User tj - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T18:28:19Z http://mathoverflow.net/feeds/user/18571 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/124462/properties-of-ring-epimorphisms-that-are-true-only-over-commutative-rings Properties of ring epimorphisms that are true only over commutative rings TJ 2013-03-13T23:03:54Z 2013-03-14T09:26:24Z <p>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.</p> <p>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. </p> http://mathoverflow.net/questions/121468/epimorphisms-and-free-submodules Epimorphisms and free submodules TJ 2013-02-11T13:21:57Z 2013-03-10T01:36:10Z <p>By inspecting the accepted answer to this question </p> <p><a href="http://mathoverflow.net/questions/120918/are-epimorphisms-from-a-division-ring-isomorphisms" rel="nofollow">http://mathoverflow.net/questions/120918/are-epimorphisms-from-a-division-ring-isomorphisms</a></p> <p>one obtains the following necessary condition for epimorphisms: </p> <blockquote> <p>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 $rs \in R$. </p> </blockquote> <p><em>Proof:</em> 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. $\blacksquare$</p> <p>However, without success I tried to drop the flatness condition. Whence my question: </p> <p><strong>Question:</strong> Is the statement above still valid, if $S$ is not supposed to be flat as right $R$-module ? </p> http://mathoverflow.net/questions/120918/are-epimorphisms-from-a-division-ring-isomorphisms Are epimorphisms from a division ring isomorphisms ? TJ 2013-02-06T00:07:05Z 2013-02-10T10:58:07Z <p>According to Corollary 1.2(3) of the paper Silver: Noncommutative Localizations and Applications. J. of Alg. 7(1964), 44-67: </p> <p>If $R$ is a (commutative) field and $\alpha: R \to S$ an epimorphism in the category of rings, then $\alpha$ is an isomorphism.</p> <p><strong>Question:</strong> Is $\alpha$ also an isomorphism if $R$ is not a field but just a division ring ? </p> <p>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: </p> <p>"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 $[R[x]:R] &lt; \infty$, 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." </p> http://mathoverflow.net/questions/116490/why-do-we-use-the-diagonal-for-diagonal-approximations Why do we use the diagonal for diagonal approximations ? TJ 2012-12-15T22:54:19Z 2012-12-16T01:47:44Z <p>First recall how the cup product is defined for the cohomology of a group $G$: </p> <p>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 <em>diagonal approximation</em>) that extends $id: \mathbb{Z} \to \mathbb{Z}$. Finally, if $M,N$ are $\mathbb{Z}G$-modules, the <em>cup product</em> is defined on cochain level by the morphism </p> <p>$$\begin{array}{lll} Hom_{\mathbb{Z}G}(P,M) \otimes Hom_{\mathbb{Z}G}(P,N) &amp; \xrightarrow{} &amp; Hom_{\mathbb{Z}(G\times G)}(P\otimes P,M\otimes N) \newline &amp; \xrightarrow{\Delta^\ast} &amp; Hom_{\mathbb{Z}G}(P,M\otimes N) \end{array}$$</p> <hr> <p>Obviously, the same construction can be made with any group homomorphism $G \to G \times G$ in place of $D$. </p> <p><strong>Question 1:</strong> What is the motivation to choose the diagonal $D$ for the definition of the cup product ? </p> <p>Or, to put it the other way round: </p> <p><strong>Question 2:</strong> What "cup product" do be get if we choose one of the group homomorphisms<br> $$G \to G \times G,\;g \mapsto (g,1) \quad\text{ or }\quad G \to G \times G,\; g \mapsto (1,g)\;\; ?$$ </p> http://mathoverflow.net/questions/115449/projective-objects-in-the-category-of-chain-complexes/115460#115460 Answer by TJ for Projective objects in the category of chain complexes TJ 2012-12-05T01:57:08Z 2012-12-05T01:57:08Z <p>In the following [K] refers to the paper <a href="http://dml.cz/bitstream/handle/10338.dmlcz/120545/ActaOstrav_07-1999-1_3.pdf" rel="nofollow">http://dml.cz/bitstream/handle/10338.dmlcz/120545/ActaOstrav_07-1999-1_3.pdf</a>. </p> <p>That a split exact complex of projectives $(P,d)$ is a projective object can be seen as follows: </p> <ol> <li><p>$im(d_n)$ is projective since it is a direct summand of $P_{n-1}$ </p></li> <li><p>By <a