User pete l. clark - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T17:26:38Z http://mathoverflow.net/feeds/user/1149 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25758/seeking-noetherian-normal-domain-with-vanishing-picard-group-but-not-a-ufd Seeking Noetherian normal domain with vanishing Picard group but not a UFD Pete L. Clark 2010-05-24T09:32:37Z 2013-05-04T03:15:36Z <p>Once again, the question says it all.</p> <p>My motivation is the article on factorization I am writing. I want to explain (as well as to understand!) why for normal Noetherian domains of dimension greater than one, the obstruction to factoriality is the nonvanishing of the (Weil) divisor class group $\operatorname{Cl}(R)$, not the Picard group $\operatorname{Pic}(R)$ (equivalently, the Cartier divisor class group).</p> <p>My understanding is that it is equivalent to find a normal Noetherian domain with vanishing Picard group which is not locally factorial. I would be especially happy to see an example among affine domains, i.e., in which the domain is finitely generated as an algebra over some field. </p> http://mathoverflow.net/questions/128194/upper-bound-on-order-of-finite-subgroups-of-gl-nz-p/128201#128201 Answer by Pete L. Clark for Upper bound on order of finite subgroups of GL_n(Z_p)? Pete L. Clark 2013-04-21T00:54:52Z 2013-04-21T02:31:21Z <p>Yes, for any fixed $p$-adic field $K$ the supremum of orders of finite subgroups of $\operatorname{GL}_n(K)$ is finite and can be explicitly bounded above. There is a beautiful discussion of this tucked away somewhere in Serre's <em>Lie Algebras and Lie Groups</em>. (What I say in the following is almost entirely derived from that, so you would do at least as well just to go back to the source.)</p> <p>The first observation is that the answer is the same or $K$ as for its ring of integers $R$: a simple compactness argument shows that any finite subgroup of $\operatorname{GL}_n(K)$ can be conjugated into $\operatorname{GL}_n(R)$. Then the idea is to show that for sufficiently large $k$, there is no torsion in the kernel of the natural surjective map $\operatorname{GL}_n(R) \rightarrow \operatorname{GL}_n(R/\pi^k R)$: this comes down to analysis of torsion in formal groups, which explains why it shows up in Serre's <em>LALG</em>. </p> <p>In fact Chapter 4 <a href="http://math.uga.edu/~pete/8410FULL.pdf" rel="nofollow">of these notes</a> is devoted to precisely this and closely related questions, for instance including a proof of <strong>Selberg's Theorem</strong>: if $K$ is any field of characteristic zero and $G$ is a finitely generated subgroup of $\operatorname{GL}_n(K)$, then $G$ has a finite index torsionfree subgroup. But the proof here is kind of antithetical to your precise question: one reduces to the case in which $K$ itself is finitely generated and applies a theorem of Cassels: any such field can be embedded in $\mathbb{Q}_p$ for some odd $p$. </p> <p>My notes give (standard) explicit upper bounds in the case of $\operatorname{GL}_n(\mathbb{Q}_p)$. </p> <p>Okay, but wait: I know how to do the general case too. You want to combine the above arguments with:</p> <blockquote> <p>Proposition: Let $K$ be a finite extension of $\mathbb{Q}_p$ with ramification index $e$, and let $R$ be its ring of integers, with maximal ideal $\mathfrak{m}$. Let $F(X,Y)$ be any formal group law (of any finite dimension; here $X$ and $Y$ are vector variables) over $R$, with associated "standard" $K$-analytic Lie group $G^1 = F(\mathfrak{m})$. Then the exponent of any finite subgroup of $G^1$ divides $p^{\gamma_p(e)}$, where for any $m \in \mathbb{Z}^+$, $\gamma_p(m) = \lfloor \log_p \left( \frac{pm}{p-1} \right) \rfloor$.</p> </blockquote> <p>(This is Proposition 9 from <a href="http://math.uga.edu/~pete/cx6.pdf" rel="nofollow">this paper</a> of myself and Xavier Xarles. It was well known to just about anyone who had worked in the area, but we couldn't find it in the literature, and in fact our paper was cited at least once for precisely this result.)</p> <p>So I think that this does exactly what you want: let me know if I'm mistaken. </p> <p>Finally, I find it striking that the answer is completely different for local fields of positive characteristic: it is not possible to bound torsion in formal groups in this setting -- somehow $e = \infty$ in the above setup -- and Selberg's Theorem is false. If I am not mistaken, there are indeed arbitrarily large finite subgroups of $\operatorname{GL}_n(\mathbb{F}_q((t)))$ for all prime powers $q$ and all $n \geq 2$. </p> http://mathoverflow.net/questions/9667/what-are-some-results-in-mathematics-that-have-snappy-proofs-using-model-theory What are some results in mathematics that have snappy proofs using model theory? Pete L. Clark 2009-12-24T09:16:01Z 2013-04-12T13:16:33Z <p>I am preparing to teach a short course on "applied model theory" at UGA this summer. To draw people in, I am looking to create a BIG LIST of results in mathematics that have nice proofs using model theory. (I do not require that model theory be the first or only proof of the result in question.)</p> <p>I will begin with some examples of my own (the attribution is for the model-theoretic proof, not the result itself).</p> <p>1) An injective regular map from a complex variety to itself is surjective (Ax).</p> <p>2) The projection of a constructible set is constructible (Tarski).</p> <p>3) Solution of Hilbert's 17th problem (Tarski?).</p> <p>4) p-adic fields are "almost C_2" (Ax-Kochen).</p> <p>5) "Almost" every rationally connected variety over Q_p^{unr} has a rational point (Duesler-Knecht).</p> <p>6) Mordell-Lang in positive characteristic (Hrushovski).</p> <p>7) Nonstandard analysis (Robinson).</p> <p>[But better would be: a particular result in analysis which has a snappy nonstandard proof.]</p> <p><b>Added</b>: The course was given in July of 2010. So far as I am concerned, it went well. If you are interested, the notes are available at</p> <p><a href="http://www.math.uga.edu/~pete/MATH8900.html" rel="nofollow">http://www.math.uga.edu/~pete/MATH8900.html</a></p> <p>Thanks to everyone who answered the question. I enjoyed and learned from all of the answers, even though (unsurprisingly) many of them could not be included in this introductory half-course. I am still interested in hearing about snappy applications of model theory, so further answers are most welcome. </p> http://mathoverflow.net/questions/127038/example-of-an-overring-of-an-integral-domain-which-is-not-a-ring-of-quotients/127042#127042 Answer by Pete L. Clark for Example of an overring of an integral domain which is not a ring of quotients? Pete L. Clark 2013-04-10T05:19:58Z 2013-04-10T05:19:58Z <p>In $\S 22.2.2$ of <a href="http://math.uga.edu/~pete/integral.pdf" rel="nofollow">my commutative algebra notes</a> I discuss overrings of Dedekind domains. In particular I discuss a beautiful theorem of Oscar Goldman: in a Dedekind domain $R$, every fractional ideal has some positive integer power which is principal -- in other words $\operatorname{Pic} R$ is a torsion group -- if and only if every overring is a localization. </p> <p>The proof is quite explicit: if $\operatorname{Pic} R$ is not torsion, there is some prime ideal $\mathfrak{p}$ of $R$ no power of which is principal, and then the argument shows that $R^{\mathfrak{p}} = \bigcap_{\mathfrak{q} \in \operatorname{MaxSpec} R, \ \mathfrak{q} \neq \mathfrak{p}} R_{\mathfrak{q}}$ is not a localization: indeed, it is a proper overring of $R$ with the same group of units.</p> http://mathoverflow.net/questions/15366/which-journals-publish-expository-work Which journals publish expository work? Pete L. Clark 2010-02-15T21:24:03Z 2013-03-29T23:50:23Z <p>I wonder if anyone else has noticed that the market for expository papers in mathematics is very narrow (more so than it used to be, perhaps). </p> <p>Are there any journals which publish expository work, especially at the "intermediate" level? By intermediate, I mean neither (i) aimed at an audience of students, especially undergraduate students (e.g. Mathematics Magazine) nor (ii) surveys of entire fields of mathematics and/or descriptions of spectacular new results written by veteran experts in the field (e.g. the Bulletin, the Notices).</p> <p>Let me give some examples from my own writing, mostly just to fix ideas. (I do <em>not</em> mean to complain.)