User simon wadsley - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T07:02:41Z http://mathoverflow.net/feeds/user/345 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130716/continuation-of-homomorphisms-of-representations/130726#130726 Answer by Simon Wadsley for Continuation of homomorphisms of representations... Simon Wadsley 2013-05-15T14:37:46Z 2013-05-15T14:37:46Z <p>You don't explicitly say your representation is complex but I think your example shows that this is the case you're interested in. If so, then $V_0$ has a $G$-invariant complement $V_1$ by Maschke's Theorem <a href="http://en.wikipedia.org/wiki/Maschke%27s_theorem" rel="nofollow">http://en.wikipedia.org/wiki/Maschke%27s_theorem</a>, and the unique linear extension of $f$ whose kernel contains $V_1$ will be a homomorphism of representations $V\to W$ extending $f$. </p> http://mathoverflow.net/questions/121881/reference-request-vector-bundles-in-rigid-analytic-geometry Reference Request: Vector bundles in rigid analytic geometry Simon Wadsley 2013-02-15T10:33:57Z 2013-03-20T23:22:00Z <p>In algebraic geometry it is well-known (see Hartshorne Exercise II.5.16 for example) that there is a 1-1 correspondence between rank $n$ (geometric) vector bundles $\pi\colon Y\to X$ on a scheme $X$ and locally free sheaves of $\mathcal{O}_X$-modules of rank $n$.</p> <p>As I would imagine is well-known to experts in the subject, there is an analogous result in rigid analytic geometry (in the sense of Tate). Here the analogue of a trivial geometric vector bundle of rank $n$ on a rigid $K$-analytic space $X$ is the fibre product $X\times_K \mathbb{A}^{n,an}$ together with its natural projection onto $X$. Here $\mathbb{A}^{n,an}$ denotes the rigid analytic space obtained by gluing together polydiscs of larger and larger radius as in Example 9.3.4.1 of Non-Archimedean analysis by Bosch, Güntzer and Remmert. One can verify that the sections of the projection map $X\times_K \mathbb{A}^{n,an}\to X$ are naturally a free $\mathcal{O}_X(X)$-module of rank $n$. Given this it is not difficult to make a definition of a general geometric vector bundle on rank $n$ on a rigid $K$-analytic space in such a way that the sections of such a bundle form a locally free sheaf of $\mathcal{O}_X$-modules (that is a vector bundle on $X$ in the only sense I can find in the literature). Moreover, as in the algebraic setting this actually defines a 1-1 correspondence between the two notions of vector bundle. </p> <p>My question is whether anyone has written this up in a form that can be easily cited. </p> http://mathoverflow.net/questions/123533/spectrum-theorem-for-p-adic-matrix-analysis/124857#124857 Answer by Simon Wadsley for Spectrum theorem for p-adic matrix analysis Simon Wadsley 2013-03-18T13:24:58Z 2013-03-18T13:24:58Z <p>All three results are true.</p> <p>Beginning with the third, it suffices to show that the set of $n\times n$-matrices in $\mathbb{C}_p$ whose characteristic polynomials have distinct roots is dense since these will certainly be diagonalizable. </p> <p>Now the characteristic polynomial of a matrix may be described as a polynomial of degree $n$ whose coefficients are themselves polynomial in the entries of the matrix (using the usual definition of the characteristic polynomial of $A$ as the determinant of $tI-A$). It follows easily that the (formal) derivative of the characteristic polynomial also has coefficients that are polynomial in the entries of the matrix. Finally using the theory of resultants <a href="http://en.wikipedia.org/wiki/Resultant" rel="nofollow">http://en.wikipedia.org/wiki/Resultant</a> one may find a single polynomial $f$ in the matrix coefficients so that $f(a_{11},a_{12},\ldots,a_{nn})\neq 0$ for $A=(a_{ij})$ if and only if the characteristic polynomial of $A$ and its derivative have no roots in common. This last happens precisely if the characteristic polynomial has distinct roots.</p> <p>To summarise the previous two paragraphs we have seen that there exists a single polynomial $f$ in $n^2$ variables such that a matrix $(a_{ij})$ is diagonalizable if and only if $f(a_{ij})\neq 0$. In fact, this works over any algebraically closed field since $f$ actually has coefficients in the integers. </p> <p>It is now easy to see that $f^{-1}(\mathbb{C}_p\backslash 0)$ is an open dense subset of the $n\times n$ matrices with coefficients in $\mathbb{C}_p$ (it is even Zariski dense) as required. </p> <p>The second follows easily from the third since any neighbourhood of a diagonal matrix will contain an invertible (and diagonal) matrix or from the comments. </p> <p>The first has already been dealt with in the comments.