href="http://mathoverflow.net/questions/103056/when-is-an-acylic-chain-complex-contractible/103071#103071" rel="nofollow">http://mathoverflow.net/questions/103056/when-is-an-acylic-chain-complex-contractible/103071#103071</a> a split exact complex is contractible, so $P$ is contractible. </p></li> <li><p>By [K], Lemma 4.4 a contractible complex (like $P$) is isomorphic to the mapping cone of the boundary subcomplex $$\cdots \to im(d_{n+1}) \xrightarrow{0} im(d_n) \to \cdots$$</p></li> <li><p>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. </p></li> </ol> http://mathoverflow.net/questions/107478/are-homogeneous-components-of-f-g-graded-modules-f-g Are homogeneous components of f.g graded modules f.g ? TJ 2012-09-18T15:32:29Z 2012-09-18T22:32:49Z <p>Suppose $M= \bigoplus_{n\in \mathbb Z} M_n$ is a finitely generated graded module over a Noetherian graded commutative ring $A=\bigoplus_{n\in \mathbb Z}A_n$. </p> <p>If $A$ is positively graded ($A_n=0$ if $n&lt;0$), then each $M_n$ is finitely generated as $A_0$-module (Atiyah, McDonald: Introduction to commutative algebra, beginning of chap. 11). </p> <p>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 ? </p> http://mathoverflow.net/questions/106386/can-the-bockstein-spectral-sequence-be-used-to-compute-cohomology-rings Can the Bockstein spectral sequence be used to compute cohomology rings ? TJ 2012-09-04T22:17:52Z 2012-09-04T22:17:52Z <p>If $G$ is a finite group then there is the so-called Bockstein spectral sequence $$E_2^n = H^n(G,\mathbb{F}_p) \Rightarrow \begin{cases} \mathbb{F}_p &amp; n =0 \newline 0 &amp; n>0\end{cases}$$ 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). </p> <p>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 ? </p> http://mathoverflow.net/questions/103553/asymptotics-for-the-coefficients-of-a-rational-function Asymptotics for the coefficients of a rational function TJ 2012-07-30T21:30:07Z 2012-08-08T19:57:56Z <p>Let $(a_n)$ be a sequence of non-negative real numbers and assume that the resulting power series defines a rational function </p> <p>$$\sum_{n=0}^\infty a_n x^n = \dfrac{f(x)}{(1-x^{k_1})\cdots (1-x^{k_d})}$$ 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 $$\frac{f(1)}{k_1 \cdots k_d}\le \limsup \frac{a_n}{n^{d-1}} \cdot (d-1)! \le f(1)$$</p> <p>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). </p> <p><strong>Questions:</strong> 1) Are there known formulas or better estimates for the $\limsup$ above in terms of $f$ and $k_1,...,k_d$ ? </p> <p>2) Are there particular techniques, that can be used to obtain good estimates (the one above is simply based on the binomial series for $(1-x)^{-d}$). </p> <p><strong>Background:</strong> 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. </p> http://mathoverflow.net/questions/101422/which-information-can-be-obtained-from-poincare-series Which information can be obtained from Poincaré series ? TJ 2012-07-05T17:53:06Z 2012-07-07T08:56:12Z <p>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 $t=1$.</p> <p><strong>Question:</strong> Are there other information about the ring structure of $A$ that can be obtained from $P(t)$ ? </p> <p>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. </p> <p>Thanks in advance.</p> http://mathoverflow.net/questions/85291/sign-in-the-product-of-the-lhs-spectral-sequence Sign in the product of the LHS spectral sequence TJ 2012-01-09T21:31:03Z 2012-01-10T18:27:59Z <p>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 $$E_2^{ip}(M) \otimes E_2^{jq}(N)\hspace{180pt}$$ $$\begin{array}{cl} = &amp; H^i(Q,H^p(H,M) \otimes H^j(Q,H^q(H,M) \newline \xrightarrow[]{\cup_Q} &amp; H^{i+j}(Q,H^p(H,M) \otimes H^q(H,N)) \newline \xrightarrow[]{\cup_H^\ast} &amp; H^{i+j}(Q,H^{p+q}(H,M \otimes N)) \newline = &amp; E_2^{i+j,p+q}(M \otimes N). \end{array}$$ 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. </p> <p><em>Question:</em> Where does this sign come from ? </p> http://mathoverflow.net/questions/79741/why-is-bg-infinite-dimensional-for-g-finite Why is BG infinite dimensional for G finite ? TJ 2011-11-01T18:18:23Z 2011-11-17T05:54:16Z <p>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: </p> <p><a href="http://mathoverflow.net/questions/64688/non-vanishing-of-group-cohomology-in-sufficiently-high-degree/64702#64702" rel="nofollow">http://mathoverflow.net/questions/64688/non-vanishing-of-group-cohomology-in-sufficiently-high-degree/64702#64702</a></p> <p>As a consequence, the CW-complex $BG$ (unique up to homotopy) can not be of finite dimension. </p> <p>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.