Let $K$ be a complete DVR of characteristic $0$, $X$ a smooth projective curve over $K$. Denote by $K^0(X)$ the Grothendieck group of locally free sheaves on $X$ and by $\mbox{det}$ the natural group homomorphism from $K^0(X)$ to $\mbox{Pic}(X)$. Suppose that for some $L \in \mbox{Pic}(X)$, there exists $[V] \in K^0(X)$ such that $\mbox{det}([V])=L$. Let $L'$ be another line bundle on $X$ of the same degree as $L$. The question is, does there exist $[V'] \in K^0(X)$ such that $V'$ is of the same rank and degree as $V$ and $\mbox{det}([V'])=L'$? If not true in general, is there any known condition/example of $K$ (other than algebraic closedness in which case the answer is known to be true via twisting $V$ by a suitable line bundle), for example if $K$ is a $C_1$field under which we have a positive answer to the question?
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$\begingroup$ Are we allowed to set $V' = L'$? $\endgroup$ – S. Carnahan♦ Sep 27 '14 at 13:47

$\begingroup$ @Carnahan: Sorry. No, we need $V'$ to have the same rank and degree as $V$. I will edit the question. $\endgroup$ – user43198 Sep 27 '14 at 13:49
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Perhaps I am misunderstanding the question, but it seems to me that $\text{deg}(V)=\text{deg}(\text{det}(V))=\text{deg}(L)$, so we are just asking if for each $n$ there is a (virtual) vector bundle $[V']$ of rank $n$ and degree $\text{deg}(L')$. But $\mathcal{O}_X^{\oplus(n1)}\oplus L'$ works, right?