</p> http://mathoverflow.net/questions/121587/polycyclic-group-not-of-type-fp-infty/121591#121591 Answer by Simon Wadsley for Polycyclic group not of type $FP_\infty$ Simon Wadsley 2013-02-12T12:20:33Z 2013-02-12T12:20:33Z <p>Polycyclic groups are certainly of type $FP_\infty$ since the integral group ring $\mathbb{Z}G$ is Noetherian (<a href="http://plms.oxfordjournals.org/content/s3-4/1/419" rel="nofollow">http://plms.oxfordjournals.org/content/s3-4/1/419</a>) and so every finitely generated module over it has a resolution by finitely generated projective modules.</p> <p>I believe it remains an open question as to whether the only groups $G$ with the property that $\mathbb{Z}G$ is Noetherian are those that are polycylic-by-finite </p> http://mathoverflow.net/questions/91610/isomorphic-but-non-conjugate-subgroups-of-gln-mathbbz/91612#91612 Answer by Simon Wadsley for Isomorphic but non-conjugate subgroups of $GL(n,\mathbb{Z})$ ? Simon Wadsley 2012-03-19T11:51:36Z 2012-03-19T13:17:45Z <p>The answer to all three questions is yes and certainly is classical. </p> <p>One simple example is the following:</p> <p>Let $C_2$ act faithfully on the set ${1,2,3,4}$ in two ways. In the first the non-trivial element of $C_2$ swaps 1,2 and also swaps 3,4. In the second the non-trivial element swaps 1,2 and fixes 3,4. </p> <p>Each action defines a representation of $C_2$ on $\mathbb{Z}^4$ via permuation matrices. In one case the trace of the non-trivial permutation matrix is $0$ in the other $2$ so the images cannot be conjugate in $GL_4(\mathbb{Z})$ or $GL_4(\mathbb{R})$ however they both generate a subgroup $C_2$ in the former. </p> <p>It is fairly clear this idea generalises to any isomorphism class of finite groups. </p> http://mathoverflow.net/questions/58094/maximal-ideals-in-formal-laurent-series-rings/58143#58143 Answer by Simon Wadsley for Maximal Ideals in Formal Laurent Series Rings? Simon Wadsley 2011-03-11T11:12:21Z 2011-09-08T09:13:46Z <p>I think that the question difficult as illustrated by Hailong's answer. I suspect that it will be hard to even find a nice parameterisation of the $H$-orbits of maximal ideals in your refined question. Certianly, as Hailong implies there will be infinitely many such orbits.</p> <p>You might find in helpful to consider a valuation theoretic approach. Konstantin Ardakov has done some work on related --- although not quite similar --- questions that I believe he has nearly finished writing up. Maybe he will appear and say more. </p> <p>Edit: For the record Konstantin's work has now appeared here <a href="http://arxiv.org/abs/1108.0371" rel="nofollow">http://arxiv.org/abs/1108.0371</a></p> http://mathoverflow.net/questions/61695/are-there-any-finitely-generated-artinian-modules-that-are-not-notherian/61700#61700 Answer by Simon Wadsley for Are there any finitely generated artinian modules that are not notherian? Simon Wadsley 2011-04-14T13:36:56Z 2011-04-14T14:29:05Z <p>Suppose you have an Artinian but not Noetherian finitely generated $R$ module $M$. Let $0\leq M_1\leq M_2\leq \cdots \leq M_n=M$ be a finite chain of $R$-modules such that each composition factor $M_i/M_{i-1}$ is cyclic for each $i$. </p> <p>Certainly each composition is Artinian since subquotients of Artinian modules are Artinian. Also one of the composition factors must be non-Noetherian since extensions of Noetherian modules by Noetherian modules are Noetherian. Thus, we may assume that $M$ is a <i>cyclic</i> $R$-module. </p> <p>Now if $R$ is commutative, $M$ is a quotient ring $R/I$ which is Artinian as such and so Noetherian also, as you say. </p> <p>If $R$ is non-commutative then I'm not sure what the answer is. <hr> Added: It seems from the wikipedia article linked from the question that Hartley showed that there are cyclic Artinian and non-Noetherian modules over certain non-commutative rings and Cohn gave another construction nearly twenty years later. See the links I give in the comments on the question for precise references.