</p> <p>1) About six years ago I submitted an expository paper "On the discrete geometry of Chicken McNuggets" to the American Mathematical Monthly. The point of the paper was to illustrate the utility of simple reasoning about lattices in Euclidean space to give a proof of Schur's Theorem on the number of representations of an integer by a linear form in positive integers. The paper was rejected; one reviewer said something like (I paraphrase) "I have the feeling that this would be a rather routine result for someone versed in the geometry of numbers." This shows that the paper was not being viewed as expository -- i.e., a work whose goal is the presentation of a known result in a way which will make it accessible and appealing to a broader audience. I shared the news with my officemate at the time, Dr. Gil Alon, and he found the topic interesting. Together we "researchized" the paper by working a little harder and proving some (apparently) new exact formulas for representation numbers. This new version was accepted by the Journal of Integer Sequences:</p> <p><a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL8/Clark/clark80.html" rel="nofollow">http://www.cs.uwaterloo.ca/journals/JIS/VOL8/Clark/clark80.html</a></p> <p>This is not a sad story for me overall because I learned more about the problem ("The Diophantine Problem of Frobenius") in writing the second version with Gil. But still, something is lost: the first version was a writeup of a talk that I have given to advanced undergraduate / basic graduate audiences at several places. For a long time, this was my "general audience" talk, and <em>it worked</em> at getting people involved and interested: people always came up to me afterward with further questions and suggested improvements, much more so than any arithmetic geometry talk I have ever given. The main result in our JIS paper is unfortunately a little technical [not deep, not sophisticated; just technical: lots of gcd's and inverses modulo various things] to state, and although I have recommended to several students to read this paper, so far nothing has come of it. </p> <p>2) A few years ago I managed to reprove a theorem of Luther Claborn (every abelian group is isomorphic to the class group of some Dedekind domain) by using elliptic curves along the lines of a suggestion by Michael Rosen (who reproved the result in the countable case). I asked around and was advised to submit the paper to <em>L'Enseignement Mathematique</em>. In my writeup, I made the conscious decision to write the paper in an expository way: that is, I included a lot of background material and explained connections between the various results, even things which were not directly related to the theorem in question. The paper was accepted; but the referee made it clear that s/he would have preferred a more streamlined, research oriented approach. Thus <em>EM</em>, despite its name ("Mathematical Education"), seems to be primarily a research journal (which likes papers taking new looks at old or easily stated problems: it's certainly a good journal and I'm proud to be published in it), and I was able to smuggle in some exposition under the cover of a new research result.</p> <p>3) I have an expository paper on factorization in integral domains:</p> <p><a href="http://math.uga.edu/~pete/factorization.pdf" rel="nofollow">http://math.uga.edu/~pete/factorization.pdf</a></p> <p>[<b>Added</b>: And a newer version: http://math.uga.edu/~pete/factorization2010.pdf.]</p> <p>It is not finished and not completely polished, but it has been circulating around the internet for about a year now. Again, this completely expository paper has attracted more attention than most of my research papers. Sometimes people talk about it as though it were a preprint or an actual paper, but it isn't: I do not know of any journal that would publish a 30 page paper giving an intermediate-level exposition of the theory of factorization in integral domains. Is there such a journal? </p> <p><b>Added</b>: In my factorization paper, I build on similar expositions by the leading algebraists P. Samuel and P.M. Cohn. I think these two papers, published in 1968 and 1973, are both excellent examples of the sort of "intermediate exposition" I have in mind (closer to the high end of the range, but still intermediate: one of the main results Samuel discusses, Nagata's Theorem, was published in 1957 so was not exactly hot off the presses when Samuel wrote his article). Both articles were published by the <em>American Mathematical Monthly</em>! I don't think the Monthly would publish either of them nowadays. </p> <p><b>Added</b>: I have recently submitted a paper to the Monthly: </p> <p><a href="http://math.uga.edu/~pete/coveringnumbersv2.pdf" rel="nofollow">http://math.uga.edu/~pete/coveringnumbersv2.pdf</a></p> <p>(By another coincidence, this paper is a mildly souped up answer to MO question #26. But I did the "research" on this paper in the lonely pre-MO days of 2008.) </p> <p>Looking at this paper helps me to see that the line between research and exposition can be blurry. I think it is primarily an expository paper -- in that the emphasis is on the presentation of the results rather than the results themselves -- but I didn't have the guts to submit it anywhere without claiming some small research novelty: "The computation of the irredundant linear covering number appears to be new." I'll let you know what happens to it.</p> <p>(<b>Added</b>: it was accepted by the Monthly.)</p> http://mathoverflow.net/questions/10934/class-number-measuring-the-failure-of-unique-factorization/10939#10939 Answer by Pete L. Clark for Class number measuring the failure of unique factorization Pete L. Clark 2010-01-06T17:43:49Z 2013-03-20T16:53:50Z <p>Theorem (Carlitz, 1960): The ring of integers $\mathbb{Z}_F$ of an algebraic number field $F$ has class number at most $2$ iff for all nonzero nonunits $x \in \mathbb{Z}_F$, any two factorizations of $x$ into irreducibles have the same number of factors.</p> <p>A proof of this (and a 1990 generalization of Valenza) can be found in $\S 22.3$ of <a href="http://math.uga.edu/~pete/integral.pdf" rel="nofollow">my commutative algebra notes</a>.</p> <p>This paper has spawned a lot of research by ring theorists on <b>half-factorial domains</b>: these are rings in which every nonzero nonunit factors into irreducibles and such that the number of irreducible factors is independent of the factorization. </p> <p>To be honest though, I think there are plenty of number theorists who think of the class number as measuring the failure of unique factorization who don't know Carlitz's theorem (or who know it but are not thinking of it when they make that kind of statement).</p> <p>Here is another try [<b>edit: this is essentially the same as Olivier's response, but said differently; I think it is worthwhile to have both</b>]: when trying to solve certain Diophantine problems (over the integers), one often gets nice results if the class number of a certain number field is prime to a certain quantity. The most famous example of this is Fermat's Last Theorem, which is easy to prove for an odd prime $p$ for which the class number of $\mathbb{Q}(\zeta_p)$ is prime to $p$: a so-called "regular" prime. </p> <p>For an application to Mordell equations $y^2 + k = x^3$, see</p> <p><a href="http://math.uga.edu/~pete/4400MordellEquation.pdf" rel="nofollow">http://math.uga.edu/~pete/4400MordellEquation.pdf</a></p> <p>Especially see Section 4, where the class of rings "of class number prime to 3" is defined axiomatically and applied to the Mordell equation. (N.B.: These notes are written for an advanced undergraduate / first year grad student audience.)</p> <p>The Mordell equation is probably a better example than the Fermat equation because:</p> <p>(i) the argument in the "regular" case is more elementary than FLT in the regular case (the latter is too involved to be done in a first course), and </p> <p>(ii) when the "regularity" hypothesis is dropped, it is not just harder to prove that there are no nontrivial solutions, it is actually very often false!</p> http://mathoverflow.net/questions/54356/nonfree-projective-module-over-a-regular-ufd Nonfree projective module over a regular UFD? Pete L. Clark 2011-02-04T20:15:09Z 2013-02-15T11:44:16Z <p>What is the simplest example of a domain $R$ which is regular (in particular Noetherian) and factorial which admits a finitely generated projective module that is not free?</p> <p>In fact I'll be at least somewhat happy with any example, since I can't think of one at the moment.</p> <p>Some brief comments: $R$ needs to have Krull dimension greater than one or else it is a PID. The module in question needs to have rank greater than one, because the hypotheses force the Picard group to be equal to the divisor class group and the divisor class group to be trivial. And famously, by work of Quillen and Suslin, one cannot take $R$ to be a polynomial ring over a field. Oh yes, and of course $R$ can't be local (or even semilocal, I suppose). I'm already out of ideas... </p> <p>P.S.: If you can get an easier example by removing the hypothesis of finite generation, I'd be interested in that as well.</p> http://mathoverflow.net/questions/13768/what-is-the-right-definition-of-the-picard-group-of-a-commutative-ring What is the right definition of the Picard group of a commutative ring? Pete L. Clark 2010-02-02T01:42:39Z 2012-12-29T00:04:58Z <p>This is a rather technical question with no particular importance in any case of actual interest to me, but I've been writing up some notes on commutative algebra and flailing on this point for some time now, so I might as well ask here and get it cleared up.</p> <p>I would like to define the Picard group of an arbitrary (i.e., not necessarily Noetherian) commutative ring $R$. Here are two possible definitions:</p> <p>(1) It is the group of isomorphism classes of rank one projective $R$-modules under the tensor product.