# Is the total space of a vector bundle over an irreducible scheme irreducible?

Let $X$ be an irreducible scheme over $\mathbb{C}$ and let $F$ be a locally free sheaf of rank $r$ on $X$.

• Is the total space $Y$ of the associated vector bundle to $F$, $Y=Spec(Sym(F^{\vee}))$, irreducible?

I know some results, like if $f: Y\rightarrow X$ is surjective with irreducible fibers, $X$ irreducible and $f$ closed, then $Y$ is irreducible.

Here i would like to apply this to $\pi: Spec(Sym(F^{\vee}))\rightarrow X$. But I only know that $\pi$ is an affine morphism. Is there some more properties in this special situation such that we can conclude the irreducibility of the total space? (We may assume $X=\mathbb{A}^n$)

If this is not true in general, are there some special situations for which this is true?

## 2 Answers

I was asked a very similar question just yesterday. The answer is yes. First the easiest case, which we will reduce to below:

If $\mathrm{Spec}(A)$ is irreducible, then so is $\mathbb{A}^r_A$.

Let $X = \mathrm{Spec}(A)$ be an irreducible affine, i.e., $A$ is a commutative ring with prime nilradical, and let $Y$ be the total space of the trivial rank $r$ vector bundle over $X$. Then $Y \cong \mathrm{Spec}(A[x_1,x_2,\dots x_r])$. The nilradical of $A[x_1,x_2,\dots x_r]$ just consists of the polynomials with nilpotent coefficients (that's some exercise in Atiyah-Macdonald, Chapter 1) and so $A[x_1,x_2,\dots x_r]_{\text{red}} = A_{\text{red}}[x_1,x_2,\dots x_r]$ is an integral domain; hence $Y$ is irreducible.

Reducing to this basic case we can prove:

Let $X$ be an irreducible scheme and $F$ a locally free $\mathcal{O}_X$-module of finite constant rank $r$. Then the total space of the associated $\mathbb{A}^r$-bundle, $\mathrm{Spec}(SF^\vee)$, is irreducible too.

In fact, it suffices to prove that $Y$ has an open cover $Y = \bigcup_i U_i$ by irreducible affine subschemes with $U_i\cap U_j\not=\emptyset$ for all indices $i,j$.

Let $X = \bigcup_i V_i$ be an affine open cover trivialising $F$. Since $X$ is irreducible, so are the $V_i$, and $V_i\cap V_j\not=\emptyset$ for all indices $i,j$. Then

1. $Y$ is covered by the affine open subschemes $U_i:=Y_{V_i} \cong\mathrm{Spec}(SF^\vee(V_i))\cong\mathbb{A}^r_{V_i}$ and
2. For each pair of indices $i,j$, since $V_i\cap V_j\not=\emptyset$, this intersection contains some non-trivial affine open $W\subset V_i\cap V_j$, so $U_i\cap U_j = Y_{V_i\cap V_j}$ contains the non-trivial affine open subscheme $Y_W \cong \mathrm{Spec}(SF^\vee(W))$.

By the easy case, the $U_i$ are irreducible, and so $Y$ is covered by irreducible affine open subschemes meeting pairwise non-trivially. This proves that $Y$ is irreducible, as claimed.

I believe the answer is yes.

Suppose that $Y = Y_1 \cup Y_2$, and pick $p \in Y_1 \cap Y_2$. Then $\mathscr O_{Y,p}$ is not an integral domain.

But it should be, since $F$ was a locally free sheaf. We should have $\mathscr O_{Y,p} \simeq \mathscr O_{X,\pi(p)}[x_1,\ldots,x_r]$ by definition of the total space, for some $r$. But this is an integral domain since $X$ is irreducible.

• Okay, that seems like a good starting point. But why can we find such a point $p$? Isn't it possible that $Y=Y_1\cup Y_2$ with $Y_1\cap Y_2=\emptyset$? – Bernie Jun 27 '15 at 13:20