13,381 reputation
11946
bio website warwick.ac.uk/alexbartel
location University of Warwick, UK
age 31
visits member for 3 years, 5 months
seen Apr 17 at 14:43

I am currently a Zeeman Lecturer at Warwick University. My research interests lie in algebraic number theory and representation theory, more specifically integral Galois module structures, the arithmetic of elliptic curves, and (integral and rational) representation theory of finite groups.


Apr
17
comment Different methods to prove $\zeta(s)=2^s\pi^{s-1}\sin\left(\frac{s\pi}{2}\right) \Gamma (1-s) \zeta (1-s)$.
@Alyosha You need to be fairly comfortable with $p$-adic numbers, and to know some basic measure theory. That would be enough to give it a go, and fill in the gaps as you progress. It doesn't require that much high-flying theory, the main difficulty for novices is that it takes a completely new point of view, compared to the classical one. Note that Tate also proves the analytic class number formula in the same big sweep, so it would help if you knew the statement of that.
Apr
4
revised The way conjugation acts on embeddings
added 40 characters in body
Apr
4
comment Integer solutions of $x^4 + 16x^2y^2 + y^4 = z^2$
So do I!$ $ $ $
Apr
4
answered The way conjugation acts on embeddings
Apr
2
comment Class number of $\Bbb Q(a)$ with $a^3-a+1 = 0$ is $1$
Clearly, an integer polynomial $f(x)$ in one variable cannot be irreducible modulo all primes: plug in any $\alpha$ for which $f(\alpha)\neq \pm 1$ and take any prime divisor $p$ of $f(\alpha)$. Then $f(\alpha)\equiv 0\pmod{p}$, so $(x-\alpha)$ is a factor of $\bar{f}$ modulo $p$.
Mar
21
comment Reducibility of a Cyclotomic Polynomial under the ring homomorphism $\mathbb{Z} \rightarrow \mathbb{F}_p$
Galois groups of irreducible polynomials over finite fields are finite, generated by Frobenius (this is also covered in the Oxford course the exam for which you are looking at). But is the Galois group of $f$ over $\mathbb{Q}$ cyclic?
Mar
21
comment Galois Theory: An automorphism fixes a field if and only if it fixes the set of generators.
This is matter of writing down all the relevant definitions.
Mar
21
answered Reducibility of a Cyclotomic Polynomial under the ring homomorphism $\mathbb{Z} \rightarrow \mathbb{F}_p$
Feb
27
comment Finding J-invariant of Legendre form of Elliptic Curve
Like Jyrki says, a transformation of the form $x'=x+r$ will do the trick.
Feb
27
revised Decomposing the tensor product representation of $S_3$ in terms of irreducibles
Corrected the formula for the idempotent e_chi
Feb
21
comment Isomorphism and crossed product
@ChristophPegel Ah, thanks, hadn't noticed that.
Feb
21
revised Isomorphism and crossed product
rolled back to a previous revision
Feb
20
answered A question about Character Degrees of G/N
Feb
19
answered Irreducible representations of finite lamplighter group
Feb
17
comment Representation theory over $\mathbb{Q}$
@anon Sorry, I see now where the confusion arose. You are right that this action fixes characters that take values in $K$. For example if $K=\mathbb{Q}$, then Galois acts trivially on the rational characters. But it still acts non-trivially on those characters that take values in a bigger number field.
Feb
16
comment Representation theory over $\mathbb{Q}$
@anon Why do you say that $\sigma\chi(g) = \chi(g)$?$$ $$ The action is almost the one you wrote down, except, as you say, to make the resulting action on the character table trivial, you want $\sigma\chi(g) = \chi(\sigma^{-1}g)$.
Feb
16
revised An analogy between group actions and group represenations
Updated link
Feb
15
comment Representation theory over $\mathbb{Q}$
@anon That's exactly right. "Compatible" means in this context that if you act simultaneously on the characters and on the group, then the character table is preserved.
Feb
14
comment Representation theory over $\mathbb{Q}$
@OliverBraun: glad you like the answer. Regarding generalisations about numbers of irreducible representations, as Jack says, and as is expanded in the linked post, there is a compatible action of the Galois group not only on the characters of a group, but also on its elements. The general result then says that the number of irreducible representations of a finite group over a number field $K$ is equal to the number of conjugacy classes of orbits of elements under this compatible Galois action. Over $\mathbb{Q}$, to be in one orbit is exactly the same as to generate the same cyclic subgroup.
Feb
14
answered Representation theory over $\mathbb{Q}$