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 $C_T$?


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 $C\times T$.


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