</p> <p>(2) It is the group of isomorphism classes of invertible $R$-modules under the tensor product, where invertible means any of the following equivalent things [Eisenbud, Thm. 11.6]:</p> <p>a) The canonical map <code>$T: M \otimes_R \operatorname{Hom}_R(M,R) \rightarrow R$</code> is an isomorphism.<br /> b) $M$ is locally free of rank $1$ [<b>edit</b>: in the weaker sense: $\forall \mathfrak{p} \in \operatorname{Spec}(R), \ M_{\mathfrak{p}} \cong R_{\mathfrak{p}}$.]<br /> c) $M$ is isomorphic as a module to an invertible fractional ideal.</p> <p>What's the difference between (1) and (2)? In general, (1) is stronger than (2), because projective modules are locally free, whereas a finitely generated locally free module is projective iff it is finitely presented. (When $R$ is Noetherian, finitely generated and finitely presented are equivalent, so there is no problem in this case. This makes the entire discussion somewhat academic.)</p> <p>So, <em>a priori</em>, if over a non-Noetherian ring one used (1), one would get a Picard group that was "too small". Does anyone know an actual example where the groups formed in this way are not isomorphic? (That's stronger than one being a proper subgroup of the other, I know.) </p> <p>Why is definition (2) preferred over definition (1)? </p> http://mathoverflow.net/questions/38780/why-do-we-care-whether-a-pid-admits-some-crazy-euclidean-norm Why do we care whether a PID admits some crazy Euclidean norm? Pete L. Clark 2010-09-15T05:02:45Z 2012-11-13T10:39:46Z <p>An integral domain $R$ is said to be Euclidean if it admits some Euclidean norm: i.e., a function $N: R \rightarrow \mathbb{N} = \mathbb{Z}^{\geq 0}$ such that: for all $x, y \in R$ with $N(y) > 0$, either $y$ divides $x$ or there exists $q \in R$ such that $N(x-qy) &lt; N(y)$. A well-known "descent" argument shows that any Euclidean domain is a PID. In fact, the argument that a Euclidean domain is necessarily a UFD is a little more direct and elementary than the argument that shows that a PID is a UFD (because, in the latter case, one needs some kind of ideal-theoretic argument to show the existence of factorizations into irreducible elements). Because of this, Euclidean domains are a familiar staple of undergraduate algebra.</p> <p>A lot of texts seem to emphasize the fact that a PID need not be a Euclidean domain. In order to show this, one has to show not only that some particular norm (and often there is a preferred norm in sight, see below) is not Euclidean, but that there is no Euclidean norm whatsoever. In general this is a very delicate question: for instance, the proof of the most standard example -- that the ring of integers of $\mathbb{Q}(\sqrt{-19})$ is a PID but does not admit <em>any</em> Euclidean norm -- is already rather intricate.</p> <p>My question is this:</p> <blockquote> <p>Given a ring $R$ that we already know is a PID, why do we care whether or not it admits some Euclidean norm?</p> </blockquote> <p>Note that in contrast, many domains admit natural norms. A class of domains which I have been thinking about recently are the infinite domains satisfying (FN): the quotient by every nonzero ideal is finite. In this case, the map $0 \mapsto 0$, <code>$x \in R \setminus \{0\} \mapsto \# R/(x)$</code> is a multiplicative norm, which I call <strong>canonical</strong>. For instance, the usual absolute value on $\mathbb{Z}$ is the canonical norm, as is the norm on any ring of integers in a number field that you meet in an algebraic number theory course. </p> <p>I have recently realized that I care quite a bit about whether certain specific norms on integral domains are Euclidean. (This has come up in my work on quadratic forms and the Davenport-Cassels theorem.) There is some very natural algebra and discrete geometry here. But why do I care if some crazy Euclidean norm exists?</p> <p>Here are three reasons that one might care about this:</p> <ol> <li><p>If a domain admits an "effective" Euclidean norm, one can give effective algorithms for linear algebra over that ring, whereas the structure theory of modules over an arbitrary PID is not <em>a priori</em> algorithmic in nature.</p></li> <li><p>(in algebraic K-theory): If $R$ is Euclidean, $\operatorname{SK}_1(R) = 0$, but there exists a PID with nonvanishing $\operatorname{SK}_1$. (Thanks to Charles Rezk for giving the precise result based on my vague allusion to it.)</p></li> <li><p>In algebraic number theory, there has been a lot of work towards proving the conjecture that if $K$ is a number field which is <em>not</em> $\mathbb{Q}(\sqrt{D})$ for $D = -19, -43, -67, -163$, then the ring $\mathbb{Z}_K$ of integers of $K$ is a PID iff it is Euclidean (for some crazy norm). In particular, disproving this would disprove the generalized Riemann hypothesis.</p></li> </ol> <p>Comments on 1: There is something to this, but I somehow doubt that it's such a big deal. For instance, the ring of integers of $\mathbb{Q}(\sqrt{-19})$ is not Euclidean, but I'm pretty sure that there are algorithms for modules over it. In particular, it seems to me that for algorithmic purposes, having a Dedekind-Hasse norm is just as good as a Euclidean norm, and every PID has a Dedekind-Hasse norm. In fact, for every PID which satisfies (FN), the canonical norm is a Dedekind-Hasse norm. (See p. 27 of <a href="http://math.uga.edu/~pete/factorization2010.pdf" rel="nofollow">http://math.uga.edu/~pete/factorization2010.pdf</a> for this.)</p> <p>Comments on 3: if I knew more about this result, I might appreciate it better. It does seem to involve some interesting geometry of numbers. But this convinces me why I should be interested in the special case of rings of integers in number fields, which, as a number theorist, I am already convinced are more worthy of scrutiny from every possible angle than an arbitrary domain.</p> <p>If there are other good reasons to care, I'd certainly like to know.</p> http://mathoverflow.net/questions/111519/why-might-andre-weil-have-named-carl-ludwig-siegel-the-greatest-mathematician-of/111529#111529 Answer by Pete L. Clark for Why might André Weil have named Carl Ludwig Siegel the greatest mathematician of the 20th century? Pete L. Clark 2012-11-05T06:31:08Z 2012-11-05T20:24:08Z <p>No one with any familiarity with his work can doubt that Siegel was <em>one of the greatest mathematicians</em> of the 20th century. Weil was a decisive, opinionated man -- just the type of person who would have an answer to this question ready at hand. And "Carl Ludwig Siegel" is a totally unsurprising answer from anyone. (Also "Andre Weil" would be a totally unsurprising answer from anyone: it might be my answer!)</p> <p>But it is especially unsurprising coming from Weil. The list of contemporary mathematicians of the Siegel-Weil caliber is short enough, and among mathematicians on that list -- e.g. Wiener, von Neumann, Kolmogorov, Godel -- the research interests of Siegel and Weil were especially close: for instance, there is a <a href="http://en.wikipedia.org/wiki/Siegel%E2%80%93Weil_formula" rel="nofollow">Siegel-Weil formula</a>. Both brought their prodigious knowledge and technique to bear on number theory, but with distinct, and distinctive, styles. To be very brief and crude, Weil had a fundamentally algebraic approach, whereas Siegel had a fundamentally analytic approach. My own approach to mathematics is rather close to Weil's (although in magnitude, microscopic compared to his): I very much appreciate that finding the right bit of "structure" can make the solution of your problems self-evident. A lot -- by no means all -- of Weil's work is like that: the finished product is so tidy and efficacious that you too easily forget to ask how he thought of any of it in the first place. To someone with this "algebraic" style, Siegel's work looks like a sequence of miracles. So it is unsurprising to me that someone like Weil would select someone like Siegel to give his top regards.