3
$\begingroup$

Let $C$ be a curve (smooth projective over a field $k$ of positive char) and $p$ a rational point on $C$. Put $\dot{C}=C-{p}$ and $T=Spec R$, where $R$ is a noetherian $k$-algebra. The question is as follows: can one always extend a vector bundle over $\dot{C}_T$ to a vector bundle over entire relative curve $C_T$?

$\endgroup$
5
$\begingroup$

I am assuming that by $C_T$ you mean $C\times_k T$. If so, then the answer is no. Here is why:

First of all, $C$ being projective has no importance since the question is local near $p\in C$. So, let's assume that $C$ is affine.

Let $C$ be any smooth affine curve and $f:C\to T$ be a morphism that's one-to-one, an isomorphism outside $p$, that is, $C\setminus\{p\}\simeq T\setminus \{f(p)\}$ via $f$, but $f$ is not an isomorphism at $p$. For instance, let $T$ be a cuspidal cubic and $f:C\to T$ the normalization.

Now let $\Gamma\subset C\times T$ be the graph of $f$. Then $\Gamma\simeq C$ is a smooth curve and since it intersects the singular locus of $C\times T$, it cannot be a Cartier divisor. (If it were, $C\times T$ would have to be smooth along $\Gamma$).

Now, let $\mathcal L=\mathcal O_{(C\setminus \{p\})\times T}(\Gamma)$, which is a line bundle on $(C\setminus \{p\})\times T$. Suppose this can be extended to $C\times T$. Then the extension has to be a line bundle and it would correspond to a Cartier divisor of the form $\Gamma + aP$ where $P=\{p\}\times T$. $P$ is a Cartier divisor, since it is just the pull back of $p$ on $C$. Therefore, this would mean that then $\Gamma$ is a Cartier divisor, but we have already seen that it is not.

So, $\mathcal L$ cannot be extended as a line bundle on $C\times T$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.