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 Yearling
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answered Prove that $|GL_n(\mathbb{F})|< q^{n^2}$.
Jun
27
comment the field fixed by inertia group is the maximal unramified field
You can reduce to the finite case: Take an arbitary $\alpha$ in $K^I$ and consider the finite extension $F(\alpha)/F$.
Jun
16
comment Visualising finite fields
Yes, that's the paper. Gassert is currently at the University of Colorado at Boulder as a postdoc. Here's the full list of his arXiv papers: arxiv.org/find/math/1/au:+Gassert_T/0/1/0/all/0/1
May
12
awarded  Yearling
Apr
1
comment $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ is Galois over $\mathbb{Q}$?
Have you tried computing the minimal polynomial for $\sqrt{2+\sqrt{2}}$?
Mar
31
comment A question regarding Kummer
I don't understand your question, and I honestly don't know the history of algebraic number theory, so I'm not sure whether ideal class groups originated with Kummer. But Kummer's example of failure of unique factorization is frequently the example used to motivate ideal class groups.
Mar
31
answered A question regarding Kummer
Mar
30
revised Probability that minimum of two numbers is less than 4
this is not linear algebra
Mar
30
suggested approved edit on Probability that minimum of two numbers is less than 4
Mar
30
answered Probability that minimum of two numbers is less than 4
Mar
23
revised How to find the matrix of the Linear Tranformation wrt the bases S and T?
Wrote v's instead of u's in the final matrix
Mar
23
answered How to find the matrix of the Linear Tranformation wrt the bases S and T?
Feb
11
comment Finitely generated algebra as a module over a subalgebra
Two polynomials (in $f_3$) are equal if and only if all of their coefficients are equal, so this is immediate.
Feb
10
answered Finitely generated algebra as a module over a subalgebra
Feb
10
comment The mod $p$ Galois representation of the Frey curve is unramified away from $2, p$
This is an excellent answer.
Feb
9
comment The mod $p$ Galois representation of the Frey curve is unramified away from $2, p$
The one-line "intuitive" answer to your question is that the primes dividing the discriminant are those for which the curve has bad reduction, i.e. reducing modulo those primes yields a singular curve over a finite field, so we expect bad things to happen at such primes. Where the curve is smooth, we expect everything to be nicely behaved.
Feb
9
revised The mod $p$ Galois representation of the Frey curve is unramified away from $2, p$
Wrote ramified when I meant unramified
Feb
9
answered The mod $p$ Galois representation of the Frey curve is unramified away from $2, p$
Jan
27
awarded  Good Answer
Dec
15
awarded  Caucus