</p> <p>I think you can also gain some insight into why Weil named Siegel by considering their ages: Siegel (born in 1896) was ten years older than Weil (born in 1906). Ten years is long enough for Siegel always to have been ahead of Weil in his career and stature, but short enough for them to be true contemporaries and competitors. Most other great mathematicians that spring to mind when I think of Weil are actually quite a bit younger, e.g. Serre (born 1926), Tate (born 1925), Shimura (born 1930); it makes sense that Weil is not going to name any of these as the greatest mathematician of the 20th century. Indeed all three are alive well into the 21st century. </p> <p>[Added: I just remembered that Chevalley (born 1909) was a contemporary of Weil of a similar stature. But Chevalley was very close to Weil, both personally and in mathematical styles and tastes. It is psychologically natural to esteem (and fear) most that which is most different from ourselves, not that which is most similar. Anyway, for Weil to name Chevalley would have sounded arrogant, as if not being able to name himself he picked the person standing right next to him.] </p> <p>By the way, I think that Shimura and Siegel are quite similar in style as well as stature. I read Shimura's autobiography, and I think he is right to be profoundly disappointed that Siegel did not take more of an interest in his work. Shimura's work is closer to being a continuation of Siegel's (including a continuation of the brilliance, creativity and orginality!) than any other mathematician I can think of, so it is natural that Shimura holds Siegel in high regard. </p> <p>There is also something "organic" in the work of both Siegel and Shimura which naturally bristles a bit at the "Bourbakistic" influence of the French school: it seems clear enough, for instance, that the modern theory of "Shimura varieties" is both an addition and a subtraction from what Shimura himself intended. I know several of Shimura's students, and though they work in what the rest of the mathematical world thinks of as parts of algebraic number theory and arithmetic geometry, in the way they actually think about mathematics they take a more analytic approach...like Siegel. I have even fewer credentials to speak for Selberg than I do for any of these others, but I imagine that he may have felt a similar kinship to Siegel, i.e., the use of an "analytic" approach to studying problems that others regard as being more algebraic. </p> http://mathoverflow.net/questions/53647/questions-about-spectra-of-rings-of-continuous-functions Questions about spectra of rings of continuous functions Pete L. Clark 2011-01-28T20:12:21Z 2012-11-01T11:17:57Z <p>I have been thinking a bit about rings of continuous functions of various kinds -- how they motivate the more modern notion of the Zariski topology on the prime spectrum as well as how they fit into a more general picture. (The commutative algebra course that I had mentioned in an earlier question has recently begun.) Here are two constructions which I have been thinking a lot about:</p> <p>For $X$ an arbitrary topological space, let $C(X)$ be the ring of continuous $\mathbb{R}$-valued functions on $X$. One has a canonical map $\mathcal{M}: X \rightarrow \operatorname{MaxSpec} C(X)$ by taking $x$ to the maximal ideal $\mathfrak{m}_x$ of all functions vanishing at $x$. This map is continuous when the codomain is given the Zariski topology (e.g. as a subspace of the Zariski topology on the prime spectrum: but really, just take the same definition and restrict to maximal ideals.) </p> <p>When $X$ is compact (by which I mean quasi-compact and Hausdorff) I proved in class that $\mathcal{M}$ is a homeomorphism. This was not as graceful as I might have hoped: I found myself having to introduce an auxiliary topology on the maximal spectrum -- the "initial", "weak" or "Gelfand" topology -- to see that it was Hausdorff and then only after having shown that $\mathcal{M}$ is a homeomorphism with the Gelfand topology did I deduce that the Gelfand topology coincides with the Zariski topology (using the characteristic property of Tychonoff spaces that any closed set is an intersection of zero sets of continuous functions). </p> <p>I mentioned that the following are true in the general case:</p> <p>(i) $\operatorname{MaxSpec} C(X)$ is compact in the Zariski topology. Therefore $X_T := \mathcal{M}(X)$, endowed with the subspace topology is Tychonoff.<br> (ii) In fact $\mathcal{M}: X \rightarrow X_T$ is the universal Tychonoff space on $X$ and the induced map $C(X_T) \rightarrow C(X)$ is an isomorphism of rings. (So we may as well assume $X$ is Tychonoff.)<br> (iii) For any Tychonoff space $X$, $\mathcal{M}: X \rightarrow \operatorname{MaxSpec} C(X)$ is nothing else than the <strong>Stone-Cech compactification</strong>.</p> <p>I am still looking for a nice, self-contained reference for these facts. Gillman and Jerison's classic text has most of them, but spread out over a fairly large number of pages. For instance, can it be so hard to see that $\operatorname{MaxSpec} C(X)$ is Hausdorff no matter what $X$ is? I am struggling even with that!</p> <p>The upshot here is that taking the ring of $\mathbb{R}$-valued functions and then the maximal spectrum has the effect of passing from an arbitrary space $X$ to its universal compact space. I find this very interesting.</p> <p>Now consider $C_2(X)$, the ring of continuous functions from $X$ to $\mathbb{F}_2$ (the latter endowed with the discrete topology: what else?). It now seems that the above discussion goes through with "universal compact space" replaced by "universal Boolean (= compact and totally disconnected) space", and that this is a slightly different (better?) take on Stone duality than the one I wrote up in my notes. </p> <p>But actually in between $\mathbb{R}$ and $\mathbb{F}_2$ I did something that is maybe silly: I threw off a remark to my students that it seemed interesting to think about the case of $\mathbb{Q}_p$-valued functions. But (without having written anything down, so I could well be mistaken) I now think that although the rings one gets in this way are of course not Boolean, their spectra still comprise precisely the Boolean spaces...and that maybe the same holds for functions with values in any totally disconnected topological field. Is this really the case?</p> <p>The above is pretty rambly, so let me end with one crisp question which is at least part of what's eating me:</p> <blockquote> <p>Can one characterize the commutative rings with Hausdorff maximal spectrum? Or with Boolean maximal spectrum? Or with maximal spectrum some other interesting class of compact spaces?</p> </blockquote> http://mathoverflow.net/questions/108543/over-which-fields-does-the-mordell-weil-theorem-hold/111024#111024 Answer by Pete L. Clark for Over which fields does the Mordell-Weil theorem hold? Pete L. Clark 2012-10-29T19:37:31Z 2012-10-30T03:18:51Z <p>This is an attempt at a relatively mild generalization of what others have said:</p> <p>Let $K$ be a field and $|\cdot|: K \rightarrow \mathbb{R}$ be a nontrivial absolute value on $K$. </p> <p>$\bullet$ If $K$ is <strong>complete</strong> for $|\cdot|$, then $E(K)$ has the structure of a $K$-analytic Lie group in the sense of Serre. In particular it is a $K$-analytic manifold so has at least continuum cardinality.</p> <p>$\bullet$ When $|\cdot|$ comes from a rank one valuation $v$, I suspect that even if $K$ is merely <strong>Henselian</strong> for $v$, then $E(K)$ cannot be finitely generated. </p> <p>Here is a proof in the case that the valuation is discrete and the residue field $k$ is infinite: standard arguments involving the formal group still give a filtration </p> <p>$E(K) \supset E^0(K) \supset E^1(K) \supset E^2(K) \supset \ldots$ </p> <p>such that (by Hensel's Lemma) for all $n \geq 1$, $E^n(K)/E^{n+1}(K) \cong (k,+)$. (Just last night I noticed that Cassels's <em>Lectures on Elliptic Curves</em> has a beautiful, elementary take on this. He works with the case $K = \mathbb{Q}_p$ but the argument holds much more generally.) If $k$ is infinite, then its additive group is not finitely generated and thus $E(K)$, having a subquotient which is not finitely generated, is itself not finitely generated.</p> http://mathoverflow.net/questions/20740/is-there-an-introduction-to-probability-theory-from-a-structuralist-categorical-p Is there an introduction to probability theory from a structuralist/categorical perspective? Pete L. Clark 2010-04-08T15:18:55Z 2012-09-23T15:05:22Z <p>The title really is the question, but allow me to explain.</p> <p>I am a pure mathematician working outside of probability theory, but the concepts and techniques of probability theory (in the sense of Kolmogorov, i.e., probability measures) are appealing and potentially useful to me. It seems to me that, perhaps more than most other areas of mathematics, there are many, many nice introductory (as well as not so introductory) texts on this subject.</p> <p>However, I haven't found any that are written from what it is arguably the dominant school of thought of contemporary mainstream mathematics, i.e., from a structuralist (think Bourbaki) sensibility. E.g., when I started writing notes on the texts I was reading, I soon found that I was asking questions and setting things up in a somewhat different way. Here are some basic questions I couldn't stop from asking myself:</p> <p>[0) Define a Borel space to be a set $X$ equipped with a $\sigma$-algebra of subsets of $X$. This is already not universally done (explicitly) in standard texts, but from a structuralist approach one should gain some understanding of such spaces before one considers the richer structure of a probability space.]</p> <p>1) What is the category of Borel spaces, i.e., what are the morphisms? Does it have products, coproducts, initial/final objects, etc? As a significant example here I found the notion of the product Borel space -- which is exactly what you think if you know about the product topology -- but seemed underemphasized in the standard treatments.</p> <p>2) What is the category of probability spaces, or is this not a fruitful concept (and why?)