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

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.

• I think that a morphism between smooth varieties is a local complete intersection morphism. So $f^{-1}$ can be constructed as the usual pull-back of cycles, and a pullback of a non-reduced cycle is non-reduced. – Francesco Polizzi May 4 '20 at 13:56
• @FrancescoPolizzi pullback of cycles along a (not-flat) lci map is not just the scheme-theoretic pullback. The construction passes through specialization to the normal cone. – user147129 May 4 '20 at 14:36
• @RizaHawkeye: oh right, I had in mind the flat case. Does the specialization in the non-flat case preserve non-reducedness of cycles? – Francesco Polizzi May 4 '20 at 14:52
• @FrancescoPolizzi moreover, the cycle associated with a subscheme does not see embedded points. – R. van Dobben de Bruyn May 4 '20 at 19:52
• @Giulio: You probably mean $f^{-1}(Z)$ is reduced at the end. – Jérôme Poineau May 6 '20 at 4:32

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.

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.