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1d
comment Do local Galois representations always lift?
You should ask this on overflow.
1d
answered Can a separable isogeny of elliptic curves have an inseparable dual?
2d
comment Kähler differentials in an inseparable field extension
I mean, the easiest way of doing this is to say that smoothness is equivalent to locally finite presentation+flat+locally free differentials of correct rank, but that seems somewhat cheating. Some more elementary ideas are as follows: 1) Reduce to purely inseparable. Reduce to primitive purely inseparable where the result is obvious. 2) Use connection between differentials and trace pairing, and the fact that the trace pairing is zero if extension is not separable.
Apr
27
comment Can a separable isogeny of elliptic curves have an inseparable dual?
It should be noted that this is, in a rigorizable sense, the only example over $\overline{\mathbb{F}_p}$. Namely, as you pointed out, we really should only be caring about isogenies of prime power order. Then, the kernel is going to be of the form $\mu_{p^m}\times(\mathbb{Z}/p^n\mathbb{Z})$ in the ordinary case and a connected-connected group scheme in the supersingular case. The latter can't lead to anything interesting (since it's connected connected) and so we focus on the former, which produces examples as above.
Apr
27
comment Can a separable isogeny of elliptic curves have an inseparable dual?
So, $\widehat{f}$ is not étale and so not separable (since separable morphisms are generically étale and this is equivalent to étale for group schemes since one can translate the generic étaleness everywhere).
Apr
27
comment Can a separable isogeny of elliptic curves have an inseparable dual?
Dear Alex: Let $E$ be an abelian variety over $\overline{\mathbb{F}_p}$ which is ordinary. Then, there exists a copy of $\mathbb{Z}/p\mathbb{Z}\subseteq E$. The isogeny $f:E\to E/(\mathbb{Z}/p\mathbb{Z})$ is separable, in fact it's étale, since its kernel is $\mathbb{Z}/p\mathbb{Z}$. That said, if one looks at the dual isogeny the kernel is the Cartier dual of the kernel of the original map. In particular, $\ker\widehat{f}=\mu_p$ which is non-étale.
Apr
24
comment Proof of Chapter 2 Proposition 2.6a in Silverman Arithmetic of Elliptic Curves
@spadey Choose any affine open neighborhood of $Q$. Then, this will give you a subring $A\subseteq K(C_2)$. Let $B$ be the integral closure in $K(C_1)$. Apply your result to this. As for your issue with the separability, this is not really an issue. If I recall, this is a common assumption in number theory so that $B$ is a finitely generated $A$-module. For varieties over a field, the analogous result is automatic (separable extension or not). See here: math.stanford.edu/~conrad/145Page/handouts/intclos.pdf
Apr
24
comment Proof of Chapter 2 Proposition 2.6a in Silverman Arithmetic of Elliptic Curves
@JyrkiLahtonen There is no canonical choice but one can work in 'charts'.
Apr
24
comment Proof of Chapter 2 Proposition 2.6a in Silverman Arithmetic of Elliptic Curves
Since $\phi$ is finite (and so affine) take any affine neighborhood $\text{Spec}(A)$ of $Q$. Then, $\phi^{-1}(\text{Spec}(A))=\text{Spec}(B)$ for some $B$. Your 'AKLB' setup is then $A K(C_2)K(C_1)B$.
Apr
20
comment On the proof that one dimensional linear algebraic groups are either isomorphic to $\mathbb{G}_m$ or $\mathbb{G}_a$.
@user062295 Those do satisfy both of the conditions that I said (at leas in characteristic not $2$), but is not one of the above examples if $k\ne\overline{k}$. (or, more operatively, doesn't contain a root of $-1$).
Apr
20
comment When exactly is a compact complex manifold algebraic?
for non-projective proper smooth varieties (and their analytifications) by Chow's lemma/GAGA. So, again, this is a necessary condition, but certainly not a sufficient one.
Apr
20
comment When exactly is a compact complex manifold algebraic?
Just a comment since some things you said seemed to imply that you may not have heard this fact. But, if $X$ is a complex Kahler manifold, then it's projective if if and only if it's Moishezon. So, any algebraic compact complex manifold which is non-projective is not Kahler. There is also some fact that algebraicity implies non-trivial $H^2_\text{sing}$. Another necessary condition for something to be the analytification of a proper smooth variety is the degeneration of the Hodge spectral sequence, this shouldn't hold for an arbitrary (non-Kahler) complex manifold, but still holds
Apr
20
comment On the proof that one dimensional linear algebraic groups are either isomorphic to $\mathbb{G}_m$ or $\mathbb{G}_a$.
I don't have the text, but I think it's worth mentioning that (unless this is implicit in Springer's assumptions) that $G$ should be connected and smooth, else it's easy to cook up counterexamples (in removing either of the hypotheses) to the claim.
Apr
20
comment Why are there no $\mathbb{R}$-valued points on a complex curve?
It might be worth noting (for the OP) that the actual input to descend (as in 'descent theory') to something over $\mathbb{R}$ is, in this case, an action of $\text{Gal}(\mathbb{C}/\mathbb{R})$. But, if you have that Galois action, you can actually get at $\mathbb{R}$-points as Galois invariant $\mathbb{C}$-points.
Apr
20
comment My strange proof of the fact that $k[\mathbb P^n]=k$
I think that it's correct, but that it's hiding a lot. In particular, I think that if one defines a regular morphism in the usual way, that it affine locally looks like polynomials, then verifying that this definition agrees with yours for projective varieties is strictly stronger than the statement that $k[X]=k$. So, I don't know which comes first, but, to me, this seems a little circular. In particular if you define morphisms as being locally polynomial its obvious that maps to $\mathbb{A}^1$ are global sections. If you define it as in your definition it's not so clear.
Apr
8
comment Geometric xample of a one object cover which does not satisfy the sheaf condition?
What about the double cover of the circle and the constant presheaf?
Apr
3
revised Reference request: Fibre functor for elliptic curves is pro-representable
edited body
Apr
1
comment Reference request: Fibre functor for elliptic curves is pro-representable
Let us continue this discussion in chat.
Apr
1
comment Reference request: Fibre functor for elliptic curves is pro-representable
@AlexSaad I'm just unreasonably intolerant of 'classical language', especially for arithmetic things where it adds so much. This is the thinking that all of Silverman is immersed in.
Apr
1
comment Reference request: Fibre functor for elliptic curves is pro-representable
PS, I hope that the extra added bit is helpful to you—I really don't like Silverman's text too much.