? For instance, a subspace of a probability space is, apparently, not a probability space: is that a problem? Is the right notion of morphism of probability spaces a measure-preserving function? </p> <p>3) What are the functorial properties of probability measures? E.g., what are basic results on pushing them forward, pulling them back, passing to products and quotients, etc. Here again I will mention that product of an arbitrary family of probability spaces -- which is a very useful-looking concept! -- seems not to be treated in most texts. Not that it's hard to do: see e.g.</p> <p><a href="http://www.math.uga.edu/~pete/saeki.pdf" rel="nofollow">http://www.math.uga.edu/~pete/saeki.pdf</a></p> <p>I am not a category theorist, and my taste for how much categorical language to use is probably towards the middle of the spectrum: that is, I like to use a very small categorical vocabulary (morphisms, functors, products, coproducts, etc.) as often as seems relevant (which is very often!). It would be a somewhat different question to develop a truly categorical take on probability theory. There is definitely some nice mathematics here, e.g. I recall an arxiv article (unfortunately I cannot put my hands on it at this moment) which discussed independence of events in terms of tensor categories in a very persuasive way. So answers which are more explicitly categorical are also welcome, although I wish to be clear that I'm not asking for a categorification of probability theory <em>per se</em> (at least, not so far as I am aware!).</p> http://mathoverflow.net/questions/46155/a-question-about-the-isometry-group-of-a-finite-metric-space/46169#46169 Answer by Pete L. Clark for A question about the isometry group of a finite metric space Pete L. Clark 2010-11-16T00:52:22Z 2012-09-13T00:04:02Z <p>As secretman says, the condition that there exist $i$ and $j$ in $X$ such that no isometry carries $i$ to $j$ is precisely to say that the action of the isometry group on $X$ is not <em>transitive</em>. If that's really your question, that seems to be all that can be said, and I doubt there any books on the subject.</p> <p>One might simply ask: what can be said about isometry groups of finite metric spaces? This is a rather broad question, but nevertheless it seems interesting. Here is what I was able to come up with via a short amount of thought and googling:</p> <p>1) Every finite group $G$ occurs up to abstract group isomorphism as the full isometry group of a finite metric space. Indeed, in <a href="http://www.math.uga.edu/~pete/Asimov76.pdf" rel="nofollow">this 1976 paper</a>, D. Asimov proved that if $G$ is finite of cardinality $k$, there exists a finite subset $X_G$ of Euclidean $k-1$-space (with the induced metric), of cardinality $k^2-k$, such that the full isometry group of $X_G$ is isomorphic to $G$.</p> <p>2) On the other hand, it is not true that every permutation group is the isometry group of a finite space: i.e., if <code>$X = \{1,\ldots,n\}$</code> and $G$ is subgroup of $S_n$, then there need not be a metric $\rho$ on $X$ such that $G = \operatorname{Aut}(X,\rho)$. For a simple example, take $n = 3$ and let $G$ be the subgroup of order $3$, i.e., $G$ is generated by $\sigma = (123)$. Since $\sigma$ is an isometry, </p> <p>$\rho(1,2) = \rho(\sigma(1),\sigma(2)) = \rho(2,3) = \rho(\sigma(2),\sigma(3)) = \rho(1,3)$. </p> <p>Thus all pairwise distances between the three elements of $X$ are equal and the isometry group is the full $S_3$.</p> <p>More generally, let $\rho$ be a metric on $X$ and $G \subset \operatorname{Sym}(X)$ the isometry group. Let $X^{(2)}$ be the set of (unordered!) two-element subsets of $X$: there is a natural action of $\operatorname{Sym}(X)$ -- and hence also of $G$ -- on $X^{(2)}$. Now the observation here is that if $G$ acts transitively on $X^{(2)}$ -- e.g. if $G$ is <em>doubly transitive</em> as a permutation group on $X$, but this condition is weaker -- then that means that all pairs of points in $X$ have the same distance, and therefore $G = \operatorname{Sym}(X)$. </p> <p><s>It strikes me that one could take this construction a step further: we may define a graph $G$ with vertex set $X^{(2)}$ by decreeing two vertices <code>$\{x_1,x_2\}$</code>, <code>$\{y_1,y_2\}$</code> to be adjacent iff $\rho(x_1,x_2) = \rho(y_1,y_2)$. Then $G$ is precisely the group of graph automorphisms of $G$. This is sort of a "combinatorialization" of the problem, although I don't know whether it's actually useful for anything.</s> This is false, as Aleksander Horawa has pointed out to me. I'm not sure what I was thinking when I wrote this. As he points out, it is at least true that when <code>$\# X \geq 3$</code>, the induced homomorphism from the isometry group of $X$ to the automorphism of the graph is injective (even this fails when <code>$\# X = 2$</code>). </p> http://mathoverflow.net/questions/59213/generating-finite-simple-groups-with-2-elements/59300#59300 Answer by Pete L. Clark for Generating finite simple groups with $2$ elements Pete L. Clark 2011-03-23T13:46:55Z 2012-08-26T06:29:43Z <p>Since I happen to know the OP is number-theoretically inclined, let me add the following remark:</p> <p>For "most" finite simple groups $G$ it is indeed the case that $G = \langle x, y \rangle$ where $x$ has order $2$ and $y$ has order $3$. Equivalently, $G$ is a quotient of the free product $\mathbb{Z}/2\mathbb{Z} * \mathbb{Z}/3\mathbb{Z} \cong \operatorname{PSL}_2(\mathbb{Z}) = \Gamma(1)$. </p> <p>This has the following geometric consequence: there is some subgroup $\Gamma_G \subset \Gamma(1)$ such that $X_G = \Gamma_G \backslash \overline{\mathcal{H}}$ is a modular curve and $X_G \rightarrow X(1) \cong \mathbb{P}^1$ is a $G$-Galois branched covering. By taking $G$ to be something else than $\operatorname{PSL}_2(\mathbb{Z}/p\mathbb{Z})$ one sees that $\Gamma(1)$ admits many <strong>non-congruence subgroups</strong>. For instance, it is well-known (<b>added</b>: I should have said "a well-known theorem of J.G. Thompson") that one can take $G$ to be the <a href="http://en.wikipedia.org/wiki/Monster_group" rel="nofollow">Fischer-Griess Monster</a>.</p> <p>I don't want to make precise what I mean by "most". Note that there are infinitely many finite simple groups with order prime to $3$ (although one has to look fairly far down the list of all finite simple groups to see them: <strong>Suzuki groups</strong>), so I definitely do not mean "all but finitely many". </p> http://mathoverflow.net/questions/36611/is-a-valuation-domain-pid-when-its-maximal-ideal-is-principal/36636#36636 Answer by Pete L. Clark for Is a valuation domain PID when its maximal ideal is principal? Pete L. Clark 2010-08-25T06:49:47Z 2012-08-16T14:56:30Z <p>I assume that by a valuation domain you mean an integral domain $R$ with fraction field $K$ such that: for all $x \in K^{\times}$, at least one of $x,x^{-1}$ lies in $R$.</p> <p>In this case, I believe the answer is <strong>no</strong>. Let $R$ be any valuation domain whose value group $K^{\times}/R^{\times}$ is isomorphic, as a totally ordered abelian group, to $\mathbb{Z} \times \mathbb{Z}$ with the lexicographic ordering. (It is known that every totally ordered abelian group is the value group of some valuation domain, e.g. by a certain generalized formal power series construction due to Neumann.) In this case, the maximal ideal consists of all elements whose valuation is strictly greater than $(0,0)$, but the valuation of any such element is at least $(0,1)$ and therefore any element of valuation $(0,1)$ gives a generator of the maximal ideal. </p> <p>For some information on valuation rings, see e.g. Section 17 of </p> <p><a href="http://math.uga.edu/~pete/integral.pdf" rel="nofollow">http://math.uga.edu/~pete/integral.pdf</a></p> http://mathoverflow.net/questions/104350/equivalent-definitions-of-invertible-modules/104503#104503 Answer by Pete L. Clark for Equivalent definitions of invertible modules Pete L. Clark 2012-08-11T17:46:03Z 2012-08-11T17:53:10Z <p>I will happily incorporate this strengthening of the statement of Proposition 19.8 of my notes. Thanks for this, and in the future please feel free to contact me directly (as well).</p> <p>As an aside, I well remember Rota's criticism of commutative algebra texts for their "hygienic theorems": see <a href="http://mathoverflow.net/questions/63715/why-are-finiteness-conditions-important-and-how-to-recognize-them" rel="nofollow">this previous MO answer</a>. In many places in my notes if I state a result which has an implication without a converse implication or a converse only under additional hypotheses, I follow it with an exercise, remark or reference about whether/why the converse is not true. I would like to do this all the time -- but I am not a real expert in the subject and I read the same textbooks as everyone else, so often (as here) I do not know enough to do so. That's why it's great to get responses like this from other mathematicians. </p> http://mathoverflow.net/questions/103731/an-extension-of-edelsteins-attraction-theorem An extension of Edelstein's Attraction Theorem? Pete L. Clark 2012-08-01T21:45:04Z 2012-08-02T11:14:04Z <p>Let $X$ be a metric space. Recall that a function $f: X \rightarrow X$ is <strong>contractive</strong> if there exists $C \in (0,1)$ such that for all $x,y \in X$, $d(f(x),f(y)) \leq C d(x,y)$; a function $f$ is <strong>weakly contractive</strong> if for all $x \neq y \in X$, $d(f(x),f(y)) &lt; d(x,y)$.