5,624 reputation
1245
bio website researchinpractice.wordpress.…
location
age
visits member for 3 years, 5 months
seen 57 mins ago

Dec
10
comment Atiyah-MacDonald Ch. 4 exercise 20: what's the module analogue of $\sqrt{\mathfrak{a}+\mathfrak{b}} = \sqrt{\sqrt{\mathfrak{a}}+\sqrt{\mathfrak{b}}}$?
A few questions: (1) I believe the lemma, but I am having trouble following the overall line of thought. Can you make more explicit how the results you are claiming answer my question? (E.g.: I assume that in my case you would take $M_1=M_2=M$, but what is your $N$ supposed to be in my scenario? My $N+N'$? Why?) (2) Atiyah and MacDonald do not assume that $M$ or $N,N'$ are finitely generated. Does the result depend on this assumption?
Dec
9
asked Atiyah-MacDonald Ch. 4 exercise 20: what's the module analogue of $\sqrt{\mathfrak{a}+\mathfrak{b}} = \sqrt{\sqrt{\mathfrak{a}}+\sqrt{\mathfrak{b}}}$?
Dec
8
comment Subgroup Lattice of D14 - normal and centre
Note that the heptagon has 7 corners so there are 7 reflections in the lines that you describe.
Dec
8
comment Subgroup Lattice of D14 - normal and centre
There are $n$ reflections. $r$, $or$, $o^2r$, etc.
Dec
4
comment Subgroup Lattice of D14 - normal and centre
If you meant $r$ to run through all the reflections, then you've got it! If $r$ represents just one of the reflections, then you're missing some subgroups.
Dec
4
revised is the natural numbers can be a valid subspace over some field?
added 308 characters in body
Dec
4
comment is the natural numbers can be a valid subspace over some field?
@KevinCarlson - good point; I will edit. I was thrown by the word "group."
Dec
3
comment Embedding $G$ in its holomorph
@spin - yes, true, if $\operatorname{Aut}G \hookrightarrow G$, there will be lots of embeddings, at least $|G|$ many conjugate embeddings (which therefore won't be normal); the holomorph has a natural transitive action on the set of elements of $G$, with the $G$ factor acting by translations and the $\operatorname{Aut}G$ factor acting in the obvious way; the stabilizer of any element of $G$ for this action is isomorphic to $\operatorname{Aut}G$.
Dec
3
comment How do you compute group cohomology in practice?
+1 Thanks! I am seeming to gravitate toward MAGMA...
Dec
3
accepted How do you compute group cohomology in practice?
Dec
3
answered How do you compute group cohomology in practice?
Dec
3
accepted Schemes to the rescue?
Dec
3
accepted Alias/alibi in permutation groups
Dec
3
answered is the natural numbers can be a valid subspace over some field?
Dec
3
asked The “muscle” behind the fact that ergodic measures are mutually singular
Dec
3
asked Embedding $G$ in its holomorph
Dec
3
answered arithmetic modulo field with real numbers vector space
Dec
3
answered Bounded polynomials are Lipschitz
Nov
26
comment Prime ideals of the ring of integers of an algebraic number field
@baltazar - proofs for everything I've said would be in any standard book on algebraic number theory, but I have in mind the proof in Marcus' Number Fields, which I believe is in chapter 3.
Nov
26
comment Prime ideals of the ring of integers of an algebraic number field
@baltazar - back to the theorem I was describing, though: in fact, it's even better than "all the primes over $p$ are obtainable this way": if $p\nmid[\mathcal{O}_K:\mathbb{Z}[\alpha]]$, then the factorization of $f$ into irreducible factors mod $p$ exactly gives you the factorization of the ideal $(p)\triangleleft\mathcal{O}_K$ into primes of $\mathcal{O}_K$.