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Apr
4 |
asked | A criterion for freeness over a local ring |
Mar
27 |
answered | Galois descent for semilinear endomorphisms |
Jan
25 |
awarded | Nice Answer |
Dec
23 |
answered | Describing the ratio of uniformizers in B_dR |
Dec
17 |
comment |
Commuting invariants and duals of C_p vector spaces
Concerning your first question: a $G_K$-equivariant map from V to $C_p$ exists, for example, whenever $V$ is of Hodge-Tate type with one weight equal to $0$. This happens for lots of repns for which $V^{G_K} = 0$. |
Dec
4 |
comment |
Algebraic maximal extension and algebraic closure
$\mathbf{C}_p$ has plenty of immediate extensions, that is extensions that have the same value group and the same residue field. The compositum of all these is its spherical completion. Now lots of intermediate fields between $\mathbf{C}_p$ and its spherical completion would not be algebraic maximal. |
Nov
23 |
comment |
I know that you know…
There is a (not serious) mention of this in "A canticle for Leibowitz", but sadly I don't remember where exactly. Can somebody find the exact quotation? |
Nov
21 |
answered | Fastest way to factor integers < 2^60 |
Nov
17 |
comment |
Why is Gauss credited with his connection?
Ah, so the story about Gauss inventing the least squares method to compute orbital parameters of asteroids is bogus? |
Nov
17 |
awarded | Nice Answer |
Nov
16 |
answered | Absolute Galois group of the field of Puiseux series over $\overline{\mathbb{F}}_p$ |
Nov
16 |
comment |
Absolute Galois group of the field of Puiseux series over $\overline{\mathbb{F}}_p$
@Spice: if you replace $\overline{\mathbb{F}}_p$ iwth $\mathbb{C}$, then the field you get is algebraically closed. |
Oct
29 |
awarded | Good Answer |
Oct
24 |
revised |
Is there any theorem like implicit function theorem in $\mathbb{Q}$ ?
deleted 25 characters in body |
Oct
24 |
revised |
Is there any theorem like implicit function theorem in $\mathbb{Q}$ ?
added 169 characters in body; added 36 characters in body |
Oct
24 |
answered | Is there any theorem like implicit function theorem in $\mathbb{Q}$ ? |
Oct
24 |
comment |
Slick ways to make annoying verifications
@Charles : yes, sorry, I forgot to say "free", so the modules should be free of the same rank. Thanks. |
Oct
23 |
answered | Slick ways to make annoying verifications |
Oct
23 |
comment |
Slick ways to make annoying verifications
And the closed graph criteria also works in a compact space! |
Oct
21 |
awarded | Nice Answer |