User tj - MathOverflow

Properties of ring epimorphisms that are true only over commutative rings

2013-03-13T23:03:54Z

I'm interested in knowing/collecting some properties of epimorphisms of rings (with identity) that are true over commutative rings but are false in the non-commutative case.

Example: I learned from MO that if $R \hookrightarrow S$ is an epimorphism of commutative rings then $S/R$ is a torsion left $R$-module. But there are counter-examples to this property if $R$ is permitted to be non-comutative.

Epimorphisms and free submodules

2013-02-11T13:21:57Z

By inspecting the accepted answer to this question

http://mathoverflow.net/questions/120918/are-epimorphisms-from-a-division-ring-isomorphisms

one obtains the following necessary condition for epimorphisms:

Let $R \le S$ be rings with identity $1\neq 0$ such that $S$ is flat as right $R$-module. If the inclusion $R \hookrightarrow S$ is an epimorphism, then for each $s \in S$ there is $r\in R, r\neq 0$ such that $rs \in R$.

Proof: The statement says that $S/R$ has no left $R$-submodule isomorphic to $R$. If it is wrong, we have an embedding $R \hookrightarrow S/R$ and hence $0 \neq S = S \otimes_R R \hookrightarrow S\otimes_R S/R$ But by the quoted answer, $S\otimes_R S/R=0$ if $R \to S$ is epi. $\blacksquare$

However, without success I tried to drop the flatness condition. Whence my question:

Question: Is the statement above still valid, if $S$ is not supposed to be flat as right $R$-module ?

Are epimorphisms from a division ring isomorphisms ?

2013-02-06T00:07:05Z

According to Corollary 1.2(3) of the paper Silver: Noncommutative Localizations and Applications. J. of Alg. 7(1964), 44-67:

If $R$ is a (commutative) field and $\alpha: R \to S$ an epimorphism in the category of rings, then $\alpha$ is an isomorphism.

Question: Is $\alpha$ also an isomorphism if $R$ is not a field but just a division ring ?

For convenience, I repeat the short proof from Silver (which doesn't seem to work for division rings): First it is noted (Prop. 1.1) that a homomorphism $\alpha: R \to S$ of (not necessarily commutative) rings with identity is epi iff multiplication $S \otimes_R S \to S$ is an isomorphism (of abelian groups). Then:

"For $x\in S$, consider the subring $R[x]$ of $S$ generated by $R$ and $x$. Since $R$ is a field, one can easily see that $R \to R[x]$ is an epimorphism using 1.1. If $x$ is transcendental over $R$, then $\beta: R[x] \to R[x]$ defined by $\beta(f)=f(0)$ agrees on $R$ with the identity map of $R[x]$. So $x$ cannot be transcendental over $R$ by definition of an epimorphism. Finally, if $[R[x]:R] < \infty$, then by 1.1, $[R[x]:R]^2=[R[x]:R]$, so $[R[x]:R]=1$ and $x \in R$. Hence $\alpha$ is an isomorphism as desired."

Why do we use the diagonal for diagonal approximations ?

2012-12-15T22:54:19Z

First recall how the cup product is defined for the cohomology of a group $G$:

Fix a projective resolution $P \to \mathbb{Z}$ over $\mathbb{Z}G$. Then $P \otimes P \to \mathbb{Z} \otimes \mathbb{Z} = \mathbb{Z}$ is a projective resolution of $\mathbb{Z}$ over $\mathbb{Z}G \otimes \mathbb{Z}G=\mathbb{Z}[G \times G]$. Since the diagonal $$D: G \to G \times G,\;g \mapsto (g,g)$$ is a group homomorphism, $P\otimes P$ can be considered as (acyclic) complex of $\mathbb{Z}G$-modules via $D$. By standard homological algebra there is a $\mathbb{Z}G$-linear map $\Delta: P \to P \otimes P$ (called a diagonal approximation) that extends $id: \mathbb{Z} \to \mathbb{Z}$. Finally, if $M,N$ are $\mathbb{Z}G$-modules, the cup product is defined on cochain level by the morphism

$$\begin{array}{lll} Hom_{\mathbb{Z}G}(P,M) \otimes Hom_{\mathbb{Z}G}(P,N) & \xrightarrow{} & Hom_{\mathbb{Z}(G\times G)}(P\otimes P,M\otimes N) \newline & \xrightarrow{\Delta^\ast} & Hom_{\mathbb{Z}G}(P,M\otimes N) \end{array}$$

