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Jun
6
comment Kähler differentials, define valuation?
I guess you mean that $R$ is a DVR, so in particular integral. In this case the module of Kahler differentials is trivial, so you can define a valuation on it after choosing some embedding $H^0(X, \Omega_X) \to k(X)$ where $X = \mathrm{Spec} R$, which is tantamount to picking a trivialization. There is no canonical trivialization, but all the valuations obtained this way will be the same since all trivializing sections do not vanish anywhere.
Oct
25
awarded  Nice Question
Sep
28
comment independence of Galois conjugates
Dear Gerry, you are right. I am sorry for the confusion. I have a certain more complicated setting in mind, and I was trying to isolate the question that I am trying to answer, but apparently I was too sloppy formulating it.
Sep
28
revised independence of Galois conjugates
added 26 characters in body
Sep
28
comment independence of Galois conjugates
you are right. I was looking for less trivial examples, though, i.e. something more complicated than Kummer extensions. I guess I should have asked for K to be algebraically closed.
Sep
28
asked independence of Galois conjugates
Sep
26
comment why is there no non-degenerate 2-forms on 4-sphere?
But he claim is for $n\neq 1,3$. Non-degenerate means that restriction of the form to any fibre of the ctangent bundle is non-degenerate as a bilinear form, i.e. induces an isomorphism $T_x M \to T_x^*M$.
Sep
26
revised why is there no non-degenerate 2-forms on 4-sphere?
added 2 characters in body
Sep
26
asked why is there no non-degenerate 2-forms on 4-sphere?
Jul
2
awarded  Curious
Feb
12
comment graph of the compostion of morphisms category-theoretically
Dear Qiaochu, thanks a lot for this answer. I am afraid for me the fact that in a span $X \leftarrow F \to Y$ the "graph" $F$ embeds into $X \times Y$ is crucial. In this case you need to use the push forward in the definition of the composition of graphs and that's what causes confusion in the proof (at least for me), I think. (Actually, not so much confusion, I think I have an idea now about using some kind of projection formula, I am figuring out the details.)
Feb
11
comment Scheme theoretic image of a base change of a morphism of schemes
is $g^{-1}(f(X))$ a closed subscheme of $Y'$? what is its scheme structure and how is the closed embedding defined?
Feb
11
revised graph of the compostion of morphisms category-theoretically
added 156 characters in body
Feb
11
comment graph of the compostion of morphisms category-theoretically
oh, I should have remarked that all morphisms are proper and image is the minimal closed subcheme through which a morphism factors.
Feb
11
asked graph of the compostion of morphisms category-theoretically
Jan
29
comment a closed subset of an algebraic group with a constant tangent space is a coset
as per the bounty, I hope that the discussion is not over, so let us wait.
Jan
29
comment a closed subset of an algebraic group with a constant tangent space is a coset
For non one-dimensional $Z$ one uses the fact that Albanese is of dimension $h^{0,1}$ and this equals $\mathrm{dim} Z$ if the canonical bundle is trivial. I think the statement also works for vector groups and one-dimensional subsets (it looks plausible that there are counterexamples with $Z$ of dimension $> 1$). It would be nice to know what is the most general statement that is still true.
Jan
29
comment a closed subset of an algebraic group with a constant tangent space is a coset
Jesko, I also have some other considerations. For example, the statement is still true for Abelian varieties. The simplest case is when $Z$ is one-dimensional, then we can just notice that a curve with a trivial tangent bundle is an elliptic curve, and by a little argument involving functoriality of taking Albanese, an ellpitic curve embedded into an Abelian variety is a coset.
Jan
29
comment a closed subset of an algebraic group with a constant tangent space is a coset
these considerations in principle should lead to a counterexample, but I still cannot construct one.
Jan
29
comment a closed subset of an algebraic group with a constant tangent space is a coset
... moreover, if it turns out that such manifold does not touch any constant vector field, then it obviously cannot be a closed subgroup.