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 Yearling
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Feb
7
revised If $M$ isn't Noetherian, $M$ has a submodule maximal with respect to being not finitely generated.
deleted 6 characters in body
Feb
7
revised If $M$ isn't Noetherian, $M$ has a submodule maximal with respect to being not finitely generated.
I simplified some parts of the reasoning following some (now disappeared) comments.
Feb
7
revised If $M$ isn't Noetherian, $M$ has a submodule maximal with respect to being not finitely generated.
Edited to make the proof independent of the choice of an order on the index set.
Feb
6
comment If $M$ isn't Noetherian, $M$ has a submodule maximal with respect to being not finitely generated.
@user26857 well, in the question they are taking a chain... which means that the modules are already ordered. Why don't you want to assume their labels are ordered? I see no restriction in that...
Nov
15
awarded  Yearling
Sep
7
revised If $x \geq 0$, $y \geq 0$, $x \leq D$, and $0 \leq 1/(1+x) - 1/(1+y) \leq C$, then is there an upper bound on the value of $y - x$?
edited the title
Sep
7
suggested approved edit on If $x \geq 0$, $y \geq 0$, $x \leq D$, and $0 \leq 1/(1+x) - 1/(1+y) \leq C$, then is there an upper bound on the value of $y - x$?
Jun
5
comment Topological Entropy of $T$, on a disjoint union?
if, say, $A$ is not $T$-invariant, what do you mean by $h(A,T_{|A})$? You need $A$ to be $T$-invariant for $T_{|A}$ to define a self-map of $A$. So, I think the question makes sense just under the assumption that your three spaces are invariant, so that you have well-defined restrictions.
Jun
3
comment Using Exchange Lemma in a decomposition
I do not think that this is even true: take a local ring $R$, let $M=R$, so that $M^{(3)}\cong R\oplus R\oplus R$, here $B=R\oplus R$. Then, $R\oplus R$ is not a direct summand of $R$...
Jun
1
revised Quotients vs equivalence relations
added 123 characters in body
Jun
1
asked Quotients vs equivalence relations
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