</p> http://mathoverflow.net/questions/59824/use-of-traces-in-physics/59856#59856 Answer by Simon Wadsley for Use of traces in physics Simon Wadsley 2011-03-28T16:01:54Z 2011-03-28T16:01:54Z <p>For a more general notion of the same kind see <a href="http://ncatlab.org/nlab/show/span+trace" rel="nofollow">http://ncatlab.org/nlab/show/span+trace</a></p> http://mathoverflow.net/questions/58826/how-many-semi-direct-products-are-there/58829#58829 Answer by Simon Wadsley for how many semi direct products are there? Simon Wadsley 2011-03-18T12:08:02Z 2011-03-18T12:08:02Z <p>Each Sylow $p$-subgroup of the copy of $\mathbb{Z}/n\mathbb{Z}$ will be normal (actually characteristic) in the semi-direct product. Moreover any element of the semi-direct product not in $\mathbb{Z}/n\mathbb{Z}$ will induce the same automorphism of each Sylow $p$-subgroup since $\mathbb{Z}/n\mathbb{Z}$ is abelian. </p> <p>Thus two of your semi-direct products will be isomorphic precisely if each of the subproducts $\mathbb{Z}/p^r\mathbb{Z}$ by $\mathbb{Z}/2\mathbb{Z}$ are isomorphic where $p^r$ is the maximal power of $p$ dividing $n$.</p> <p>In the square-free case this shows that there are $2^k$ non-isomorphic semi-direct products of the form you require, where $k$ denotes the number of odd prime factors of $n$. </p> <p>In the general case this argument atill reduces you to the case $n$ is a prime power. I guess you'll still get two choices for each odd prime dividing $n$ and two choices for the prime $2$ if $4$ divides $n$ but only one otherwise.</p> http://mathoverflow.net/questions/58448/virtual-chain-conditions-in-groups/58449#58449 Answer by Simon Wadsley for virtual chain conditions in groups Simon Wadsley 2011-03-14T17:29:33Z 2011-03-14T17:29:33Z <p>The virtual DCC doesn't seem so different from the notion of Krull dimension $1$ that I explained in answer to this <a href="http://mathoverflow.net/questions/2525/different-definitions-of-the-dimension-of-an-algebra/2588#2588" rel="nofollow">http://mathoverflow.net/questions/2525/different-definitions-of-the-dimension-of-an-algebra/2588#2588</a> question. </p> http://mathoverflow.net/questions/56929/d-modules-on-rigid-analytic-spaces/56983#56983 Answer by Simon Wadsley for D-modules on rigid analytic spaces Simon Wadsley 2011-03-01T09:49:04Z 2011-03-01T09:49:04Z <p>Yes. Although it is only beginning to be developed.</p> <p>You probably want to start with Berthelot: D-modules arithmétiques I : Opérateurs différentiels de niveau fini and Introduction à la théorie arithmétique des D-modules, and other papers that can be found at <a href="http://perso.univ-rennes1.fr/pierre.berthelot/" rel="nofollow">http://perso.univ-rennes1.fr/pierre.berthelot/</a> Section 5 of the second paper I mentioned is perhaps most relevant.</p> <p>There is also a recent paper of Caro which I cannot find online called 'Holonomie sans structure de Frobenius et criteres d'Holonomie' which removes the necessity of the Frobenius action from Berhelot's work. I suppose he would send you a copy of upon request.</p> <p>Finally, in a piece of shameless self-advertising, Konstantin Ardakov and I recently put a preprint on the arXiv <a href="http://arxiv.org/abs/1102.2606" rel="nofollow">http://arxiv.org/abs/1102.2606</a> part of which seeks to find a framework to further develop the theory. </p> http://mathoverflow.net/questions/49391/how-exotic-can-dvrs-be-in-the-ring-of-rational-functions-over-a-local-field How exotic can DVRs be in the ring of rational functions over a local field? Simon Wadsley 2010-12-14T14:28:48Z 2010-12-15T17:40:32Z <p>Suppose that $R$ is a complete DVR with field of fractions $K$, uniformiser $\pi$ and residue field $k$.</p> <p>Let $B$ be a subring of the ring $K(t)$ of rational functions over $K$. Moreover assume that $B$ is a discrete valuation ring such that $B[\pi^{-1}]=K(t)$. </p> <p>Can the residue field of $B$ be an algebraic extension of $k$? If yes then can it be an infinite algebraic extension of $k$?</p> <p>I am only really interested in the case where $K$ has characteristic $0$ and $k$ has characteristic $p>0$. </p> http://mathoverflow.net/questions/23001/is-there-a-good-account-of-d-affinity-and-localization-theorem-for-partial-flag-v/46146#46146 Answer by Simon Wadsley for Is there a good account of D-affinity and localization theorem for partial flag varieties? Simon Wadsley 2010-11-15T20:09:30Z 2010-11-16T09:21:35Z <p>The answer is now yes, I think </p> <p><a href="http://arxiv.org/PS_cache/arxiv/pdf/1011/1011.0896v2.pdf" rel="nofollow">http://arxiv.org/PS_cache/arxiv/pdf/1011/1011.0896v2.pdf</a></p> <p>Edit: as requested: <a href="http://arxiv.org/abs/1011.0896" rel="nofollow">http://arxiv.org/abs/1011.0896</a></p> http://mathoverflow.net/questions/38984/dimension-of-fibres-of-moment-maps-in-characteristic-p Dimension of fibres of moment maps in characteristic $p$ Simon Wadsley 2010-09-16T15:37:29Z 2010-09-17T08:14:15Z <p>Suppose $G$ is a connected semisimple linear algebraic group with Lie algebra $\mathfrak{g}$ and $X$ is a homogeneous $G$-space with isotropy subgroup $H$ (associated Lie algebra $\mathfrak{h}$) that is normalised by a parabolic subgroup $P$. Write $A=P/H$.