595 reputation
112
bio website
location
age
visits member for 1 year, 5 months
seen Apr 10 at 12:55

Feb
3
comment Sequence where the sum of digits of all numbers is 7
I do not think that the problem is clearly stated... it is not written anywhere that your sequence contains ALL numbers whose sum of the digits is 7...
Jan
15
awarded  Revival
Dec
17
comment If D is an Integral Domain and has finite characteristic P, prove P is prime. Is my proof correct?
As Praphulla Kounshik points out you cannot conclude as you did. Anyway, notice that, in your notation, $Pa=R(Sa)=(R1)(Sa)$. You obtain that either $R1=0$ or $Sa=0$. Of course, if $R1=0$, then also $Ra=(R1)a=0$. Thus, either $Ra=0$ or $Sa=0$. With this you can conclude. Let me also say that the argument below given by Bill Dubuque is much more elegant!
Dec
9
awarded  Critic
Nov
19
revised Submodules of $M/G$
added 9 characters in body
Nov
15
awarded  Yearling
Nov
13
awarded  Tumbleweed
Nov
6
asked Minimal spectrum of graded rings
Nov
4
comment Product varieties with the constructible topology
@Cantlog: Of course for schemes it is less trivial... any guess for that case? For example when X and Y are ring spectra?
Nov
4
comment Product varieties with the constructible topology
Well... I think this is a trivial question, since points are closed (and so constructible) in the Zariski topology, then they are open in the constructible topology, which then coincides with the discrete topology... so the answer is "Yes, trivially."
Nov
4
asked Product varieties with the constructible topology
Aug
13
comment Definition of local maxima, local minima
Well, it is not exactly as you say... if a sequence $(x_n)_n$ has a limit point $a$, you can very well have $x_n=a$ for some $n$. The point is that it may happen but it is not necessary that the $x_n$ never coincide with $a$. The idea is, as you say, that the $x_n$ go closer and closer to $a$. More precisely, for any open ball $B$ around $a$, $x_n\in B$ for all but finite $n\in\mathbb N$. You can easily see by the definition I gave you that the sequence $(y_n)_n$ with $y_n=a$ for all $n\in\mathbb N$ has $a$ as limit point.
Aug
4
accepted Amenable group rings embeddable in skew fields
Jul
13
comment Projective Resolution of $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$
yep, you are right...
Jul
12
answered Is the unit disk in $\Bbb R^2$ a subspace?
Jul
12
comment Projective Resolution of $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$
@user49685: as Jack Schmidt noticed there is probably a typo and what Relativeo whants to compute is $Ext^n(B,\mathbb Z)$. Interpreting litteraly the question, your comment completely answers the question...
Jul
12
answered Translation from Spanish - Simple differentiation exercise
Jul
12
comment Projective Resolution of $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$
(a part $n=0$ when $Ext^0(\mathbb Z,B)=Hom(\mathbb Z,B)\cong B$). The interesting thing is to compute $Hom(B,\mathbb Z)$, for which you can use the injective resolution I have written in my answer.
Jul
12
comment Projective Resolution of $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$
To construct the derived functors of this second functor you shuold construct a projective resolution of $A$ and then proceed as you know. Now, it is a standard result (but not completely trivial... you should check it) that the two constructions give you the same out-put when you use them to construct $Ext^{n}(A,B)$. As Jack Schmidt was telling you in his comment, there is a mistake in your question, in the sense that, being $\mathbb Z$ a projective object, the $Ext$-groups $Ext^n(\mathbb Z,B)$ that you want to construct are always trivial [...]
Jul
12
comment Projective Resolution of $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$
Well, in principle $Hom_{\mathbb Z}(-,-)$ is a functor from $Ab\times Ab$ to $Ab$, in particular it sends an ordered pair $(A,B)$ of Abelian groups to an Abelian group $Hom(A,B)$. Now, in order to construct its derived functors you just fix one of the two entries (say $A$) obtaining a functor $Hom(A,-):Ab\to Ab$. This gives you a left exact functor, to compute its derived functors you have to take an injective resolution of $B$ and proceed as you probably know. Of course there is another possibility, that is, fixing $B$ you have a right exact contravariant functor $Hom(-,B):Ab\to Ab$. [...]