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Let $k$ be an algebraically closed field of positive characteristic and $X$ be a smooth projective curve over $k$ of genus $g \ge 2$. Fix a polarization $L$ on $X$. Does there exist a semi-stable vector bundle on $X$ of rank $r$ and degree $d$ with gcd$(r,d)=1$?

I know that this result is true in characteristic zero. I have heard this to be true in positive characteristic but have not been able to find a good reference for this fact or a counterexample.

Any reference/hint would be very helpful.

EDIT Assume further that $k$ is countable.

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  • $\begingroup$ Is your field uncountable? If so, I believe you can just choose a first-order deformation to which a maximal destabilizing bundle does not extend. $\endgroup$ – Jason Starr Oct 6 '15 at 10:29
  • $\begingroup$ @JasonStarr Sorry, but the field I have in mind in $\overline{\mathbb{F}_p}$, which is countable. I will edit the question. $\endgroup$ – Ron Oct 6 '15 at 15:42
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It seems relatively easy to construct such a bundle by induction. Namely:

Step 0 If $r=1$, there are clearly line bundles of given degree $d$ on $X$, since $k$ is assumed to be algebraically closed.

Step 1 Choose $(r',d')$ such that $0<r'<r$, $d'/r'>d/r$, and there are no integral points within (or on the edges) of the triangle with vertices $(0,0)$, $(r',d')$, and $(r,d)$. For instance, can take $(r',d')$ satisfying the first two inequalities whose distance from the line through $(0,0)$ and $(r,d)$ is minimal.

Step 2 The condition that there are no integral points on the boundary of the triangle implies that $gcd(r',d')=gcd(r-r',d-d')=1$. Let $E_1$ and $E_2$ be stable bundles such that $rk(E_1)=r'$, $deg(E_1)=d'$, $rk(E_2)=r-r'$, $deg(E_2)=d-d'$, which exist by the induction hypothesis. Choose a non-trivial extension $$0\to E_2\to E\to E_1\to 0,$$ which exists by the Riemann-Roch Theorem (here we need that the genus of $X$ is at least one).

Step 3 It is easy to see that $E$ is stable. Indeed, assume it is not, and $F\subset E$ is destabilizing. Then the slope of $F$ is at most $d'/r'$, and dually the slope of $E/F$ is at least $(d-d')/(r-r')$. From the condition that there are no integral points in the triangle, we see that $rk(F)=r'$ and $deg(F)=d'$. Because of stability of $E_1$ and $E_2$, this implies that the map $F\to E\to E_1$ is an isomorphism, which would split the exact sequence for $E$. Contradiction.

Remark The only place where we need the field to be algebraically closed is Step 0. So as long as you know that line bundles of every degree exist on $X$, the statement holds. For instance, it works if $X$ has a $k$-point, or if the base field is finite (see this question), but fails if the field is $\mathbb{R}$.

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  • $\begingroup$ I do not understand why the slope of $F$ is at most $d'/r'$ as mentioned in step $3$. Could you please suggest some reference or give a hint. $\endgroup$ – Ron Oct 7 '15 at 14:02
  • $\begingroup$ Consider the filtration on $F$ induced by the filtration on $E$; that is, consider the exact sequence $0\to(F\cap E_2)\to F\to F/(F\cap E_2)\to 0$. The left (resp. the right) term of the sequence is isomorphic to a subsheaf of $E_2$ (resp. $E_1$), and therefore its slope cannot exceed $(d-d')/(r-r')$ (resp. $d'/r'$). This gives the desired bound on the slope of $F$. This also shows that the slope of $F$ is equal to $d'/r'$ if and only if it projects isomorphically onto $E_1$. $\endgroup$ – t3suji Oct 7 '15 at 14:36
  • $\begingroup$ Thank you for the hint. I understand the proof now. Although you say that this is an easy solution, I think it is a very interesting solution. I have also not seen this solution anywhere. Since I will reproduce the solution in an upcoming article, I would either like to mention a reference for the solution or acknowledge you for this. In case there is no known reference, you should tell me your identity. $\endgroup$ – Ron Oct 8 '15 at 11:46
  • $\begingroup$ Well, I don't know a reference... The name is Dima Arinkin (math.wisc.edu/people/faculty-directory) $\endgroup$ – t3suji Oct 8 '15 at 21:17

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