Obviously, the same construction can be made with any group homomorphism $G \to G \times G$ in place of $D$.

Question 1: What is the motivation to choose the diagonal $D$ for the definition of the cup product ?

Or, to put it the other way round:

Question 2: What "cup product" do be get if we choose one of the group homomorphisms
$$G \to G \times G,\;g \mapsto (g,1) \quad\text{ or }\quad G \to G \times G,\; g \mapsto (1,g)\;\; ? $$

Answer by TJ for Projective objects in the category of chain complexes

2012-12-05T01:57:08Z

In the following [K] refers to the paper http://dml.cz/bitstream/handle/10338.dmlcz/120545/ActaOstrav_07-1999-1_3.pdf.

That a split exact complex of projectives $(P,d)$ is a projective object can be seen as follows:

$im(d_n)$ is projective since it is a direct summand of $P_{n-1}$
By http://mathoverflow.net/questions/103056/when-is-an-acylic-chain-complex-contractible/103071#103071 a split exact complex is contractible, so $P$ is contractible.
By [K], Lemma 4.4 a contractible complex (like $P$) is isomorphic to the mapping cone of the boundary subcomplex $$ \cdots \to im(d_{n+1}) \xrightarrow{0} im(d_n) \to \cdots$$
By [K], Theorem 3.1, the mapping cone of a complex of projectives with zero differentials is a projective object. Hence $P$ is a projective object by 1. and 3.

Are homogeneous components of f.g graded modules f.g ?

2012-09-18T15:32:29Z

Suppose $M= \bigoplus_{n\in \mathbb Z} M_n$ is a finitely generated graded module over a Noetherian graded commutative ring $A=\bigoplus_{n\in \mathbb Z}A_n$.

If $A$ is positively graded ($A_n=0$ if $n<0$), then each $M_n$ is finitely generated as $A_0$-module (Atiyah, McDonald: Introduction to commutative algebra, beginning of chap. 11).

But what happens, if $A$ is not assumed to be positively graded, like $A= \mathbb{Q}[t,t^{-1}]$ ? Is $M_n\;(n \in \mathbb Z)$ also finitely generated over $A_0$ in this case ?

Can the Bockstein spectral sequence be used to compute cohomology rings ?

2012-09-04T22:17:52Z

If $G$ is a finite group then there is the so-called Bockstein spectral sequence $$E_2^n = H^n(G,\mathbb{F}_p) \Rightarrow \begin{cases} \mathbb{F}_p & n =0 \newline 0 & n>0\end{cases}$$ that can be used to compute the integral cohomology out of the mod-$p$ cohomology. In more detail, each non-zero element of $d_r(E_r^{n-1})\subseteq E_r^n$ corresponds to a direct $\mathbb{Z}/p^r$-summand of $H^n(G,\mathbb{Z})$ (Corollary 5.9.12 in Weibel's Homological Algebra book).

Now my question is whether it is possible to compute the integral cohomology ring of $G$ if the mod-$p$ cohomology rings for the primes dividing $|G|$ and the differentials in the associated Bockstein spectral sequences are known ?