</p> <p>Working over $\mathbb{C}$, Borho and Brylinski explain in Proposition 2.8 of "Differential Operators on Homogeneous Spaces I" how to use the moment map $T^\ast X\rightarrow \mathfrak{g}^\ast$ to induce a map from $\pi\colon T^\ast X/A\rightarrow \mathfrak{g}^\ast$. We may also understand this map as a map from the vector bundle $G\times^P\mathfrak{h}^\perp$ to $\mathfrak{g}^\ast$. It is a generalisation of the Springer resolution (which arises in case $H=P$ is a Borel). </p> <p>The same construction can be made over an algebraically closed field of characterstic $p>0$. My question is what is known about the dimensions of the fibres of $\pi$ in this characteristic $p$ case? I am most interested in knowing the largest possible dimension of a fibre over a non-zero point in $\mathfrak{g}^\ast$ in the case where $P$ is a Borel and $H$ is its unipotent radical but more general results where $P$ is any parobolic and $H$ is its unipotent radical are also of interest. I am not interested in the case $H=P$. </p> <p>Added following request for motivation and examples:</p> <p>My motivation comes from a project aiming to understand the possible dimensions (that is the canonical dimension defined for Auslander-Gorenstein rings that corresponds intuitively but not precisely to GK-dimension for almost commutative algebras) of simple modules for the localisation of Iwasawa algebras $\mathbb{Z}_p[[G]]$ at the m.c. set generated by $p$ where $G$ is a compact $p$-adic Lie group of semisimple type. At the moment the answer we have is a simple function of the dimension of these fibres when the group $G$ in the question is the associated Lie group over the algebraic closure of $\mathbb{F}_p$. It would be good to give precise values. </p> <p>We believe that it suffices to know the answer when $P$ is a Borel and $H$ is its unipotent radical which is why I am most interested in this case. </p> http://mathoverflow.net/questions/513/generic-noether-normalisation Generic Noether Normalisation Simon Wadsley 2009-10-14T21:47:31Z 2010-09-15T10:30:25Z <p>Suppose that M is a finitely generated module over A=k[X_1,...,X_n] of Krull dimension m with k an infinite field. Then one version of Noether normalisation says there is an m-dimensional k-subspace W of the k-vector space spanned by X_1,...,X_n such that M is finitely generated over Sym(W) considered as a subring of A. </p> <p>As is surely well-known, in fact one can show that the set of m-dimensional k-vector spaces W that work is open in the appropriate Grassmannian. My question is where is there a reference for this fact in the literature?</p> http://mathoverflow.net/questions/513/generic-noether-normalisation/38802#38802 Answer by Simon Wadsley for Generic Noether Normalisation Simon Wadsley 2010-09-15T10:30:25Z 2010-09-15T10:30:25Z <p>In case anyone else has the same question and discovers this page I have just found a more explicit reference for this result: Remark 3.4.4 of A Singular introduction to commutative algebra by Greuel and Pfister. <a href="http://www.springerlink.com/content/u62645311l0h2256/" rel="nofollow">http://www.springerlink.com/content/u62645311l0h2256/</a> is a page that links to a pdf of the appropriate chapter. It is possibe that a subscription is required to open it though. </p> http://mathoverflow.net/questions/38289/comparing-lower-central-series-and-augmentation-ideal-completions/38318#38318 Answer by Simon Wadsley for Comparing lower central series and augmentation ideal completions Simon Wadsley 2010-09-10T14:43:57Z 2010-09-10T14:43:57Z <p>I don't quite follow your definition of the mod $p$ lower central series as $s$ only seems to appear once in the definition. However whatever it is the answer is no.</p> <p>If $G=\mathbb{Z}$ then the $I$-adic completion of $\mathbb{Z}/p[G]$ is isomorphic to a power series in one variable $T=x-1$ with coefficients in $\mathbb{Z}/p$ --- here $x$ is a generator of $G$. </p> <p>Thus this $I$-adic completion is a commutative Noetherian algebra that is not finitely generated over the base field so cannot be a group algebra of any group since commutativity would imply that the group is abelian and Noetherianity would imply the group has no strictly ascending chains of subgroups. Amongst abelian groups only finitely generated groups have this latter property.