 # ag.algebraic geometry – Finite decision by semi-stable bundles Answer

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ag.algebraic geometry – Finite decision by semi-stable bundles

Devlin Mallory’s strategy is actually rectify, and in reality the status is plane a miniature higher than he urged.

Let $$V$$ breathe an capricious vector bundle of rank $$n$$. Let $$W$$ breathe a steady bundle of rank $$n+1$$ (if one exists) or a semistable bundle. Fix an copious bundle $$mathcal O(1)$$.

I pretense that for $$m$$ sufficiently sizable, there’s a map $$W(-m) to V$$ that’s surjective and whose kernel is a line bundle (and thus is mechanically steady).

It suffices to take $$m$$ sizable sufficient that $$Votimes W^vee otimes mathcal O(m)$$ is globally generated. We can perceive sections of this line bundle as maps $$W(-m) to V$$, and the fiber of the part at a degree corresponds to the fiber of the map at a degree.

Maps from an $$m+1$$-dimensional vector area to an $$m$$-dimensional vector area have complete rank outdoors of a codimension $$2$$ locus. Thus amongst world sections of $$V otimes W^vee otimes mathcal O(m)^vee$$, these with out complete rank at a given level have codimension $$2$$. So these which fail to have complete rank at a number of factors of $$C$$ have codimension $$1$$. Thus we are able to discover a part which has complete rank at each level, which suggests it’s surjective and its kernel is a line bundle.

Over finite fields, this codimension dispute would not labor, however a similar counting dispute does.

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