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](https://mathoverflow.net/questions/137345/is-the-preimage-of-a-nonreduced-subscheme-nonreduced) 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.



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


