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11 votes
Accepted

One-point compactification of ample line bundle

This is EGA 2, Prop. 8.8.2. It basically says that if $L$ is ample then one can contract the zero section of the geometric realization $\mathbb V(L)$ of $L$ to a point. The result is called the affine ...
Friedrich Knop's user avatar
8 votes
Accepted

Pushforward of a very ample line bundle on a curve to $\mathbb{P}^1$

No, not in general. Take $C=\mathbb{P}^1$, $L=\mathcal{O}(1)$, $p$ to be map $x\mapsto x^2$ in affine coordinates. Then $p_*L$ has rank $2$, but $$2=h^0(L)=h^0(p_*L)=h^0(\mathcal{O}(e_1))+h^0(\mathcal{...
Donu Arapura's user avatar
  • 35.2k
7 votes
Accepted

Does a projective variety have only finitely many associated Hilbert polynomials?

Yes for the first question, by Riemann--Roch. No for the second --- even in the simplest case of a projective line, the polynomial $td + 1$ is the Hilbert polynomial (with respect to $L = O(d)$).
Sasha's user avatar
  • 39.3k
6 votes
Accepted

Ampleness verifiable over faithfully flat cover

According to stacks project tag 0B5V, if $S$ is Noetherian, $g$ is finite and surjective, and $X$ is proper over $S$ (which automatically implies properness of $X'$), then $\mathcal L$ is ample if and ...
Will Sawin's user avatar
  • 148k
5 votes
Accepted

Classifying ample line bundles over the flag manifold $G/B$

As Donu said, line bundles on $G/B$ correspond to elements of the weight lattice. The ample bundles correspond to the dominant regular weights. This means that the pairing of the weight $\lambda$ with ...
Sam Gunningham's user avatar
5 votes
Accepted

Does ampleness descend along finite maps?

There is a non-quasi-affine variety $X$ with quasi-affine normalization. See Tag 0271. Then $\mathcal{O}_X$ is a counter example.
darx's user avatar
  • 66
4 votes
Accepted

Varieties with an ample vector bundle mapping to their tangent bundle

The answer to this question is now known to be yes -- it is Corollary 1.2 of this paper of Jie Liu.
Daniel Litt's user avatar
4 votes
Accepted

Ample vector bundles and embeddings

The answer to your first question is yes. Gieseker (https://projecteuclid.org/download/pdf_1/euclid.nmj/1118798367) calls such bundles as strongly ample. ( see p 92 of above paper). To see that such ...
Priyankur Chaudhuri's user avatar
4 votes
Accepted

The ample cone of a surface with an algebraic $\mathbb C^*$ action

This is true, and follows from Corollary 3.29 of https://arxiv.org/pdf/0811.0517.pdf
demo's user avatar
  • 56
3 votes
Accepted

Making a vector bundle ample by twisting with ample line bundle

Because $L$ is ample, $E\otimes L^n$ is generated by global sections for $n\gg 0$, i.e., there is a surjective morphism $\mathscr O_X^{\oplus r} \to E\otimes L^n$, which implies that there is a ...
Sándor Kovács's user avatar
2 votes
Accepted

Fiber product of projective varieties and ample line bundles

Let $D_X$ and $D_Y$ be the Chern classes of $L_X$ and $L_Y$. Assume $X$ and $Y$ have dimension at least one. Thus $pr_{1*}pr_1^*D_X=0$ and $pr_{2*}pr_2^*D_Y=0$. Applying the projection formula we ...
Edward Teach's user avatar
1 vote

Bertini type theorem for very ample line bundle

1. General hypersurface of a normal variety is normal; see ``The hyperplane sections of normal varieties'', Seidenberg, A. (1950), Transactions of the American Mathematical Society, 69(2), 357-386. (...
Sheng Meng's user avatar

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