Reputation
5,146
Top tag
Next privilege 10,000 Rep.
Access moderator tools
Badges
1 12 27
Impact
~62k people reached

Apr
15
accepted I've proved everything about the ideal correspondence easily except $\pi ^{-1} \pi (\frak{a}) = \frak{a}$
Apr
15
comment I've proved everything about the ideal correspondence easily except $\pi ^{-1} \pi (\frak{a}) = \frak{a}$
I don't think that's a rephrasing. It's a complete rewriting. And mine used plenty of math-English o__O
Apr
15
asked I've proved everything about the ideal correspondence easily except $\pi ^{-1} \pi (\frak{a}) = \frak{a}$
Apr
11
accepted Quadratic reciprocity: $\left( \dfrac{-1}{p}\right) = (-1)^{\frac{p-1}{2}}$
Apr
11
comment Quadratic reciprocity: $\left( \dfrac{-1}{p}\right) = (-1)^{\frac{p-1}{2}}$
Thanks. I'll study this for a bit
Apr
11
comment Quadratic reciprocity: $\left( \dfrac{-1}{p}\right) = (-1)^{\frac{p-1}{2}}$
I've seen that, I guess now to learn it.
Apr
11
asked Quadratic reciprocity: $\left( \dfrac{-1}{p}\right) = (-1)^{\frac{p-1}{2}}$
Apr
5
comment is “$a^0 = 1$” a definition or there exists a proof?
What have you tried?
Apr
5
answered Rank of free group
Apr
1
revised There exists a descending chain of symmetry groups from a formal language string down to its smallest grammar.
added 24 characters in body
Apr
1
revised There exists a descending chain of symmetry groups from a formal language string down to its smallest grammar.
added 24 characters in body
Apr
1
revised There exists a descending chain of symmetry groups from a formal language string down to its smallest grammar.
deleted 74 characters in body
Apr
1
asked There exists a descending chain of symmetry groups from a formal language string down to its smallest grammar.
Mar
31
revised Are these proofs all right? (Automorphism group of a string).
added 19 characters in body
Mar
31
comment Are these proofs all right? (Automorphism group of a string).
@ThomasAndrews if $s[i] \neq s[j]$ and $\{i,j\} \cap \{u, v\} = \varnothing$, then $(i,j)$ clearly commutes with $(u,v)$. And if $(u,v) = (i,j)$ they clearly commute. And if they share an index, then $((u,v) (u,j))(u,v) = (v,j)(u,v) = (u,j)$. So where is the problem?
Mar
31
asked Are these proofs all right? (Automorphism group of a string).
Mar
29
comment Finite groups acting on strings.
what do you mean by composition factor?
Mar
29
comment Finite groups acting on strings.
Thanks. Does anyone apply it to the smallest grammar problem directly? To yield approx smallest grammars?
Mar
29
accepted Finite groups acting on strings.
Mar
29
revised Finite groups acting on strings.
added 104 characters in body