</p> http://mathoverflow.net/questions/36568/subtle-counterexample-to-m-neq-n-but-rmrn-for-some-ring-r/36570#36570 Answer by Simon Wadsley for Subtle counterexample to $m\neq n$ but $R^m=R^n$ for some ring $R$ ? Simon Wadsley 2010-08-24T17:42:01Z 2010-08-24T17:42:01Z <p>Yes. I think you are looking for the Leavitt algebras. I don't know much about them but you could start here: <a href="http://arxiv.org/PS_cache/arxiv/pdf/0905/0905.3827v1.pdf" rel="nofollow">http://arxiv.org/PS_cache/arxiv/pdf/0905/0905.3827v1.pdf</a></p> <p>The idea is that of the Leavitt algebras $R=L(1,n)$ is that for these algebras $n$ is smallest natural number bigger than $1$ so that $R\cong R^n$</p> http://mathoverflow.net/questions/27424/a-ring-such-that-all-projectives-are-stably-free-but-not-all-projectives-are-free/27445#27445 Answer by Simon Wadsley for A ring such that all projectives are stably free but not all projectives are free? Simon Wadsley 2010-06-08T08:05:40Z 2010-06-08T08:05:40Z <p>Example 1.2.2 in Chapter 1 of Weibel's book in progress on K-theory <a href="http://www.math.rutgers.edu/~weibel/Kbook.html" rel="nofollow">http://www.math.rutgers.edu/~weibel/Kbook.html</a> says that $R_2$ in the notation of your question has a stably free module that is not free.</p> <p>Intertestingly, further down the page is the "Bass Cancellation Theorem for stably free modules". This says that if $R$ is a commutative Noetherian ring of Krull dimension $d$ then every stably free module of rank $>d$ is free.</p> http://mathoverflow.net/questions/25474/primes-in-a-commutative-jacobson-ring Primes in a (commutative) Jacobson ring Simon Wadsley 2010-05-21T12:12:43Z 2010-05-23T07:23:47Z <p>Recall that a commutative ring is Jacobson if every prime ideal is the intersection of the maximal ideals that contain it. </p> <p>In the exercises of a commutative algebra course I gave I asked the students to show that a commutative ring is Jacobson if and only if every non-maximal prime ideal is the intersection of the prime ideals that strictly contain it. I now suspect that somewhere in the back of my mind I had imposed the condition that the ring should be Noetherian without actually saying this. Of course, Jacobson rings will always have this other property, and the converse is straightforward to prove if there is no strictly ascending chains of prime ideals. But is the result true in general?</p> http://mathoverflow.net/questions/23095/what-are-in-units-of-an-affinoid-algebra What are in units of an affinoid algebra? Simon Wadsley 2010-04-30T11:18:58Z 2010-04-30T17:04:30Z <p>Suppose that $K$ is a complete local field and $A$ is an affinoid $K$-algebra. Is there a known way to produce an explicit description of the units of $A$?</p> <p>Here is what I already know: write $A^\circ$ for the subring consisting of elements of $A$ of norm at most $1$ and $A^{\circ\circ}$ for the ideal of $A^\circ$ consisting of all elements of norm strictly less that $1$. Certainly every element of $1+A^{\circ\circ}$ is a unit in $A$. So is every non-zero element of $K$. Let's call the group generated by these units the group of standard units of $A$. I believe that if the ring $A^\circ/A^{\circ\circ}$ is prime then all units in $A$ are standard. </p> <p>I also know that in general there can be non-standard units. Perhaps an easier question than the one above is `must the group of units of $A$ modulo the group of standard units be finitely generated?'. </p> <p>I have a particular application in mind for a solution to this but I feel that the question is sufficiently interesting in its own right and the application sufficiently distant from the problem that it is not worth explaining it now.</p> <p>Edit: given some of the comments/answers below I probably want to modify my definition of standard units to include any non-zero element of a finite field extension of $K$ inside $A$.</p> <p>Edit 2: thanks for the help so far... I'm actually happy to consider as 'standard' anything in $A^\circ$ that is a unit as an element of $A^\circ$ if that makes things easier. </p> http://mathoverflow.net/questions/4135/relation-between-lie-algebra-cohomology-and-number-of-relations-of-a-cyclic-modul/6170#6170 Answer by Simon Wadsley for Relation between Lie Algebra Cohomology and Number of Relations of a Cyclic Module? Simon Wadsley 2009-11-19T20:55:58Z 2009-11-25T14:11:54Z <p>This looks like the kind of thing that one might be able to prove by filtering $U$ by word-length over $g$ and then passing to the associated graded ring. The idea is that this would reduce the problem to the case that $g$ is abelian, since the associated graded ring is a polynomial ring in $\dim g$ variables and the number of generators of $I$ as a left ideal will be at least as big as the number of generators of $\mathrm{gr}(I)$ its associated graded ideal.