</p> <p>Also let us say that an <strong>attracting point</strong> for $f: X \rightarrow X$ is a point $L \in X$ such that for every $x_0 \in X$, the sequence of iterates <code>$\{ (f \circ \cdots \circ f)(x_0)\}_{n=0}^{\infty}$</code> converges to $L$. (Since $f$ is continuous, an attracting point is necessarily a fixed point.) Finally, we say $f: X \rightarrow X$ is <strong>attractive</strong> if it has an attracting point. </p> <p>Recall the following two <strong>Attraction Theorems</strong>:</p> <blockquote> <ol> <li><p>Banach Attraction Theorem: If $X$ is complete, then every contractive mapping $f: X \rightarrow X$ is attractive.</p></li> <li><p>Edelstein Attraction Theorem: If $X$ is compact, then every weakly contractive mapping $f: X \rightarrow X$ is attractive. </p></li> </ol> </blockquote> <p>(Side Remark: Of course these theorems imply that $f$ has a fixed point and are usually called "fixed point theorems". In the case of Banach's Theorem the stronger conclusion is well known: in fact it is well known that the sequences of iterates converge with exponential speed. But in the case of Edelstein's Theorem, most of the literature I have found states it as a fixed point theorem only, despite the fact that this weaker conclusion is at the level of a homework or exam problem: consider $\min_{x \in X} d(x,f(x))$...One exceptional reference which treats this topic very nicely is <a href="http://www.math.uconn.edu/~kconrad/blurbs/analysis/contraction.pdf." rel="nofollow">Keith Conrad's note on contraction mappings</a>.)</p> <p>I started thinking about these results in the context of teaching a "Spivak Calculus course" this past academic year. In the context of that course, functions were defined on <em>intervals</em>, not metric spaces, but not necessarily intervals that are closed and/or bounded. Thus one is not always working in a complete metric space. It is clear that in an incomplete case a contraction mapping need not have a fixed point: one need look no further than</p> <p>$f: (0,1) \rightarrow (0,1), \ x \mapsto \frac{x}{2}$.</p> <p>However this counterexample doesn't look "serious", because $f$ continuously extends to $[0,1)$ and has $0$ as an attracting point there. Thinking over this phenomenon, I realized it was rather general:</p> <blockquote> <p>$3$. Let $I \subset \mathbb{R}$ be an(y) interval, and let $f: I \rightarrow I$ be continuous.<br> a) At least one of the following occurs:<br> (i) $f$ has a fixed point in $I$.<br> (ii) $\sup I$ is an attracting point for $f$. (If $I$ is unbounded above, this means that every sequence of iterates diverges to $\infty$.)<br> (iii) $\inf I$ is an attracting point for $f$. (If $I$ is unbounded below...)<br> b) If $f$ is moreover weakly contractive, then it has an attracting point in $[\inf I,\sup I]$. </p> </blockquote> <p>I expect many readers will find a proof immediately, but see $\S 10.4$ of <a href="http://math.uga.edu/~pete/2400full.pdf" rel="nofollow">these notes</a> if you like.</p> <p>The case $I = \mathbb{R}$ of Theorem 3a) was the subject of <a href="http://www.math.uga.edu/~pete/Beardon06.pdf" rel="nofollow">a short Monthly note of A.F. Beardon</a>.</p> <p>Is there an extension of Theorem 3 to $\mathbb{R}^n$? If you look at the proof of Theorem 3, you see that it really uses the order-theoretic, rather than the metric properties of $\mathbb{R}^n$: when $I$ is a bounded interval the order-theoretic completion and the metric completions coincide, but not in the unbounded case. So let's restrict to bounded sets and ask the following question:</p> <blockquote> <p>Question 1: Let $X \subset \mathbb{R}^n$ be a bounded subset. Does every weakly contractive map $f: X \rightarrow X$ have an attracting fixed point in $\overline{X}$ (closure in $\mathbb{R}^n$ = metric completion)?</p> </blockquote> <p>Here is a reformulation / generalization of the question using the concepts of metric spaces. If I have a contraction mapping $f$ on an incomplete metric space $X$, I can't expect it to have a fixed point, but I can ask for a fixed point in the unique extension of $f$ to a continuous mapping $\overline{f}: \overline{X} \rightarrow \overline{X}$ on the metric completion $\overline{X}$. But it is easy to see that if $C$ is a contraction constant for $f$, it is also a contraction constant for $\overline{f}$, so the following result is not really any more general than Theorem 1.</p> <blockquote> <p>$1'$ Every contractive map on a metric space extends to an attractive map on the metric completion.</p> </blockquote> <p>As for 2., it is clear that the compactness hypothesis cannot just be thrown out entirely: for instance $f: \mathbb{R} \rightarrow \mathbb{R}, \ x \mapsto \sqrt{x^2+1}$ is weakly contractive but has no (finite!) fixed point. However, recalling that a metric space is compact iff it is complete and <a href="http://en.wikipedia.org/wiki/Totally_bounded_space" rel="nofollow">totally bounded</a> there is a way to "remove only the completeness hypothesis". So I ask:</p> <blockquote> <p>Question 2: Let $X$ be a totally bounded metric space. Must every weakly contractive map on $X$ have an attracting point (or even a fixed point) in the compact space $\overline{X}$? </p> </blockquote> <p>Note that bounded sets in $\mathbb{R}^n$ are totally bounded, so an affirmative answer to Question 2 indeed implies an affirmative answer to Question 1.</p> <p>Finally, is <em>a priori</em> conceivable -- although I am rather doubtful about it -- that Question 2 may follow trivially from Theorem 2 in the same way that Theorem 1'. follows trivially from Theorem 1. That is:</p> <blockquote> <p>Question 3: Let $f$ be a weakly contractive map on a metric space $X$, and let $\overline{f}$ be the unique continuous extension to the metric completion $\overline{X}$. Must $\overline{f}$ be weakly contractive? </p> </blockquote> http://mathoverflow.net/questions/103676/maximal-subfields-in-a-division-algebra-over-a-local-field/103734#103734 Answer by Pete L. Clark for Maximal subfields in a division algebra over a local field Pete L. Clark 2012-08-01T22:07:21Z 2012-08-01T22:07:21Z <p>To address 2.: for any central simple algebra $A$ over a field $k$, there is a well-developed theory describing the relations between finite splitting fields $l/k$ for $A$ and fields which are sub-$k$-algebras of $A$. The most important result is probably this one:</p> <blockquote> <p>Theorem: Let $A$ be a central simple algebra over a field $k$, of dimension $n^2$. For a field extension $l/k$ of degree $n$, the following are equivalent:<br> (i) There is a $k$-algebra embedding $l \hookrightarrow A$.<br> (ii) $l$ is a splitting field for $A$.</p> </blockquote> <p>A proof of this result can be found, for instance, in $\S 6.2$ of <a href="http://math.uga.edu/~pete/noncommutativealgebra.pdf" rel="nofollow">these notes on noncommutative algebra</a>. Citations to more substantial treatments are given in $\S 6.10$. </p> http://mathoverflow.net/questions/102763/the-use-of-parentheses-to-mean-i-wont-tell-you-this-again/103060#103060 Answer by Pete L. Clark for the use of parentheses to mean "I won't tell you this again" Pete L. Clark 2012-07-25T01:18:33Z 2012-07-25T01:18:33Z <p>I think it might be beneficial to see the actual context in which the comments were made (by me; not as a referee, but just someone that Jim wrote to and asked for comments on his nice paper, which by the way, has a fair bit of its provenance in various MO threads). </p> <p>The work in question is on the arxiv <a href="http://arxiv.org/pdf/1204.4483v2.pdf" rel="nofollow">here</a>. Various properties of an ordered field $R$ are being considered and compared. The last two are:</p> <blockquote> <p>(17) The Shrinking Interval Property: suppose $I_1 \supset I_2 \supset \ldots$ are (bounded) closed intervals in $R$ with lengths decreasing to zero. Then the intersection of the $I_n$'s is nonempty.</p> </blockquote> <p>and</p> <blockquote> <p>(18) The Nested Interval Property: Suppose $I_1 \supset I_2 \supset \ldots$ are (bounded) closed intervals in $R$. Then the intersection of the $I_n$'s is nonempty. </p> </blockquote> <p>I was not thrilled with the use of (bounded) in (17), but I let it go. I objected to the use of (bounded) in (18).</p> <p>Note that "(bounded)" is playing different roles in the two statements. In (17), it <em>is</em> a superfluous hypothesis: if the lengths of the intervals are decreasing to zero then necessarily all but finitely many of them are bounded. In (18) it certainly isn't. I found this lack of parallelism especially confusing: so confusing that the first time I read it I honestly did arrive at the (ridiculous) conclusion that Jim Propp was unaware that e.g. $\bigcap_{n=1}^{\infty} [n,\infty) = \varnothing$. </p> http://mathoverflow.net/questions/103055/notation-for-formal-laurent-series/103058#103058 Answer by Pete L. Clark for notation for formal Laurent series Pete L. Clark 2012-07-25T00:52:55Z 2012-07-25T00:52:55Z <p>I agree with the mathematician of your acquaintance -- well, okay, I am the mathematician of your acquaintance.