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visits member for 1 year, 7 months
seen Jul 21 at 5:42

Sorry, I am not good at English. Sorry, I didn't know about "accepted answer". I will click.


Jul
2
awarded  Curious
May
26
awarded  Critic
May
17
revised simple modules over non commutative polynomial rings
added 2 characters in body
May
17
asked simple modules over non commutative polynomial rings
Feb
28
awarded  Popular Question
Dec
15
awarded  Yearling
Dec
6
comment Are two different prime ideals relatively prime?
I got it, thanks!
Dec
5
comment Are two different prime ideals relatively prime?
oh..., I forgot that $(0)$ is prime. Thanks!
Dec
5
asked Are two different prime ideals relatively prime?
Nov
3
awarded  Tumbleweed
Oct
29
revised numbers on circles and combination of sums of some neighboring numbers
edited body
Oct
27
asked numbers on circles and combination of sums of some neighboring numbers
Oct
27
asked Does a rectangle exists on any closed curve?
Sep
26
comment a base for a Galois group
Let $(G_{i})_{i\in I}$ be a projective system in which the $G_{i}$ are finite groups and endowed with the discrete topology. Every subsets of $G_{i}$ are open. The topology of $\prod G_{i}$ is defined as the product topology. The open sets of this set looks like $\prod S_{i}$, $S_{i}=G_{i}$ without finitely many $i\in I$. The topology on $\varprojlim G_{i}$ is the induced topology. This group is closed in $\prod G_{i}$.
Sep
25
comment a base for a Galois group
I forgot to insert "open subset". I understand the topology as the following. Gal$(L/K)\cong \varprojlim_{E\in I}$Gal$(E/K)$ as groups. The right side is considered as a profinite group and this defines the topology of the left side.
Sep
25
comment a base for a Galois group
Let $U\subset G$ and $x\in U$. I want to find $\sigma$ and $F$ such that $x\in U_{\sigma, F}\subset U$ but I can't do this. Denote by $I$ the set of subfields $E$ of $L$ which $E$ is a finite Galois extension of K. Since $U$ is open, there exists a finite subset $I'$ of $I$ and $U=\prod_{E} U_{E}$, if $E\in I-I',U_{E}=Gal(E/K)$, if $E\in I'$, $U_{E}\subset Gal (E/K)$. I expect that $F$ is determined by $\{U_{E}\}_{E\in I'}$ but I don't come up with the way to find $F$.
Sep
25
comment a base for a Galois group
I can't prove this and I have little intuition that $U_{\sigma, F}$ form a base for the open sets of $G$.
Sep
25
asked a base for a Galois group
Sep
11
accepted field extension $K\subset L$ and Uniqueness of $G$ such that $K=L^{G}$
Sep
11
asked field extension $K\subset L$ and Uniqueness of $G$ such that $K=L^{G}$