</p> http://mathoverflow.net/questions/78287/examples-of-tate-cohomology-rings Examples of Tate cohomology rings TJ 2011-10-16T20:39:23Z 2011-10-19T12:13:42Z <p>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 </p> <ul> <li><p>Cyclic group: </p> <p>$\hat{H}^\ast(C_n,\mathbb{Z}) = \mathbb{Z}[z,z^{-1}]/(nz), |z|=2$</p></li> <li><p>Quaternion group:</p> <p>$\hat{H}^\ast(Q_{2^n},\mathbb{Z}) = \mathbb{Z}[z,z^{-1},a,b]/(2^nz, 2a,2b), |z| =4, |a| = |b| =2$ </p></li> <li><p>Binary icosahedral group: </p> <p>$\hat{H}^\ast(I,\mathbb{Z}) = \mathbb{Z}[z,z^{-1}]/(120z), |z|=4$ </p></li> </ul> <p>The same formular holds in mod-$p$ cohomology, if the cohomology of $G$ is $p$-periodic. </p> <p><em>Question:</em> Are there computations of integral or mod-$p$ Tate cohomology rings of finite groups with non-periodic cohomology in the literature ? </p> http://mathoverflow.net/questions/101546/epimorphisms-with-artinian-domain/101552#101552 Comment by TJ TJ 2013-03-18T19:58:44Z 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! http://mathoverflow.net/questions/124462/properties-of-ring-epimorphisms-that-are-true-only-over-commutative-rings/124471#124471 Comment by TJ TJ 2013-03-14T06:18:34Z 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. http://mathoverflow.net/questions/124462/properties-of-ring-epimorphisms-that-are-true-only-over-commutative-rings/124471#124471 Comment by TJ TJ 2013-03-14T06:12:10Z 2013-03-14T06:12:10Z Mark, thanks for the answer. Do you have a counter-example for the non-commutative case ? http://mathoverflow.net/questions/124462/properties-of-ring-epimorphisms-that-are-true-only-over-commutative-rings Comment by TJ TJ 2013-03-14T06:10:55Z 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). http://mathoverflow.net/questions/124294/question-on-an-exercise-on-homological-algebra/124299#124299 Comment by TJ TJ 2013-03-12T18:23:11Z 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. http://mathoverflow.net/questions/124294/question-on-an-exercise-on-homological-algebra/124299#124299 Comment by TJ TJ 2013-03-12T14:49:56Z 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. http://mathoverflow.net/questions/124294/question-on-an-exercise-on-homological-algebra Comment by TJ TJ 2013-03-12T13:50:36Z 2013-03-12T13:50:36Z Can you tell where the exercise is taken from ? http://mathoverflow.net/questions/124294/question-on-an-exercise-on-homological-algebra/124299#124299 Comment by TJ TJ 2013-03-12T13:48:41Z 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). http://mathoverflow.net/questions/124112/putting-objects-into-boxes-so-that-each-box-gets-2-objects/124185#124185 Comment by TJ TJ 2013-03-10T23:52:23Z 2013-03-10T23:52:23Z From the OP's question: &quot;It's relatively easy to express this as a sum [...], but can we get a closed formula?&quot; http://mathoverflow.net/questions/121468/epimorphisms-and-free-submodules/123539#123539 Comment by TJ TJ 2013-03-09T12:21:33Z 2013-03-09T12:21:33Z Nice counter-example. Thanks. http://mathoverflow.net/questions/121468/epimorphisms-and-free-submodules/123579#123579 Comment by TJ TJ 2013-03-09T12:08:16Z 2013-03-09T12:08:16Z @Ralph: Nice proof, especially the construction of $r_0$ in the last part is clever. http://mathoverflow.net/questions/121468/epimorphisms-and-free-submodules/123579#123579 Comment by TJ TJ 2013-03-05T16:55:09Z 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. http://mathoverflow.net/questions/121468/epimorphisms-and-free-submodules Comment by TJ TJ 2013-03-04T15:28:19Z 2013-03-04T15:28:19Z @Manny: No, as explained by Torsten's example, $rs=0$ is possible even if $r,s\neq 0$. http://mathoverflow.net/questions/121468/epimorphisms-and-free-submodules Comment by TJ TJ 2013-02-11T17:17:30Z 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. http://mathoverflow.net/questions/120918/are-epimorphisms-from-a-division-ring-isomorphisms/120921#120921 Comment by TJ TJ 2013-02-07T11:42:35Z 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.