bio | website | |
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location | ||
age | ||
visits | member for | 3 years, 11 months |
seen | yesterday | |
stats | profile views | 192 |
MathSE chatroom co-owner alongside robjohn.
Feb 9 |
awarded | Commentator |
Sep 24 |
awarded | Autobiographer |
Apr 9 |
revised |
Domains with prime ideal theorems
added 40 characters in body |
Apr 9 |
asked | Domains with prime ideal theorems |
Feb 15 |
comment |
Bijection between irreducible representations and conjugacy classes of finite groups
Could there more generally be some kind of duality between irreducible representations of $G$ over a field $K$ and the $K$-conjugacy classes of $K$-regular elements of $G$? |
Oct 21 |
comment |
Norm on the number field?
Yes. If $v$ is some extension of $v_p(\cdot)$ then the maximal ideal of the valuation subring of $K$ is a prime ideal of ${\frak O}_K$ lying above $p$ with ramification index $e\le m$. The value group is $\frac{1}{e}\Bbb Z$. |
Oct 7 |
awarded | Constituent |
Oct 7 |
awarded | Caucus |
Aug 24 |
awarded | Critic |
Jul 12 |
awarded | Talkative |
Jun 25 |
awarded | Citizen Patrol |
Feb 8 |
comment |
A zeta function using half of the primes
Every other prime seems to be an arbitrary choice. In my opinion, a more interesting question borne of the same motivation would be: for which subsets $S$ of the primes does the corresponding Euler product admit an analytic continuation? |
Jan 28 |
comment |
Counting square-free numbers smoothly
Perhaps we want f(t)=0 in the first case so the sum is always finite? |
Jan 28 |
revised |
Differences in tree picture of ${\bf Q}_p$, $\overline{{\bf Q}_p}$, ${\bf C}_p$, $\Omega_p$
added 611 characters in body |
Jan 26 |
asked | Differences in tree picture of ${\bf Q}_p$, $\overline{{\bf Q}_p}$, ${\bf C}_p$, $\Omega_p$ |
Apr 4 |
awarded | Supporter |
Aug 21 |
awarded | Scholar |
Aug 21 |
accepted | What might the (normalized) pair correlation function of prime numbers look like? |
Aug 18 |
awarded | Editor |
Aug 18 |
comment |
What might the (normalized) pair correlation function of prime numbers look like?
@David Speyer: I'm not sure. I've added in a note at the bottom on how I got the expression, and while I believe I did the adaptation correctly, it may still be a naive substitution and in need of change. |