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Let $\mathcal{E}$ be a coherent sheaf on an irreducible scheme $S$ ($S$ can be pretty nice, say noetherian of finite type), and let $\mathbf{P}(\mathcal{E}):=\mathrm{Proj}(\mathrm{Sym}(\mathcal{E}))$ be the projective bundle associated to $\mathcal{E}$. Is it true that if the structural map $\mathbf{P}(\mathcal{E})\to S$ is dominant (or even surjective), then $\mathbf{P}(\mathcal{E})$ is irreducible?

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2 Answers 2

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This can fail when $\mathscr E$ has torsion:

Example. Let $k$ be a field, let $R = k[t]$ with maximal ideal $\mathfrak m = (t)$, and let $S = \operatorname{Spec} R$. Take $\mathscr E = R\oplus R/\mathfrak m = Rx \oplus Ry/(ty)$. Then $$\operatorname{Sym}^*(\mathscr E) = R[x,y]/(ty),$$ so taking $\mathbf{Proj}$ gives the locus $V(ty) \subseteq \mathbf P^1_R = \mathbf P^1_{[x:y]} \times \mathbf A^1_t$. This has two components: a horizontal component $y=0$ mapping isomorphically down to $S = \mathbf A^1$, and a vertical component $t=0$ living entirely above the origin $0 \in \mathbf A^1$.

Remark. I do not know a clean criterion for irreducibility of $\mathbf P_S(\mathscr E)$ for a coherent sheaf $\mathscr E$ on a locally Noetherian scheme $S$. Clearly a necessary criterion is that $\operatorname{Supp}(\mathscr E)$ is irreducible, since this is the image of $\mathbf P_S(\mathscr E) \to S$ (of course the image is closed, either since $\mathbf P_S(\mathscr E) \to S$ is proper or since it equals the support). So, as in your question, let me restrict to the case $S$ irreducible and $\mathbf P_S(\mathscr E) \to S$ surjective. For irreducibility questions, we may replace $S$ by $S_{\text{red}}$ to assume $S$ is integral.

Then there is a positive result if all $\operatorname{Sym}^n\mathscr E$ all torsion-free. In fact, in this case $\mathbf P_S(\mathscr E)$ and even $\mathbf A_S(\mathscr E)$ is integral: if $\eta \in S$ is the generic point, then torsion-freeness means that the natural map $\operatorname{Sym}^*_{\mathcal O_S}(\mathscr E) \to \operatorname{Sym}^*_{\kappa(\eta)}(\mathscr E_\eta)$ is injective, so $\operatorname{Sym}^*(\mathscr E)$ embeds into a polynomial algebra over a field. Conversely, if $\mathbf A_S(\mathscr E)$ is integral, clearly we need all $\operatorname{Sym}^n \mathscr E$ to be torsion-free.

However, this turns out to be a subtle condition on $\mathscr E$. For instance, when $\mathscr E = \mathcal I$ is a sheaf of ideals, then there is a natural map $\operatorname{Sym}^n \mathcal I \twoheadrightarrow \mathcal I^n$, which is an isomorphism if and only if $\operatorname{Sym}^n\mathcal I$ is torsion-free. This is a well-studied problem; for instance if $\mathcal I = \mathfrak m$ is the ideal of a closed point, then $\operatorname{Sym}^* \mathfrak m$ is torsion-free if and only if $R$ is regular [Vas, Cor. 2.1.4].

Example. This gives a very transparent set of examples: if $S = \mathbf A^n$ and $I = \mathfrak m$ is the ideal of the origin, then $\operatorname{Sym}^* I = \bigoplus_n I^n$ is the Rees algebra by the discussion above, so $\mathbf P(I)$ is the blowup of $\mathbf A^n$ in the origin (which of course we know is integral).

(This also very clearly shows the issue with the wrong argument in an earlier version of this answer, as pointed out in the comments below.)

On the other hand, if $I = \mathfrak m$ is the ideal of a singular point, then $\mathbf P(I)$ does not agree with the Rees algebra, and at least $\mathbf A(I)$ is not integral.

Two things I couldn't figure out:

  • If $S$ is integral and $\mathbf P_S(\mathscr E)$ is irreducible, then is $\mathbf P_S(\mathscr E)$ integral?
  • If $S$ is integral and $\mathbf P_S(\mathscr E)$ is integral, then is $\mathbf A_S(\mathscr E)$ integral, i.e. are all symmetric powers of $\mathscr E$ torsion-free?

