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May
13
accepted Compact subgroups are contained in open compact subgroups in locally profinite groups
May
13
comment Compact subgroups are contained in open compact subgroups in locally profinite groups
Now I see, thanks! I think your argument perfectly works now. There is a misprint in your answer: on the 5th line you claim that $C$ is open, I think you mean either $N$ or $CN$, any of the two would work.
May
12
comment Compact subgroups are contained in open compact subgroups in locally profinite groups
I do not see why the proof in your comment works, it seems a circular argument: to show that $y^{-1}Ny\subseteq N$ for a given $y\in C$, you use that $(yx)^{-1}N(yx)\subseteq N$ for all $x\in C$. But for $x=1$ this was the statement you wanted to prove...
May
12
comment Compact subgroups are contained in open compact subgroups in locally profinite groups
Thanks for posting this proof. Let me ask for some details: why is $CN$ a subgroup? Consider two elements $ck^x$ and $dh^y$, with $c,d,x,y\in C$, $h,k\in K$ and $k^x,h^y\in N$. Then, there exists $k'\in K$ such that $k^x=(k')^{yd^{-1}}$, so that $ck^x dh^y=c(k')^{yd^{-1}}dh^y=(cd)(k')^y$. Still, why is this $(k')^y$ belonging to $N$?
May
11
asked Compact subgroups are contained in open compact subgroups in locally profinite groups
Apr
29
comment $R/I^n$ is a local ring
you are right! There is something missing in my proof...
Apr
29
answered $R/I^n$ is a local ring
Apr
29
comment Choice of a skeleton
OK, thanks anyway :-)
Apr
29
comment Choice of a skeleton
Not just that. It is a full subcategory where any two isomorphic objects are equal. The part which is not clear to me is especially the second part of the question, the existence of that specific functor
Apr
29
comment Choice of a skeleton
Of course, me neither... this is why I am asking here...
Apr
29
asked Choice of a skeleton
Apr
17
answered Definition of multiplicity
Apr
4
comment Definition of multiplicity
See the limit formula of Lech for multiplicities
Jan
29
comment Verifying a Vector Space Via Given Axioms
The proof of 3) is the following: we have to show that there exists a neutral element for the sum in our candidate vector space. We affirm that $\{z_n\}$, with $z_n=0$ for all $n$, is our neutral element. We have already seen that addition is commutative so it is enough to verify that $\{\alpha_n\}+\{z_n\}=\{\alpha_n\}$ for all $\{\alpha_n\}$ (otherwise one should also check that $\{z_n\}+\{\alpha_n\}=\{\alpha_n\}$). So, $\{\alpha_n\} + \{z_n\} = \{\alpha_n + z_n\} = \{\alpha_n\}$, since $z_n=0$ is the neutral element for the addition in $\mathbb K$.
Jan
29
comment Verifying a Vector Space Via Given Axioms
The proof of 2) goes like this: $\{\alpha_n\} + \{\beta_n\}=\{\alpha_n+\beta_n\}\overset{(*)}{=}\{\beta_n+\alpha_n\} = \{\beta_n\} + \{\alpha_n\}$. Where $(*)$ holds by the commutativity of addition in $\mathbb K$.
Jan
28
comment Verifying a Vector Space Via Given Axioms
For property 7: I do not see the utility of writing $(\lambda)(\mu)$, instead of $\lambda\mu$, they mean exactly the same. What you should use here is the associativity of multiplication in the base field, do you see how? Property 8 seems OK.
Jan
28
comment Verifying a Vector Space Via Given Axioms
(3) Your third property is the existence of zero element... from your proof it is not clear who is the zero... of course it is the sequence all of whose elements are zero, it is not clear from what you wrote. your properties 4, 5 and 6 seem OK.
Jan
28
comment Verifying a Vector Space Via Given Axioms
Yes, there are possible improvements. Let me list some: (1) In property 1 you are actually using, implicitly, the associativity of addition in your base field, maybe you should make it explicit using parenthesis, (2) the second property does not hold for the reason you wrote, but it holds for the commutativity of addition in your base field. In fact, it is actually false that $\{\alpha_n\}+\{\beta_n\}=0$ for all $\{\alpha_n\}\,,\{\beta_n\}$...
Jan
28
revised Verifying a Vector Space Via Given Axioms
added 333 characters in body
Jan
28
answered Verifying a Vector Space Via Given Axioms