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comment Does Nagata theorem hold in a field that is not algebraically closed?
First of all, isn't this Hilbert's theorem? Also, isn't $R^G\otimes_{\mathbb R} \mathbb C$ the same as the $G$-invariants of $R\otimes_{\mathbb R}\mathbb C$? That shows that $R^G\otimes_{\mathbb R}\mathbb C$ is finitely generated over $\mathbb C$ and implies that $R^G$ is finitely generated over $\mathbb R$.
Apr
1
comment Hartshorne 4.1.6 Gonality of a curve
@HeitorFontana Yes, you have inclusions of vector spaces $H^0(X,\mathcal O_X) \subset H^0(X,\mathcal O_X(P)) \subset \ldots \subset H^0(X,\mathcal O_X((g+1)P)$. It could very well be that $f$ lies in some proper subspace. The reason it works with "g+1" is because it is guaranteed in this case that the Riemann-Roch space is of dimension at least two. O the other hand, there are certainly better bounds for the gonality of a genus $g$ curve (by Brill-Noether theory). I think $g/2+1$ is one.
Mar
31
comment Example of a complete non-projective variety
I see. The point of that exercise is indeed to give an example of a complete non-projective variety, but your question has nothing to do with that a priori. Anyway, Exercise 6.9 shows that $Pic(C)$ fits into a short exact sequence. You might want to write down the analogous short exact sequence for $Pic(C\times \mathbb A^1)$ and then show that this sequence splits. That would give you the first isomorphism. I don't think any part of Exercise 6.9 asks you to prove that $Pic(C) \cong\mathbb G_m$ (as it is false).
Mar
31
answered Hartshorne 4.1.6 Gonality of a curve
Mar
31
comment Example of a complete non-projective variety
Did you miswrite the title of the question?
Mar
27
comment Can the quotient by a nonabelian group yield an abelian singularity?
@BenBlum-Smith that's good. You can post Jason's comment as an answer here, I guess.
Mar
27
comment Can the quotient by a nonabelian group yield an abelian singularity?
@BenBlum-Smith Good to hear you're getting closer to your goal.
Mar
20
comment Smooth fibration on a smooth curve
Then it might fail. Consider the singular curve X given by $xy =0$ over $\mathbb A^1$ via the map $(x,y)\mapsto x$. This is not flat. All its fibers are smooth though.
Mar
19
answered Smooth fibration on a smooth curve
Mar
19
comment Can the quotient by a nonabelian group yield an abelian singularity?
I remember now where I saw this example. It is indeed given by an irr. rep'n. Have a look at the paper linked in my answer.
Mar
19
answered Can the quotient by a nonabelian group yield an abelian singularity?
Mar
19
comment Isotrivial family: different definitions
If the fibers of the curve are of genus at least two, then the two conditions are equivalent. It has to do with Isom-schemes of such curves being finite. To see that 1 and 2 are not equivalent in the case of a genus zero curve, try looking at some projective bundles.
Mar
19
comment Isotrivial family: different definitions
I would say that the right definition of isotrivial would be 2 in this case. It might actually even be better to take $U$ to be the complement of the union of a countable number of closed subsets in $B$.
Mar
19
answered Isotrivial family: different definitions
Mar
19
comment Can the quotient by a nonabelian group yield an abelian singularity?
I guess you can take $G$ to act via its abelianization. Or do you want the action to be faithful? In that case, if I recall correctly, there is an example of an $A_5$-action which gives an abelian quotient singularity.
Mar
10
answered Definition of a smooth variety for arbitrary field?
Mar
10
answered Pullback of an invertible sheaf through an isomorphism
Mar
10
comment if the fibers of all points are finite then will the map be finite?
If $f:X\to Y$ is a quasi-finite finite type morphism of schemes, then $f$ is finite if and only if $f$ is proper. If your schemes happen to be affine, then $f$ is separated, so that $f$ is finite if and only if $f$ is universally closed.
Mar
10
comment Does blow up of subscheme in special fiber change the generic fiber?
@mqx No. Blow-ups do not commute with any base-change. Suppose that you blow-up a point $p$ in $X$. What would the base-change along $p \to X$ be?
Mar
10
comment Picard group of a generic point
It looks like (and it is) the trivial homomorphism, as the Picard group of a field is the trivial group. (Note that $\eta =$ Spec $K(X)$.)