</p> <p>Here are some references for the notation $K((x))$ for the field of formal Laurent series $\sum_{n \geq n_0} a_n x^n$ over $K$:</p> <blockquote> <p>The <a href="http://en.wikipedia.org/wiki/Formal_Laurent_series#Formal_Laurent_series" rel="nofollow">wikipedia article on formal power series</a>.</p> </blockquote> <p>$\textbf{}$</p> <blockquote> <p>Jacobson's <em>Basic Algebra II</em>, $\S 9.12$.</p> </blockquote> <p>$\textbf{}$</p> <blockquote> <p>Lam's <em>Introduction to Quadratic Forms Over Fields</em>, $\S VI.1$</p> </blockquote> <p>$\textbf{}$</p> <blockquote> <p>Neukirch's <em>Algebraic Number Theory</em>, $\S II.4$.</p> </blockquote> <p>$\textbf{}$</p> <blockquote> <p>Serre's <em>Corps Locaux</em>, $\S 1.1$.</p> </blockquote> http://mathoverflow.net/questions/43464/complex-analysis-applications-toward-number-theory/43465#43465 Answer by Pete L. Clark for Complex Analysis applications toward Number Theory Pete L. Clark 2010-10-25T03:18:46Z 2012-06-27T18:58:39Z <p>I think basic is on the right track. The two big classical theorems in analytic number theory whose classical proofs use some complex analysis are <a href="http://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions" rel="nofollow">Dirichlet's Theorem on primes in arithmetic progressions</a> and the <a href="http://en.wikipedia.org/wiki/Prime_Number_Theorem" rel="nofollow">Prime Number Theorem</a>. (It is also useful to learn about the combination of the two: the <a href="http://en.wikipedia.org/wiki/Prime_Number_Theorem#The_prime_number_theorem_for_arithmetic_progressions" rel="nofollow">Prime Number Theorem for Arithmetic Progressions</a>.) </p> <p>For the former, I can recommend my own lecture notes:</p> <p><a href="http://math.uga.edu/~pete/4400dirichlet.pdf" rel="nofollow">http://math.uga.edu/~pete/4400dirichlet.pdf</a></p> <p><a href="http://math.uga.edu/~pete/4400DT.pdf" rel="nofollow">http://math.uga.edu/~pete/4400DT.pdf</a></p> <p>The second part is explicitly a digested version of the proof Serre presents in his <em>Course in Arithmetic</em>. I don't have a similarly canonical reference to give you for the proof of the Prime Number Theorem (i.e., I don't have any notes on it!), but it can be found in many analytic number theory books, for instance in Apostol's <em>Introduction to Analytic Number Theory</em>, Davenport's <em>Multiplicative Number Theory</em> or G.J.O. Jameson's <em>The Prime Number Theorem</em>.</p> http://mathoverflow.net/questions/25663/elementary-proof-wanted-every-local-principal-ideal-ring-is-a-quotient-of-a-pid Elementary proof wanted: every local principal ideal ring is a quotient of a PID Pete L. Clark 2010-05-23T13:45:13Z 2012-06-18T05:43:12Z <p>I am looking for a more elementary proof of the following result:</p> <p>Theorem (Hungerford, 1968): Let $R$ be a principal ideal ring. Then $R \cong \prod_{i=1}^n R_i$, where each $R_i$ is a homomorphic image of a principal ideal <em>domain</em> (PID).</p> <p>Hungerford's article is available free online at:</p> <p><a href="http://projecteuclid.org/euclid.pjm/1102986148" rel="nofollow">http://projecteuclid.org/euclid.pjm/1102986148</a></p> <p>What do I mean by "more elementary"? Hungerford uses the Cohen structure theory of complete local rings, which I would like to avoid (because I have notes on commutative algebra which do not discuss such things). </p> <p>Note that Hungerford's theorem is a refinement of a previous result of Zariski and Samuel, which asserts that a principal ideal ring is isomorphic to a finite direct product of rings, each of which is either a PID or a "special principal ideal ring", i.e., a local Artinian principal ideal ring. The proof of this result uses primary decomposition, which is acceptable to me (in fact I put a section on primary decomposition into my notes for exactly this application). </p> <p>Given the theorem of Zariski-Samuel, Hungerford's result is plainly equivalent to the fact that every Artinian local principal ideal ring is the quotient of a PID. Now doesn't that sound like you should be able to prove it without invoking the structure theory of complete local rings? </p> http://mathoverflow.net/questions/28892/which-pair-of-mathematicians-has-the-most-joint-papers Which pair of mathematicians has the most joint papers? Pete L. Clark 2010-06-20T23:47:53Z 2012-06-05T14:24:12Z <p>I was searching on MathSciNet recently for a certain paper by two mathematicians. As I often do, I just typed in the names of the two authors, figuring that would give me a short enough list. My strategy was rather dramatically unsuccessful in this case: the two mathematicians I listed have written 80 papers together!</p> <p>So this motivates my (rather frivolous, to be sure) question: which pair of mathematicians has the most joint papers? </p> <p>A good meta-question would be: can MathSciNet search for this sort of thing automatically? The best technique I could come up with was to think of one mathematician that was both prolific and collaborative, go to their "profile" page on MathSciNet (a relatively new feature), where their most frequent collaborators are listed, alphabetically, but with the wordle-esque feature that the font size is proportional to the number of joint papers. </p> <p>Trying this, to my surpise I couldn't beat the 80 joint papers I've already found. Erdos' most frequent collaborator was Sarkozy: 62 papers (and conversely Sarkozy's most frequent collaborator was Erdos). Ronald Graham's most frequent collaborator is Fan Chung: 76 papers (and conversely).</p> <p>I would also be interested to hear about triples, quadruples and so forth, down to the point where there is no small set of winners.</p> <hr> <p><b>Addendum</b>: All right, multiple people seem to want to know. The 80 collaboration pair I stumbled upon is Blair Spearman and Kenneth Williams. </p> http://mathoverflow.net/questions/9641/how-far-is-lindelof-from-compactness/9642#9642 Answer by Pete L. Clark for How far is ﻿Lindelöf from compactness? Pete L. Clark 2009-12-23T21:47:38Z 2012-04-11T01:12:48Z <p>I've never heard of that result (which is not to say that I doubt its truth -- I have no opinion either way), but it reminds me of the following</p> <p>Theorem (N. Noble): If each power of a $T_1$-space is normal, then the space is compact. </p> <hr> <p>See</p> <p>MR0283749 (44 #979) Noble, N. Products with closed projections. II. Trans. Amer. Math. Soc. 160 1971 169--183</p> <p>and for a simpler proof,</p> <p>MR0415571 (54 #3656) Franklin, S. P.; Walker, R. C. Normality of powers implies compactness. Proc. Amer. Math. Soc. 36 (1972), 295--296. </p> <hr> <p>I wonder if there is any actual connection here?</p> http://mathoverflow.net/questions/17846/existence-of-maximal-totally-ramified-extensions-of-an-arbitrary-cdvf Existence of maximal totally ramified extensions of an arbitrary CDVF Pete L. Clark 2010-03-11T09:30:34Z 2012-01-27T15:18:40Z <p>Let $K$ be a complete, discretely valued field with (let's say) perfect residue field $k$. We have a unique maximal unramified extension $K^{unr}$ of $K$ and a unique maximal tamely ramified extension $K^{tame}$ of $K$ and hence short exact sequences</p> <p>$1 \rightarrow Gal(K^{sep}/K^{unr}) \rightarrow Gal(K^{sep}/K) \rightarrow Gal(K^{unr}/K) \rightarrow 1$</p> <p>and </p> <p>$1 \rightarrow Gal(K^{tame}/K^{unr}) \rightarrow Gal(K^{tame}/K) \rightarrow Gal(K^{unr}/K) \rightarrow 1$.</p> <p>In the second case, the normal subgroup is abelian and I know exactly what the action of the quotient on it is: the tame cyclotomic character. Therefore if it splits, I know its structure as an explicit semidirect product.</p> <p>In the most famous case, $k$ is finite, so <code>$Gal(K^{unr}/K) = Gal(k^{sep}/k) \cong \widehat{\mathbb{Z}}$</code> is a projective profinite group, so both sequences certainly split. This means that I (and lots of other people) do know the structure of the tame Galois group explicitly: it is $\prod_{\ell \neq p} \mathbb{Z}_{\ell}(1) \rtimes \widehat{\mathbb{Z}}$. Similarly the first sequence splits so there is a totally ramified extension $L/K$ such that $K^{sep}/L$ is unramified. Moreover, this is a very useful fact: it follows for instance that any abelian variety over $K$ with potentially good reduction acquires good reduction over a totally ramified base extension.</p> <p>What is known in general? We have $Gal(K^{unr}/K) = Gal(k^{sep}/k)$, but if $k$ is almost anything else reasonable -- e.g. a local or global field, or the function field of a variety -- then its absolute Galois group certainly will not be projective. What is known about the splitting of these two short exact sequences in general, and especially about the class $\eta \in H^2(Gal(K^{unr}/K),Gal(K^{tame}/K^{unr}))$ defined by the second sequence? Is there information on how the analogues of the above results do / do not work out if these sequences do not split? </p> http://mathoverflow.net/questions/8167/notation-name-for-artin-schreier-roots Notation/name for "Artin-Schreier roots"? Pete L. Clark 2009-12-08T06:45:06Z 2011-12-20T14:16:13Z <p>If x is an element of a field K and n is a positive integer, we have both a symbol and a name for a root of the polynomial t^n - x = 0: we denote it by x^{1/n} and call it an nth root of x.