</p> <p>I'm not sure whether the invariant dimension will be preserved when you pass to the associated graded though. You might expect it to be as a similar invariant, the grade of the module M, j(M) is preserved when you do this. (Recall that j(M) is the smallest integer j such that $Ext^j_U(M,U)\neq 0$)</p> <p>Edit: Expansion of strategy:</p> <p>Proposition 7.1 of <a href="http://www.informaworld.com/smpp/content~db=all?content=10.1080/00927879808826267" rel="nofollow"> Auslander-Gorenstein Rings by Ajitabh, Smith and Zhang</a> tells us that $\mathrm{Ext}^j_U(M,k)$ is isomorphic to the $k$-vectorspace dual of $\mathrm{Ext}^{d-j}_U(k,M)$ where $d=\dim g$. Thus invariant dimension is $d-min(j|\mathrm{Ext}^j_U(M,k)\neq 0)$. If one can relate the non-vanishing of $\mathrm{Ext}^j_U(M,U)$ and $\mathrm{Ext}^j(M,k)$ then it might be possible to reduce the problem to something relating the grade of $M$ to the number of relations of $M$. This can be done by passing to the associated graded ring.</p> http://mathoverflow.net/questions/6776/how-to-construct-pair-of-adjoint-functors-from-category-a-to-category-a-dcategor/6813#6813 Answer by Simon Wadsley for How to construct pair of adjoint functors from category A to category A_D(category of diagrams) Simon Wadsley 2009-11-25T13:49:55Z 2009-11-25T14:08:52Z <p>There are several pairs of adjoint functors of the kind you desire but it isn't clear to me if any (or all) of them will give you enough projectives in $A^D$.</p> <p>For example for each $d$ in $D$ $\iota_d$ which sends an object of $A$ to the diagram that is $A$ at $d$ and zero elsewhere and does the obvious thing on morphisms is left adjoint to the functor $\pi_d$ that sends a diagram to its value at $d$.</p> <p>Also if $D$ is pointed and $A$ has $D$-colimits then $\mathrm{colim}\colon A^D\rightarrow A$ is left-adjoint to the constant functor that sends $X$ in $A$ to the diagram that is $X$ everywhere and all morphisms are $\mathrm{id}_X$ (see <a href="http://en.wikipedia.org/wiki/Adjoint_functors#Colimits_and_diagonal_functors" rel="nofollow">Wikipedia</a>). </p> <p>These left adjoints will all map projectives to projectives. </p> http://mathoverflow.net/questions/6263/can-an-infinite-conjugacy-class-in-a-group-split-into-more-than-one-conjugacy-cla/6265#6265 Answer by Simon Wadsley for Can an infinite conjugacy class in a group split into more than one conjugacy class in some subgroup of finite index? Simon Wadsley 2009-11-20T13:48:59Z 2009-11-20T13:48:59Z <p>I'm not sure what you mean by an infinite length conjugacy class. Most likely you mean that the cardinality is infinite.</p> <p>Consider the Heisenberg group generated by 3 elements $x,y$ and $z$ with relations so that $z$ is central and $xyx^{-1}y^{-1}=z$. Then the conjugacy class containing $y$ consists of all elements of the form $yz^n$ for integers $n$. </p> <p>If we pass to the finite index subgroup generated by $x^k,y$ and $z$ for some natural number $k$ this splits into $k$ distinct classes represented by $yz^i$ for $i=0,\ldots k-1$.</p> http://mathoverflow.net/questions/5800/generalization-of-the-two-bucket-puzzle/5806#5806 Answer by Simon Wadsley for Generalization of the two bucket puzzle Simon Wadsley 2009-11-17T09:31:35Z 2009-11-17T09:31:35Z <p>Yes. The answer follows from Bezout's theorem which says that given integers A,B and C, C can be written as XA+YB if and only if C is a multiple of the highest common factor of A and B. Euclid's algorithm tells you how to compute X and Y.</p> <p>It is not too hard to see that the only volumes you can get are ones of the that are integer linear combinations of A and B and you can get every positive volume that arises in this way (as long as you have a large enough additional container to store it all).</p> http://mathoverflow.net/questions/4699/examples-of-left-reversible-semigroups/4850#4850 Answer by Simon Wadsley for Examples of left reversible semigroups Simon Wadsley 2009-11-10T12:51:37Z 2009-11-10T12:51:37Z <p>I suppose that the non-zero elements of a left Ore domain would work --- presumably this is why they are sometimes called Ore semigroups.