I suspect the answer to both questions is negative, but I have not yet found counterexamples. Maybe Sasha's answer gives an example where $\mathbf P_S(\mathscr E)$ is integral (since it is irreducible, generically reduced, and Cohen–Macaulay) but $\mathbf A_S(\mathscr E)$ is not.

Here is an example showing that $\mathbf P_S(\mathscr E)$ is not always reduced if we drop the irreducibility assumption:

Example. Let $k$ be a field, $R = k[t]$ with maximal ideal $\mathfrak m = (t)$, and let $S = \operatorname{Spec} R$. Take $\mathscr E = R \oplus R/\mathfrak m^2 = Rx \oplus Ry/(t^2y)$. Then $$\operatorname{Sym}^*(\mathscr E) = R[x,y]/(t^2y),$$ whose $\mathbf{Proj}$ is not reduced. However, as in the first example, it is also not irreducible.


References.

[Vas] W. V. Vasconcelos, Arithmetic of blowup algebras. London Mathematical Society Lecture Note Series 195. Cambridge University Press (1994). DOI:10.1017/CBO9780511574726.

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    $\begingroup$ Now, I might be missing something, but what happens with the Abel-Jacobi map from the $dth$ symmetric product of $X$ to $Pic(X)$, when $X$ is a curve for example? By people.math.harvard.edu/~bejleri/teaching/math259xfa19/…, we should get a projective bundle but not all the fibers are of the same dimension, something that would happen if the associated coherent sheaf is locally free. Am I missing something? $\endgroup$
    – rfauffar
    Dec 1, 2022 at 4:17
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    $\begingroup$ @rfauffar: The mistake in the proof is in the reduction to the case of a DVR: if say $S$ is of dimension $> 1$ and $s$ is a closed point and $t$ the generic point, then the DVR $R$ has to be chosen so that its residue field is a transcendental extension of the residue field of $s$ and this will cause the dimension computation to go wrong. $\endgroup$
    – naf
    Dec 1, 2022 at 6:05
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    $\begingroup$ @rfauffar: You can deduce from the argument that irreducibility implies that the locus in $S$ over which the rank jumps, i.e., increases, must be of codimension at least two and this can be refined to give upper bounds on the dimensions of the loci over which the rank is bounded below by a fixed number $\endgroup$
    – naf
    Dec 1, 2022 at 6:07
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    $\begingroup$ @rfauffar: No, that's not true: take your example, pick in $S=\operatorname{Pic}^d(X) $ a general point $t$ and a point $s$ where the dimension jumps, and a curve $T\subset \operatorname{Pic}^d(X) $ passing through $t$ and $s$. Then, as in the (wrong) answer, the pull back of $\operatorname{Sym}^d(C) $ to $T$ has at least two components. $\endgroup$
    – abx
    Dec 1, 2022 at 13:41
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    $\begingroup$ @abx aha, your comment clarifies where my error was. Is it then true that $\mathscr E$ must be reflexive or torsion-free on some irreducible closed subscheme, or is it even more tricky? $\endgroup$ Dec 1, 2022 at 15:29
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Let me mention a positive result. Assume

  1. $S$ is Cohen-Macaulay and irreducible;

  2. $\mathcal{E}$ has projective dimension $1$ and generically has rank $r > 0$;

  3. $\mathrm{codim}(\{s \in S \mid \dim(\mathcal{E}_s) \ge r + i \}) \ge i + 1$ for all $i > 0$.

Then $\mathbf{P}(\mathcal{E})$ is irreducible.

To prove this, first, use the projective dimension assumption to write a locally free resolution $$ 0 \to \mathcal{F}_1 \to \mathcal{F}_0 \to \mathcal{E} \to 0 $$ (maybe locally over $S$). Then observe that the epimorphism induces an embedding $$ \mathbf{P}(\mathcal{E}) \subset \mathbf{P}(\mathcal{F}_0) $$ as the zero locus of a global section of the vector bundle $p^*\mathcal{F}_1^\vee \otimes \mathcal{O}(1)$ on $\mathbf{P}(\mathcal{F_0})$. Then use the codimension assumption to check that $\mathbf{P}(\mathcal{E})$ has at most one component of dimension $\dim(S) + r - 1$. Then use the Cohen-Macaulay assumption to check that $\mathbf{P}(\mathcal{E})$ is Cohen-Macaulay, hence doesn't have components of dimension less than $\dim(S) + r - 1$.

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  • $\begingroup$ Cool answer! This really helps. Thanks a lot. $\endgroup$
    – rfauffar
    Dec 6, 2022 at 18:52

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