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May
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comment Intersection of affine varieties is affine
I think you know to prove it for opens (use separatedness), and I think you know how to do it for closed (the closed embed into an affine)--your intersection is then just going to be some disjoint union of such intersections (take the components, which are still affine).
May
16
comment Construction of a line bundle with lower degree and lower dimension of global sections
@Omega I think Ted is suggesting to look at the sequence involving O, O(-p), and a skyscraper sheaf. Tensor this with L, and notice that it's still exact since L is O_C-flat.
May
16
comment Reference request: Cohomology of Elliptic Curves
A couple questions. I assume you don't want torsion points defined over K--you don't want a trivial action, yeah? So what do you mean? Do you want E[p^n](K^ab)? Also, have you tried the obvious like of attack using inflation-restriction sequence? Also, have you tried the only tenable case (eg. K=Q)? Finally, where did this question come from?
May
16
comment Whether a functor is exact?
I think you may be overthinking this. Can't you also think of $C^n(G,A)$ in terms of inhomogenous cochains? i.e. maps from $G^n\to A$ with a weird action? I think then it's pretty clear it's exact--unless I am being silly. :)
May
16
comment Classification of all subrings
Just to comment, the only tenable case I can imagine to this (although, this might just be a lack of knowledge on my part) is to classify, in the geometric case, plane curves with the desired normalization. One could just take your plane curve $X$ and modify any set of points with a singularity, the set of which is well-understood for plane curves (e.g. $n$-nodes, etc.).
May
16
comment Classification of all subrings
This seems like a reasonable place to start: ams.org/mathscinet-getitem?mr=MR0195922 That said, this seems like a pretty hopelessly difficult problem. Even if you wanted something simpler like "classify all orders in $\mathcal{O}_K$" then you'd geometrically be asking for "list all singular curves with normalization $X$" (for some given non-singular curve $X$). This seems very hard. So, it seems like a first beginning question one might ask is "what are the affine curves $X$ which are birational to $\mathrm{Spec}(\mathcal{O}_K)$?"
May
16
answered How can I prove this: “$\mathbb{P}^{n} \times \mathbb{A}^{m}$ is not affine variety$”
May
16
answered Test for a $G$-torsor to be trivial?
May
16
comment Surjectivity of ring homomophism induced by Frobenius endomorphism
@Hebe No problem. :) You should answer your own question below, so people in the future can benefit from your insight.
May
16
comment Why is this scheme $Y$ an affine bundle?
I mean, $Y$ needn't be affine in general. Take the trivial bundle over a non-affine scheme. The morphism $\varphi$ is affine though--that's just because definitionally, locally on the base it looks like $\mathrm{Spec}(A)\times\mathbb{A}^n\to\mathrm{Spec}(A)$. Am I misunderstanding your question?
May
16
comment Dimension is an invariant of isomorphism class of projective varieties
@hert3583 An isomorphism is defined to be a morphism with an inverse. Any morphism $f$ is continuous, so if it's an isomorphism there is an inverse map $g$ (which must also be continuous, since it's a morphism) so that $f\circ g$ and $g\circ f$ is the identity. In particular, $f$ has a continuous inverse so is definitionally a homeomorphism. Thinks about it as algebro-geometric morphisms are stricter than topological ones--they preserve even more data (i.e. they take polynomial functions to polynomial functions). Thus, an algebro-geometric isom. is a top. isom.
May
16
comment Surjectivity of ring homomophism induced by Frobenius endomorphism
@Hebe I said that if it's the absolute Frobenius (i.e. the Frobenius on the coordinate ring) then it's NOT an automorphism. For example, the map $\mathbb{F}_q[T]\to\mathbb{F}_q[T]$ sending $f\mapsto f^p$ is not surjective. This is clear, and is related (equivalent) to the fact that $\mathbb{F}_q(T)$ is not perfect. Look at this page which lists all the various types of Frobenii, maybe it will be helpful en.wikipedia.org/wiki/Frobenius_endomorphism ALSO, be sure to ping people in the future with (e.g. @Hebe) so they know you've responded.
May
16
answered k-schemes examples
May
16
comment Surjectivity of ring homomophism induced by Frobenius endomorphism
I don't quite understand all the words you are using (e.g. does an $\mathbb{F}_q$-structure mean it comes from a variety of $\mathbb{F}_q$?). So, I can't tell if you're looking at the Frobenius on $V\otimes\overline{\mathbb{F}_q}$ just acting on the second coordinate, or the 'absolute Frobenius' acting on the coordinate ring of $V$. In the former, it's clearly an automorphism. In the latter, it's not surjective in general--i.e. if $V=\mathbb{A}^1_{\mathbb{F}_q}$. Could you help me make sure I am understanding your question?
May
16
comment Smoothness for morphism of schemes
You mean checking only at closed points on the base (i.e. $X$)? If so, note that definitionally if $f$ is smooth at $x\in X$, it's smooth in a neighborhood of $x$. Thus, with mild conditions this is true (e.g. if $X$ is 'Jacobson'=all locally closed subsets contain a closed point). You might be able to get away with the conditions on $X$ if you also assume that $f$ is flat, so you can check smoothness on fibers.
May
16
answered An element of $f$ of a function field such that $P$ is the only pole of $f$.
Apr
28
revised Best Sets of Lecture Notes and Articles
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Apr
25
comment norm map and local class field theory
They are abstractly isomorphic, but the restriction map is not an isomorphism. You should, as good practice, write down why this is true, but it's analagous to the fact that :$G_{\mathbb{F}_{p^r}}\to G_{\mathbb{F}_p}$ is not an isomorphism, even though the two groups are isomorphic (the map is multiplication by $r$). You'll find that your restriction map is surjective though, which is good since the norm surjects for unramified extensions.