</p> <p>To expand: a ring is a left Ore domain if it has no non-trivial zero-divisors and for every non-zero element s of the ring and every other element r of the ring one can find r' in the ring and s' non-zero and in the ring such that rs'=sr'. </p> <p>Goldie's theorem says every left Noetherian ring without zero-divisors is a left Ore domain so the non-zero elements will form a left-reversible cancellative semigroup. In fact Goldie's theorem says a little more than this but I don't have time to check if the non-zero divisors will always give what you want in any left Goldie ring. </p> <p>(It is possible I have my left and rights mixed-up here if so just swap them around!). </p> http://mathoverflow.net/questions/4827/how-does-one-identify-properties-of-objects-with-good-inheritance/4833#4833 Answer by Simon Wadsley for How does one identify properties of objects with good "inheritance"? Simon Wadsley 2009-11-10T09:31:30Z 2009-11-10T09:31:30Z <p>My interests largely lie in ring theory and related areas so I only feel qualified to comment on part of this question and I would be interested to be corrected on what I say, but it seems to me that Hilbert's basis theorem really is a genuine piece of ring theory rather than something that should fit into some more general framework of inheritance properties. </p> <p>It is true that there are some facts about Noetherian rings like "a quotient of a Noetherian ring is Noetherian", "the localisation of a Noetherian ring is Noetherian", even "an algebra over a Noetherian ring that is finitely generated as a module over the base ring is Noetherian" that do seem to be part of some wider framework of the kind you seem to be looking for. But Hilbert's basis theorem is different. I think that we should think of it as a surprising fact that we happen to be able to prove and makes the theory of Noetherian rings interesting rather than something we should have expected a priori because the Noetherian hypothesis is well chosen. </p> http://mathoverflow.net/questions/2525/different-definitions-of-the-dimension-of-an-algebra/2588#2588 Answer by Simon Wadsley for Different definitions of the dimension of an algebra Simon Wadsley 2009-10-26T09:05:24Z 2009-10-26T09:05:24Z <p>Often the most useful dimension in non-commutative algebra is the length of the minimal injective resolution of the ring as a module over itself. In many important cases this is the same as the global dimension when the latter is finite, but it is more robust in that it is finite more often. In a commutative Noetherian ring it is the same as the Krull dimension when it is finite.</p> http://mathoverflow.net/questions/2525/different-definitions-of-the-dimension-of-an-algebra/2586#2586 Answer by Simon Wadsley for Different definitions of the dimension of an algebra Simon Wadsley 2009-10-26T08:30:38Z 2009-10-26T08:30:38Z <p>In non-commutative algebra Krull dimension has been generalised by Gabriel &amp; Rentschler. A decent account of it can be found in Chapter 6 of McConnell and Robson's book on non-commutative Noetherian rings. </p> <p>The basic idea is as follows: An artinian module has Krull dimension 0. </p> <p>A module that does not have Krull dimension 0 has Krull dimension 1 if in every infinite descending chain of submodules all but finitely many composition factors have Krull dimension 0. </p> <p>A module that does not have Krull dimension 0 or 1 has Krull dimension 2 if in every infinite descending chain of submodules all but finitely many composition factors have Krull dimension 0 or 1.</p> <p>The definition continues for all finite ordinals (and can be extended to all ordinals). Then the Krull dimension of a ring R is the Krull dimension of R as a module over itself.</p> http://mathoverflow.net/questions/1939/separable-and-fin-gen-projective-but-not-frobenius/2070#2070 Answer by Simon Wadsley for Separable and Fin. Gen. Projective but not Frobenius? Simon Wadsley 2009-10-23T11:18:20Z 2009-10-23T11:18:20Z <p>Theorem 4.2 of <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.ojm/1200691953" rel="nofollow">this paper</a> says that the answer is always yes.</p> http://mathoverflow.net/questions/130716/continuation-of-homomorphisms-of-representations/130726#130726 Comment by Simon Wadsley Simon Wadsley 2013-05-16T09:40:19Z 2013-05-16T09:40:19Z That's right. http://mathoverflow.net/questions/128678/motivating-geometric-representation-theory Comment by Simon Wadsley Simon Wadsley 2013-04-25T07:55:43Z 2013-04-25T07:55:43Z Kazhdan--Lusztig conjectures? <a href="http://en.wikipedia.org/wiki/Kazhdan%E2%80%93Lusztig_polynomial" rel="nofollow">en.wikipedia.org/wiki/&hellip;</a> http://mathoverflow.net/questions/125725/another-reference-request-about-dualizing-sheaves-for-nodal-surfaces Comment by Simon Wadsley Simon Wadsley 2013-03-27T14:24:39Z 2013-03-27T14:24:39Z Why should the age of a paper harm its quality as a reference? http://mathoverflow.net/questions/125075/hochschild-cohomology-and-kahler-differentials