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?



Basic question on projective bundles
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](https://mathoverflow.net/a/435703/82179) 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.




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**References.**


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


