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seen Dec 19 at 13:36

Dec
16
awarded  Caucus
Dec
9
accepted Complete but not cocomplete category
Nov
15
awarded  Yearling
Oct
23
comment Complete but not cocomplete category
The question was answered on mathoverflow, giving an example of a cocomplete but not complete Abelian category.
Sep
11
comment Uniqueness Of Linear Transformation
I can probably try to interpret your question... but I think you should try to formulate it in a precise way yourself... so... how do you define $T$? I deduce it is not an arbitrary morphism otherwise you would not ask if it is 1-1 (it could be the $0$ map for example)...
Sep
1
comment Is every subspace of a normed linear space which is not closed a hyperspace.
let $x_1$ and $x_2\in \mathfrak {X\setminus B}$. Then $x_1-x_2\notin B$ but clearly $f(x_1-x_2)=1-1=0$... so the kernel of your map may be bigger than $B$. Where are you using that your subspaces is not closed?
Aug
28
comment Does every free $R$-module have a maximal proper submodule?
Take the quotient $M/(\mathrm{Span}(\mathfrak{B}\setminus \{x\}))$, this will be isomorphic to $R$, while a submodule $N$ of $M$ is maximal if and only if $M/N$ is simple. Your idea works iff $R$ is simple as an $R$-module.
Jul
29
comment On finite exponent abelian $p$-groups
I see, I never found this convention but it is not hard to believe that in some context this could be useful. What I found is a different definition for $p$-groups, if you have a $p$-group $G$ such that $\sup\{|g|:g\in G\}=p^k$ you say that $k$ is the exponent, while sometimes you take directly $p^k$. Of course these definitions produce the same classes of "groups with finite exponent".
Jul
29
comment On finite exponent abelian $p$-groups
@user1729 the correct thing is to use "supremum" instead of maximum. Indeed, the set $\{p^n:n\in\mathbb N\}$ does not have a maximum, but it does have a supremum (i.e., $\infty$). In fact, $\mathbb Z(p^{\infty})$ has exponent, the fact is that it is not finite.
Jul
28
comment On finite exponent abelian $p$-groups
@user1729 what do you mean by "finite but unbounded" ??? If the set of the orders of the elements is unbounded, then its sup (which by definition is the exponent of the group) is infinite. Of course the Prüfer groups do have infinite exponent, as they are unbounded.
Jul
28
comment On finite exponent abelian $p$-groups
page 88 in the edition of 1970. The idea of the proof is that any bounded group equals its basic subgroup, which is a direct sum of cyclics.
Jul
28
comment On finite exponent abelian $p$-groups
Exactly, I mean that there is a common (finite) upper bound for the orders of the elements of the group.
Jul
28
answered On finite exponent abelian $p$-groups
Jul
28
comment Noetherian group rings
thanks for your example! Regarding the second part, it may very well be true that any group ring embeds in a Noetherian ring but at least this is certainly not know (e.g., Kaplasky's stable finiteness conjecture would be trivial in that case)
Jul
27
comment Noetherian group rings
I do not know if it is interesting enough, maybe such examples are well-known... but if you think it is the case, if is fine with me
Jul
27
asked Noetherian group rings
Jul
17
revised Homotopy limits
deleted 1329 characters in body
Jul
15
revised Homotopy limits
I have included part of a possible solution, making more precise the part that I do not understand.
Jul
15
comment Homotopy limits
Now I see! Since fibrations in the usual injective model structure are (suitable) epimorphisms, the cone and the kernel of the fibrant replacement have to be quasi-isomorphic! Thanks!
Jul
15
comment Homotopy limits
@ZhenLin, I have tried to understand this for a while but I do note see why, given a homomorphism $f:X\to Y$ of cochain complexes, the complex $cone(f)$ should be quasi-isomorphic to the kernel of the induced map $f':X'\to Y'$ between to fibrant replacements of $X$ and $Y$ respectively. Is this what you mean?