Comment by Simon Wadsley Simon Wadsley 2013-03-20T14:49:36Z 2013-03-20T14:49:36Z <a href="http://ncatlab.org/nlab/show/Hochschild-Kostant-Rosenberg+theorem" rel="nofollow">ncatlab.org/nlab/show/&hellip;</a> http://mathoverflow.net/questions/125018/classification-of-hopf-algebra-with-exactly-two-1-dimensional-modules Comment by Simon Wadsley Simon Wadsley 2013-03-20T11:12:50Z 2013-03-20T11:12:50Z Not really engaging with your question, but a family of examples over a field $k$ of characteristic $p$ (for $p$ odd) is the group algebra $kG$ for $G$ any group of order $2p^n$. The Sylow $p$-subgroup of $G$ must be normal as it has index $2$ and must act trivially on any simple module. Thus the simple modules factor through $kC_2$ which obviously has two (1-dimensional) simple modules. http://mathoverflow.net/questions/125018/classification-of-hopf-algebra-with-exactly-two-1-dimensional-modules Comment by Simon Wadsley Simon Wadsley 2013-03-19T21:21:34Z 2013-03-19T21:21:34Z Over $\mathbb{C}$? http://mathoverflow.net/questions/124419/projectives-in-the-category-of-discrete-g-modules/124426#124426 Comment by Simon Wadsley Simon Wadsley 2013-03-13T15:40:57Z 2013-03-13T15:40:57Z Perhaps even easier, take $G$ to be $\mathbb{Z}_p$, the $p$-adic integers. Then a discrete $G$-module is a $p$-torsion abelian group, etc. http://mathoverflow.net/questions/121187/reference-for-rigid-analytic-gaga/121215#121215 Comment by Simon Wadsley Simon Wadsley 2013-02-21T13:53:24Z 2013-02-21T13:53:24Z Thanks. It wasn't so much that I would expect them to be there as that I thought it plausible that they might be. http://mathoverflow.net/questions/121881/reference-request-vector-bundles-in-rigid-analytic-geometry/122470#122470 Comment by Simon Wadsley Simon Wadsley 2013-02-21T09:03:13Z 2013-02-21T09:03:13Z Looking again I guess you're meaning Proposition 4.7.2(1) which does point in the right direction in that it explains the relationship between the locally free sheaf and the gluing data of the trivial geometric sub-bundles. I'd like something more explicit though. http://mathoverflow.net/questions/121881/reference-request-vector-bundles-in-rigid-analytic-geometry/122470#122470 Comment by Simon Wadsley Simon Wadsley 2013-02-21T08:57:26Z 2013-02-21T08:57:26Z Sadly not. They do talk about locally free sheaves and call them vector bundles but they don't discuss what I call geometric vector bundles in my question. http://mathoverflow.net/questions/121187/reference-for-rigid-analytic-gaga/121215#121215 Comment by Simon Wadsley Simon Wadsley 2013-02-20T22:08:29Z 2013-02-20T22:08:29Z Could I ask whether this paper of K&#246;pf answers my question <a href="http://mathoverflow.net/questions/121881/reference-request-vector-bundles-in-rigid-analytic-geometry" rel="nofollow" title="reference request vector bundles in rigid analytic geometry">mathoverflow.net/questions/121881/&hellip;</a>? It doesn't seem to be so easy to obtain a copy. http://mathoverflow.net/questions/121881/reference-request-vector-bundles-in-rigid-analytic-geometry Comment by Simon Wadsley Simon Wadsley 2013-02-19T18:10:35Z 2013-02-19T18:10:35Z No, I believe that there are no extra difficulties once you have proved that the sections of the trivial geometric vector bundle naturally form a free module over the global sections. A citeable reference to a general statement for ringed spaces (when the base space is equipped with a Grothendieck rather than usual topology) would also be appreciated if one does not exist for this specific case --- the fewer things I would have to explicitly check the better. http://mathoverflow.net/questions/2525/different-definitions-of-the-dimension-of-an-algebra/2588#2588 Comment by Simon Wadsley Simon Wadsley 2013-01-13T12:40:27Z 2013-01-13T12:40:27Z I see 'self injective dimension' more often, I think. But yes, it is just the injective dimension of the ring as a module over itself. http://mathoverflow.net/questions/112564/finitely-generated-resolutions Comment by Simon Wadsley Simon Wadsley 2012-11-16T10:33:25Z 2012-11-16T10:33:25Z $R$ (left) Noetherian is both necessary and sufficient: if $R$ has a non-f.g. left ideal $I$ then the kernel of $R\to R/I$ will be $I$ and so not be finitely generated. The converse is not difficult. http://mathoverflow.net/questions/107237/what-is-this-deformed-group-algebra-named Comment by Simon Wadsley Simon Wadsley 2012-09-15T07:12:29Z 2012-09-15T07:12:29Z $G$-graded ring? <a href="http://en.wikipedia.org/wiki/Graded_algebra#G-graded_rings_and_algebras" rel="nofollow">en.wikipedia.org/wiki/&hellip;</a>