bio  website  ucs.louisiana.edu/~avm1260 

location  Lafayette, LA, USA  
age  45  
visits  member for  4 years, 11 months 
seen  10 hours ago  
stats  profile views  3,128 
With the move of MathOverflow into the SE network, this account is now associated with dormant accounts in math.SE and other sites in the network. While I plan to continue my (generally lowlevel) participation in MO, my current plans do not include restarting my participation in those other sites. Therefore, I will be ignoring any comments or pings that reach me from those sites, unless and until I resume my active participation there.
I remain "gone for the foreseeable future" from math.SE, tex.SE, and meta.SE.
Please do not send me private email to call my attention to comments, questions, or other matters related to those sites. Thank you. Also, as I no longer participate in those sites, I do not wish to be sent, by private email, questions that you can just as well ask on those sites. I would have thought it was obvious, but apparently I need to say so explicitly.
22h

revised 
Ext of Skyscraper sheaf
spelling, punctutation 
2d

comment 
Categorical proof subgroups of free groups are free?
(My comment was not in the linked question, but it is in this one ). 
2d

comment 
Categorical proof subgroups of free groups are free?
(continued) One would expect a truly "categorical proof" to be something you could do in the context of any variety of groups; but since the conclusion does not hold in most of them, I don't expect you would be able to find such a proof in the first place. 
2d

comment 
Categorical proof subgroups of free groups are free?
As I believe I said in math.SE, I don't think you can have a true "categorical proof" (though you can probably have proofs cast in the language of categories, or proofs that have a categorical flavor), because the fact is not true "categorically". If you look at the category of all groups in a variety of groups (which is a reflective subcategory of Groups and hence generally speaking wellbehaved), the only varieties in which "subgroup of free is free" holds are the variety of all groups, all abelian groups, and all abelian groups of exponent $p$ (a prime). (cont) 
Jan 28 
comment 
Categorical proof subgroups of free groups are free?
It's not even true that subobjects of relatively free groups are relatively free groups (in the corresponding variety). In fact the only Schreier varieties of groups are the variety of all groups, all abelian groups, and of all abelian groups of exponent $p$ ($p$ a prime). 
Jan 26 
reviewed  Approve Second duals of Grothendieck spaces 
Jan 25 
comment 
Circle squarer and solution of polynomial equations
A circular arc can be constructed if and only if the central angle can be constructed. A central angle of $x$ radians corresponds to an arc of length $x\pi$, and a central angle of $x$ radians can be constructed if and only if $\cos(x)$ can be constructed, if and only if $\cos(x)$ lies in the quadratic closure of your ground field. So it still seems to me like you aren't getting that much more out of it, nor do I see how you would get $\sqrt[3]{2}$. 
Jan 23 
comment 
Circle squarer and solution of polynomial equations
The squarer gives you $\pi$, so you are essentially working in the quadratic closure of $\mathbb{Q}(\pi)$; does it contain $\sqrt[3]{2}$? 
Jan 23 
revised 
Circle squarer and solution of polynomial equations
spelling 
Jan 22 
comment 
Elements of order 3 normalizing no nonidentity 2subgroups in Almost Simple Groups
Should that be "Such a group $G$ has $S\leq G\leq \mathrm{Aut}(S)$" in line $3$? 
Jan 22 
revised 
Increasing order
not set theory 
Jan 22 
comment 
Increasing order
math.Overflow is for research level questions. This would work well in math.stackexchange.com, though. 
Jan 20 
comment 
Irreducible/prime/indivisible elements
I misread the condition for "indivisible". 
Jan 20 
comment 
Irreducible/prime/indivisible elements
In a commutative ring, "indivisible" and "prime" are the same thing for nonzero elements. For in a commutative ring, prime and completely prime are equivalent for ideals. If $p$ is prime, and $pab$, then $ab\in (p)$, hence $a\in (p)$ or $b\in (p)$, hence $pa$ or $pb$. Conversely, if $p$ is indivisible and $ab\in (p)$, then $pab$, hence $pa$ or $pb$, hence $a\in (p)$ or $b\in (p)$, thus $(p)$ is completely prime, and hence prime; therefore, $p$ is prime. 
Jan 20 
comment 
Irreducible/prime/indivisible elements
Q2 and Q3: Fields are noetherian domains and have no primes and no irreducible elements. However, they have no zero divisors, so your parenthetical comment is false. In particular, it is false that Noetherian domains must have indivisble elements. 
Jan 18 
comment 
A categorical method to, say, determine the cardinality of a group
What do you mean, $\mathbb{Z}$ has "two idempotents"? As an additive group, $\mathbb{Z}$ has a unique idempotent, same as any other group: because $x^2=x$ implies $x=e$ (the identity). And multiplicatively, it is not a group. 
Jan 16 
comment 
A categorical method to, say, determine the cardinality of a group
I was wondering... In any case: there is a nice, simple proof that all group epimorphisms are surjective due to Linderholm; you can see the proof here 
Jan 16 
comment 
A categorical method to, say, determine the cardinality of a group
I don't see how the epimorphisms allow you to recover the lattice of all subgroups. If $G$ is a simple group, then the only epimorphisms with domain $G$ are those into the trivial group and those into $G$ itself. In the former case, there is but one epimorphism; in the latter case, you get $\mathrm{Aut}(G)$, and all maps are equivalent since given two automorphisms $f,g\colon G\to G$, we can let $i=gf^{1}$ to get $if=g$, hence $f\leq g$ holds for any pair. So my lattice contains two elements: how did I recover the lattice of subgroups of $G$ from this? 
Jan 12 
reviewed  Approve Solving gradient of an especial heat equation 
Jan 8 
awarded  Nice Answer 