Asymptotics for the coefficients of a rational function

2012-07-30T21:30:07Z

Let $(a_n)$ be a sequence of non-negative real numbers and assume that the resulting power series defines a rational function

$$\sum_{n=0}^\infty a_n x^n = \dfrac{f(x)}{(1-x^{k_1})\cdots (1-x^{k_d})}$$ where $k_1,...,k_d>0$ are integers and $f(x)$ is a real polynomial s.t. $f(1) \neq 0$. It is not hard to show that $$\frac{f(1)}{k_1 \cdots k_d}\le \limsup \frac{a_n}{n^{d-1}} \cdot (d-1)! \le f(1)$$

As a special case we obtain for example $\limsup = f(1)$ if $k_1=\cdots k_d = 1$ (this is in fact not only the limsup but even the limit of the sequence).

Questions: 1) Are there known formulas or better estimates for the $\limsup$ above in terms of $f$ and $k_1,...,k_d$ ?

2) Are there particular techniques, that can be used to obtain good estimates (the one above is simply based on the binomial series for $(1-x)^{-d}$).

Background: Such rational functions occur as Poincaré series of graded Noetherian algebras where $a_n$ is the dimension of the subspace of homogeneous lements of degree $n$. I'm trying to relate this quantity to the rational function.

Which information can be obtained from Poincaré series ?

2012-07-05T17:53:06Z

If $A= \bigoplus_{i\ge 0}A_i$ is a graded commutative Noetherian algebra over a field, its Poincaré series is given by $P(t) = \sum_{i\ge 0} \dim(A_i)t^i$. Although the definition of $P(t)$ only depends on the graded vector space underlying $A$, the Krull dimension of the ring $A$ can be obtained from the Poincaré series as the order of the pole of $P(t)$ at $t=1$.

Question: Are there other information about the ring structure of $A$ that can be obtained from $P(t)$ ?

Since I'm particularly interested in cases where $A$ is the cohomology ring of a finite group, I'm also looking for an example of finite groups whose mod-p cohomology rings are not isomorphic but have identical Poincaré series.

Thanks in advance.

Sign in the product of the LHS spectral sequence

2012-01-09T21:31:03Z

Given an extension of groups $$ 1 \to H \to G \to Q \to 1,$$ there is a spectral sequence $$E^{ip}_2(M) = H^i(Q,H^p(H,M)) \Rightarrow H^{i+p}(G,M).$$ I understand that the composition of the cup products for $Q$ and $H$ defines a pairing $$ E_2^{ip}(M) \otimes E_2^{jq}(N)\hspace{180pt}$$ $$\begin{array}{cl} = & H^i(Q,H^p(H,M) \otimes H^j(Q,H^q(H,M) \newline \xrightarrow[]{\cup_Q} & H^{i+j}(Q,H^p(H,M) \otimes H^q(H,N)) \newline \xrightarrow[]{\cup_H^\ast} & H^{i+j}(Q,H^{p+q}(H,M \otimes N)) \newline = & E_2^{i+j,p+q}(M \otimes N). \end{array}$$ But according to section 7.3 in L. Evens' book (The Cohomology of Groups), the sign $$(-1)^{pj}$$ is needed in order to make this pairing a proper product in the spectral sequence.

Question: Where does this sign come from ?

Why is BG infinite dimensional for G finite ?

2011-11-01T18:18:23Z

If $G \neq \lbrace 1 \rbrace$ is a finite group with classifying space $BG$ then there are infinitely many i such that $H^i(BG,\mathbb{Z}) \neq 0$. This can be found, for example, there:

http://mathoverflow.net/questions/64688/non-vanishing-of-group-cohomology-in-sufficiently-high-degree/64702#64702

As a consequence, the CW-complex $BG$ (unique up to homotopy) can not be of finite dimension.

Question: Are there alternative proofs for this observation. In particular, I would be interested in knowing if there is a purely topological proof without homological algebra.

