this is a refining of this question, asked by Shende and answered by Sawin. In the answer, $Z$ was supported on the singular locus of $Y$. In my question, I am adding smoothness and properness hypotheses.

2- The following I think is an important example. Let $f\colon X\to Y$ be the blow-up of a smooth point $p$ on a surface. Let $Z$ be isomorphic to $Spec(k[x]/x^2)$, supported at $p$, and $x$ goes to a tangent direction $v$. Then $f^{-1}(Z)$ is the exceptional divisor (with reduced scheme structure) union an embedded point at $v$, so it is indeed nonreduced. However, if you remove $v$, then the map is still surjective but $f^{-1}(Z)$ is reduced. So properness is important.

Answer 1

If $Z$ is not generically reduced, then its pullback is not reduced. (The argument can show that the pullback is not generically reduced, but not the way I wrote it.)

Lemma: To show that the pullback of $Z$ is not reduced, it suffices to check that there is a smooth curve $C$ mapping to $Y$ such that the pullback of $\mathcal I_Z$ and $\sqrt{\mathcal I_Z}$ to $C$ differ, where $\mathcal I_Z$ is the ideal sheaf of $Z$ and $\sqrt{\mathcal I_Z}$ is its radical.

Proof: By the valuative criterion, you can map some ramified cover of $C$ to $X$, lifting the map to $Y$, and then pulling back to this ramified cover shows that $f^* \sqrt{\mathcal I_Z}\neq f^* \mathcal I_Z$. Now because $f^* \mathcal I_Z$ contains a power of $f^* \sqrt{\mathcal I_Z}$, it follows that $f^* \mathcal I_Z$ is not radical and thus $f^* Z$ is not reduced.

Now let's check that if $Z$ is generically non-reduced, there exists such a $C$. To do this, work locally near the generic point of $Z$, so that the radical just becomes the maximal ideal at this point. Since $Z$ is not reduced, the map from the ideal of $Z$ to the Zariski cotangent space at the generic point (i.e. generators of this maximal ideal) is not surjective, so there exists a nonzero vector in the Zariski tangent space which is perpendicular to the image of the ideal of $Z$.

Pick a smooth curve $C$ whose tangent vector is that nonzero vector. The pullback of the maximal ideal to $C$ will have multiplicity $1$ while the pullback of $I_Z$ will have multiplicity $>1$, so they will be distinct.

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However, this approach (showing equality between $\mathcal I_Z$ and $\sqrt{\mathcal I_Z}$) will not work in general. Let $Z$ be the vanishing locus of $$x_1 x_3, x_2 x_4, (x_1 x_2 - x_2 x_3), (x_2 x_3 - x_3 x_4), (x_3 x_4 - x_1 x_4) $$

as well as $x_i x_j x_k$ for all triples $i,j,k$, not all equal. The point is that the induced reduced subscheme of $Z$ is the vanishing locus of $$x_1x_2,x_1x_3,x_1x_4, x_2x_3,x_2x_4, x_3,x_4$$ and $Z$ differs from this by an embedded point.

I claim the pullback of $Z$ and its induced reduced subscheme to the blowup of $\mathbb A^4$ at $0$ are equal (but neither is reduced since they each contain a double neighborhood of the exception divisor).

In a typical affine chart $x_1 =a_1 z, x_2 =a_2 z, x_3 =a_3 z, x_4 = z$, the pullback of the induced reduced is the vanishing locus of $z^2 (a_1,a_2,a_3)$ and the pullback of $Z$ is the vanishing locus of $$z^2 (a_1a_3, a_2, (a_1a_2-a_2a_3), (a_2a_3 - a_3), (a_3 - a_1)) $$ which equals $z^2(a_1,a_2,a_3)$ since we can cancel $a_2a_3$ with $a_2$ and get $a_3$ then cancel $a_3$ and get $a_1$.

By the symmetry rotating the four variables, all the affine charts look like this, so the pullbacks are equal as ideal sheaves.