Ralph
Reputation
797
Next privilege 1,000 Rep.
Create tags
 Jan7 awarded Yearling Jul8 comment Homology groups of $SL(2,\mathbb Z)$ For solving the exercise it's enough to compute $H_n A \to H_nG_1 \oplus H_nG_2$ induced by $A \hookrightarrow G_i$. By using the standard periodic resolution for cyclic groups $A, G_i$ it's easy to show that it's the canonical inclusion $A \hookrightarrow G_1 \oplus G_2$ in odd degrees (and zero in positive even degrees). Jul2 awarded Curious Jan7 awarded Yearling Dec10 answered Why is $\hbox{Ext}_R^* (M,M) = H^*(\hbox{Hom}_R^*(P^*,P^*))$? Oct30 answered Subgroups containing kernel of group morphsm to an abelian group are normal. Oct14 answered Hochschild homology - motivation and examples Oct14 comment Hochschild homology - motivation and examples Yes, but I was of course wrong because your DGA isn't concentrated in a single degree as long as $d> 0$. Oct6 comment Cohomology of finite groups with finite coefficients I'm happy to do so: A part of the long exact cohomology sequence is the exact sequence $H^n(G,M) \xrightarrow[]{p} H^n(G,M) \to H^n(G,M/pM) \to \cdots$. Now suppose $H^n(G,M/pM)=0$. Using $H^n(G,M)=\mathbb{Z}/|G|$ we have the exact sequence $\mathbb{Z}/|G| \xrightarrow[]{p} \mathbb{Z}/|G| \to 0$, i.e. multiplication by $p$ is surjective. However, this isn't possible since $p$ divides $|G|$. Oct6 comment Cohomology of finite groups with finite coefficients Well, in ring theory a finite module usually means finitely generated. Anyway. Let $M \le F$ be f.g. as above with $H^n(G,M)=\mathbb{Z}/|G|$ and let $p$ be a prime divisor of $|G|$. Since $M$ is (as abelian group) a finitely generated abelian group, $M/pM$ is a finite abelian group and a $G$-module. Now the long exact cohomology sequence corresponding to the short exact sequence of $G$-modules $0 \to M \xrightarrow[]{p} M \to M/pM\to 0$ shows $H^n(G,M/pM)\neq 0$. Oct6 comment Cohomology of finite groups with finite coefficients Note that the construction above yields a finitely generated $M$ if one starts with $I_G$: If $N$ is supposed to be f.g. then $F$ can be choosen of finite rank and hence $M$ is f.g. because $\mathbb{Z}G$ is Noetherian. Oct6 answered Cohomology of finite groups with finite coefficients Oct3 comment Hochschild homology - motivation and examples Writing down the Hochschild homology (HH) of the abstract $k$-algebra $k[x]/(x^{n+1})$ shouldn't be to hard. But I don't know how the HH of a DGA is defined (maybe it equals the HH of an abstract algebra if the DGA is concentrated in a single degree like yours ?). So can you please give the definition of HH of a DGA ? Jul11 comment Bases that intersect to bases @Ross: By intersection I just mean the intersection of sets as described in the question. Jul11 comment Bases that intersect to bases @Curufin: If the answer is "no", can you please give an explicit example ? Jul11 asked Bases that intersect to bases May13 comment Prime norm ideals that are also principal From en.wikipedia.org/wiki/Landau_prime_ideal_theorem: "... so that the prime ideal theorem is dominated by the ideals of norm a prime number." Therefore I guess the quantity you are looking for is still $X/\log(X)$. May9 comment Intersection of kernels and linear dependence of linear maps Thanks for your counterexample. I included the condition $\dim V \ge \dim W$. May9 revised Intersection of kernels and linear dependence of linear maps added 146 characters in body May9 asked Intersection of kernels and linear dependence of linear maps