Examples of Tate cohomology rings

2011-10-16T20:39:23Z

If $G$ is a finite group with periodic cohomology then the Tate cohomology ring can be easily computed to be the localization $\hat{H}^\ast(G,\mathbb{Z}) = H^\ast(G,\mathbb{Z})_{(z)}$ where $z$ is a unit of minimal positive degree. Examples are

Cyclic group:

$\hat{H}^\ast(C_n,\mathbb{Z}) = \mathbb{Z}[z,z^{-1}]/(nz), |z|=2$
Quaternion group:

$\hat{H}^\ast(Q_{2^n},\mathbb{Z}) = \mathbb{Z}[z,z^{-1},a,b]/(2^nz, 2a,2b), |z| =4, |a| = |b| =2$
Binary icosahedral group:

$\hat{H}^\ast(I,\mathbb{Z}) = \mathbb{Z}[z,z^{-1}]/(120z), |z|=4$

The same formular holds in mod-$p$ cohomology, if the cohomology of $G$ is $p$-periodic.

Question: Are there computations of integral or mod-$p$ Tate cohomology rings of finite groups with non-periodic cohomology in the literature ?

Comment by TJ

2013-03-18T19:58:44Z

@Steven: I understand your proof if $R$ is local Artinian, but I think it's false/incomplete if $R$ isn't local (and neither the question nor your answer makes this assumption). The problem is that you suppose $M$ is nilpotent, but this isn't true in general: Take a direct product of fields $R=F\times F$. Then $F\times 0$ is a maximal ideal of $R$ that isn't nilpotent!

Comment by TJ

2013-03-14T06:18:34Z

@Moderators: Why has this answer (along with my question) been downvoted ? I've seen the discussion on downvotes on meta, and perhaps it's possible to find out some details about the downvoter by help of this new sample.

Comment by TJ

2013-03-14T06:12:10Z

Mark, thanks for the answer. Do you have a counter-example for the non-commutative case ?

Comment by TJ

2013-03-14T06:10:55Z

I consider it as left module, because basically all my modules are left modules (unless I really need a right module).

Comment by TJ

2013-03-12T18:23:11Z

@Boris: I don'' t understand the argument in your last comment. Can you please elaborate why $E:= Ext(N,F)\neq 0$ and $E=E\oplus L$ imply $L\neq 0$ ? Also it makes me wonder that you don't need $n=gldim(R)$ in your proof.

Comment by TJ

2013-03-12T14:49:56Z

Then, what is $R^k$ ? Note that $Ext(N,-)$ commutes with direct products but in general not with direct sums.

Comment by TJ

2013-03-12T13:50:36Z

Can you tell where the exercise is taken from ?

Comment by TJ

2013-03-12T13:48:41Z

$F$ need not have finite rank. Therefore it would be better to make clear that $F \cong R^k$ is an additional assumption you made (and not a conclusion from the OP's assumptions).

Comment by TJ

2013-03-10T23:52:23Z

From the OP's question: "It's relatively easy to express this as a sum [...], but can we get a closed formula?"

Comment by TJ

2013-03-09T12:21:33Z

Nice counter-example. Thanks.

Comment by TJ

2013-03-09T12:08:16Z

@Ralph: Nice proof, especially the construction of $r_0$ in the last part is clever.

Comment by TJ

2013-03-05T16:55:09Z

@Todd: You're right. I could also have said: If $S$ is flat as $R$-module (either from the left or the right) then for each $s \in S$ there are $r,r' \neq 0$ in $R$ such that $rsr' \in R$. But I think which version one more likes is a matter of taste.

Comment by TJ

2013-03-04T15:28:19Z

@Manny: No, as explained by Torsten's example, $rs=0$ is possible even if $r,s\neq 0$.

Comment by TJ

2013-02-11T17:17:30Z

Torsten, thanks, you are right, I just need that $S/R$ has no $R$-submodule isomorphic to $R$. But this leads to an even nicer necessary condition.

Comment by TJ

2013-02-07T11:42:35Z

Are you sure it has been downvoted ? The last time I looked in, your answer had 1pt - now it has 3pts.