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May
2
comment About morphism from elliptic curve to projective space and pullback divisor
@user322778 I would be happy to try to answer, of course.
Apr
27
comment Every irreducible representation is either even or odd.
@YasirMahsud It means that $p(-I)$ is of the form $\lambda I$, for $\lambda \in \mathbb{C}$. Hence, acting like a scalar. And Schur's lemma says that any endomorphism of an irreducible representation, a linear map commuting with the action of the group, is a scalar. en.wikipedia.org/wiki/Schur%27s_lemma
Apr
26
comment Proof of algebraic set involving dimension
On phone now. Just suggesting that you break this idea down into two steps: how do you describe the cotangent space as a cokernel, hence by rank of some matrix, and how do ranks of a matrix of polynomials change as you move about and evaluate at different points. There are some inequalities to be careful with.
Apr
26
answered Proof of algebraic set involving dimension
Apr
26
comment Every irreducible representation is either even or odd.
@AlexMathers No problem, it was basically the right idea. Now I am shamelessly putting in my answer :)
Apr
26
revised Every irreducible representation is either even or odd.
added 1 character in body
Apr
26
answered Every irreducible representation is either even or odd.
Apr
26
comment Every irreducible representation is either even or odd.
@AlexMathers There are many square roots of $I$ in $GL(V)$. One also needs to use Schurs lemma here.
Apr
25
comment Why do the characters of an abelian group form a group?
There is the dual representation.
Apr
25
revised If $\pi : X \to Y$ is a flat, proper $O$-connected morphism of locally Noetherian schemes, then is $h^0(X_q, O_{X_q}) = 1$?
edited body
Apr
25
comment If $\pi : X \to Y$ is a flat, proper $O$-connected morphism of locally Noetherian schemes, then is $h^0(X_q, O_{X_q}) = 1$?
@Hoot Wait... no I am confused still. Of course $A$ (say Spec A is $Y$ or something) is flat over $A/m$, but the question is whether the morphism $Spec (A \to A/m)$ is flat, and that isn't clear. The morphism should have been from $Spec k(q) \to Y$, because I am trying to pullback along it. It was a typo.
Apr
25
answered tangent space of a curve in projective space
Apr
25
comment If $\pi : X \to Y$ is a flat, proper $O$-connected morphism of locally Noetherian schemes, then is $h^0(X_q, O_{X_q}) = 1$?
@Hoot Oh you are right... thanks. I was getting confused, because I felt that closed subschemes where not usually flat. But of course everything is flat over a field...
Apr
25
asked If $\pi : X \to Y$ is a flat, proper $O$-connected morphism of locally Noetherian schemes, then is $h^0(X_q, O_{X_q}) = 1$?
Apr
21
awarded  Nice Question
Apr
19
comment Universal property of relative proj in simple situations?
@KeenanKidwell Yeah, that is great, thanks! It still seems like the relative proj constructions is much easier to work with by gluing patches and studying transition maps, at least for the examples I have been studying so far... can you think of some situation in which the universal property is easier to use?
Apr
18
asked Universal property of relative proj in simple situations?
Apr
18
comment A morphism from a projective curve $X$ to a curve $Y$ is either constant or surjective
Would you be happy with arguments that avoided Riemann-Roch? I really don't see the connection, but my eyes are not very good. If you don't need to use Riemann-Roch, then you can try to use the fact that the image of a projective curve under a regular map is closed. This is sometimes known as properness of projective curves (and varieties). It is very handy. (Also, your curves should be irreducible.)
Apr
13
comment Cocycle condition on bundles as some equivalence relation?
@popo About your first question, if I understand it, it is just a matter of where you put indices in your notation. For the second, the equivalence relationship is that gluing is defined via an equivalence relationship. These compatibility conditions are here to make sure those gluings don't mess things up locally.
Apr
12
comment Systematic way to find the quotient field of a ring?
@user323507 You should take a close look at user26857s answer, which is more complete. (Note that $X^2 + Y^2 + 1 = (X + Y + 1)^2$ in characteristic 2, so the thing on the left isn't even a domain, hence doesn't have a field of fractions.) I didn't answer what appears to be your question, since I didn't see that you were asking about the fraction field, but I'm glad what I wrote helped you understand something.