</p> <p>Of course nth roots play a vital role in field theory, e.g. in the characterization of solvable extensions in characteristic 0. However, in characteristic p > 0, the extraction of a p-power root is a much different business: it gives rise to purely inseparable extensions, not composition factors of solvable Galois extensions. </p> <p>To repair the characterization of solvable extensions in characteristic p as those being attainable as a tower of "radical" extensions, one needs to include the operation of taking roots of Artin-Schreier polynomials: t^q - t - x = 0, for q = p^a a power of the characteristic. </p> <p>Finally my question: do we have a name for an element t solving the equation t^q - t = x and/or a special notation for it? I do not know one. Similarly, whereas classically we often speak of x as being "an nth power", in this case I find myself writing "x is in the image of the Artin-Schreier isogeny \rho". Is there something better than this?</p> http://mathoverflow.net/questions/12072/what-is-the-prime-spectrum-of-a-cauchy-series-ring What is the prime spectrum of a Cauchy series ring? Pete L. Clark 2010-01-17T04:15:30Z 2011-11-30T09:28:50Z <p>Let $k$ be a field, and let $| \ |$ be a norm on $k$. The norm induces a metric. To construct the completion $\hat{k}$ as a normed field, the standard recipe is to take the quotient of the ring $\mathcal{C}(k)$ of all Cauchy sequences in $k$ -- viewed as a subring of $k^{\infty} = \prod_{i=1}^{\infty} k$ -- by the maximal ideal $\mathfrak{m}_0$ of all sequences converging to $0$.</p> <p>This brings up the following [idle] question: what are the maximal ideals of $\mathcal{C}(k)$? The prime ideals?</p> <p>My vague recollection had been that <code>$\mathfrak{m}_0$</code> was the unique maximal ideal of $\mathcal{C}(k)$, but this is evidently not the case: for every $n$, there is a maximal ideal <code>$\mathfrak{m}_n$</code> consisting of sequences whose $n$th coordinate is $0$, the residue field being $k$ again. It is easy to see though that <code>$\mathfrak{m}_0$</code> is the unique maximal ideal containing the <s>prime</s> ideal $\mathfrak{c} = \bigoplus_{i=1}^{\infty} k$. (Edit: $\mathfrak{c}$ is prime iff the norm is trivial.)</p> <p>Now this reminds me of filters. The prime ideals of the (zero-dimensional) ring $k^{\infty}$ correspond precisely to the ultrafilters on $\mathbb{Z}^+$. The principal ultrafilter of all sets containing $n$ pulls back to the maximal ideal <code>$\mathfrak{m}_n$</code>. Since every nonprincipal ultrafilter contains the Frechet filter of cofinite sets, it follows that it pulls back to $\mathfrak{m}_0$. But is it true that every maximal ideal of $\mathcal{C}(k)$ is pulled back from a prime (= maximal) ideal of $k^{\infty}$? If so, is this an instance of a general theorem? </p> <p><hr /></p> <p>Addendum: </p> <p>Note that in the case that the norm is trivial -- so that the induced metric is the discrete metric -- a sequence converges iff it is eventually constant, so a sequence converges to $0$ iff it has only finitely many nonzero terms: $\mathfrak{c} = \mathfrak{m}_0$. The converse also holds: for any nontrivial norm there exist nowhere zero sequences converging to $0$, e.g. <code>$\{x^n\}$</code> for any $x \in k$ with $0 &lt; |x| &lt; 1$.</p> <p><hr /></p> <p>Once the original question is worked out, I am also curious about generalizations. What is the analogue for the ring of minimal Cauchy filters in an arbitrary topological ring? </p> http://mathoverflow.net/questions/23571/number-fields-with-same-discriminant-and-regulator/23572#23572 Answer by Pete L. Clark for Number fields with same discriminant and regulator? Pete L. Clark 2010-05-05T11:43:02Z 2011-11-09T22:23:41Z <p>Yes, see e.g. the paper "Arithmetically equivalent number fields of small degree" (Google for it) by Bosma and de Smit. </p> <p>In brief: two number fields $K$ and $K'$ are said to be <strong>arithmetically equivalent</strong> if they have the same Dedekind zeta function. A famous group-theoretic construction of Perlis (Journal of Number Theory, 1977) gives many nontrivial (i.e., non-isomorphic) pairs of arithmetically equivalent number fields. Remarkably, this construction works equally well to construct isospectral, non-isometric Riemannian manifolds, as was later shown by Sunada.</p> <p>Arithmetically equivalent number fields necessarily share many of the simplest invariants, for instance they have equal discriminants. </p> <p>As the aformentioned paper explains, for arithmetically equivalent $K$ and $K'$, comparing zeta functions gives</p> <p>$h(K)r(K) = h(K')r(K')$,</p> <p>where $h$ is the class number and $r$ is the regulator. Therefore, to get an affirmative answer to your question you want a nontrivial pair of arithmetically equivalent number fields $K$ and $K'$ with $h(K) = h(K')$. The paper by Bosma and de Smit gives such examples.</p> http://mathoverflow.net/questions/80359/which-small-finite-simple-groups-are-not-yet-known-to-be-galois-groups-over-q Which small finite simple groups are not yet known to be Galois groups over Q? Pete L. Clark 2011-11-08T05:04:19Z 2011-11-09T05:53:23Z <p>The subject line pretty much says it all. To expand just a little bit:</p> <p>1) What is the smallest simple group that is not yet known to occur as a Galois group over $\mathbb{Q}$? (Variants: not known to occur <strong>regularly</strong> over $\mathbb{Q}$; not known to occur <strong>regularly</strong> over every Hilbertian field.)</p> <p>2) Is there a convenient list of the simple groups of order, say, at most $10,000$ together with information about which of them are known to be Galois groups over $\mathbb{Q}$?</p> <p>3) Is there perhaps some nice website keeping track of information like this? (There should be!)</p> <p><b>Added</b>: The table in $\S 8.3.4$ of Serre's <em>Topics in Galois Theory</em> (the 1992 edition; I don't know if this matters) lists <code>$\operatorname{SL}_2(\mathbb{F}_{16})$</code> -- which has order $4080$ -- as the smallest simple group which is not known to occur regularly over $\mathbb{Q}$. On the other hand <a href="http://www.warwick.ac.uk/~masjam/papers/sl2f16.pdf" rel="nofollow">this 2007 paper</a> of Johan Bosman shows that <code>$\operatorname{SL}_2(\mathbb{F}_{16})$</code> occurs as a Galois group over $\mathbb{Q}$, but mentions that the problem of realizing it regularly is still open. Thus the answer to the "regular" variant of the question seems to be <code>$\operatorname{SL}_2(\mathbb{F}_{16})$</code>. But to be clear, I am really looking for as much data as possible, not just "records".</p> http://mathoverflow.net/questions/103556/a-unified-description-of-zeta-functions-of-a-curve-over-mathbbf-q-and-rieman/103558#103558 Comment by Pete L. Clark Pete L. Clark 2013-05-16T15:53:25Z 2013-05-16T15:53:25Z Fun fact: today I am thinking about counting ideals in the coordinate ring of an affine curve over a finite field. Searching the internet for the fact I needed about the zeta function, I came to your answer, which I found helpful. Then I clicked on the link to my own set of notes, which was further help! Thanks for the answer. http://mathoverflow.net/questions/128194/upper-bound-on-order-of-finite-subgroups-of-gl-nz-p Comment by Pete L. Clark Pete L. Clark 2013-04-23T03:31:13Z 2013-04-23T03:31:13Z Actually, now that I say that it is tempting to look for a cheap model-theoretic proof that every finite subgroup of $\operatorname{GL}_n(\mathbb{C})$ embeds in $\operatorname{GL}_n(\mathbb{Q}_p)$ for infinitely many primes $p$! http://mathoverflow.net/questions/128194/upper-bound-on-order-of-finite-subgroups-of-gl-nz-p Comment by Pete L. Clark Pete L. Clark 2013-04-23T03:29:13Z 2013-04-23T03:29:13Z @Francois: If I am not mistaken, the descent from $\mathbb{C}$ to $\overline{\mathbb{Q}}$ follows easily from the completeness of the theory of algebraically closed fields of characteristic $0$. But anyway this is a good first step towards the solution. http://mathoverflow.net/questions/128318/embeddings-of-finite-groups-into-gln-q-p Comment by Pete L. Clark Pete L. Clark 2013-04-23T03:25:02Z 2013-04-23T03:25:02Z I'm glad you posted the question. I think though you should also refer the reader interested in CET to either Cassels's book <i>Local Fields</i> or to: J.W.S. Cassels, An embedding theorem for fields. Bull. Austral. Math. Soc. 14 (1976), 193-198. http://mathoverflow.net/questions/128267/are-there-modular-elliptic-curves-over-a-field-extension-of-mathbbzi/128424#128424 Comment by Pete L. Clark Pete L. Clark 2013-04-23T03:21:03Z 2013-04-23T03:21:03Z @Robert: It sounds like you are looking for, among other things, the definition of the <b>Hasse-Weil Zeta Function</b> of a variety defined over a number field. Knowing the name will allow you to see that there is quite a large literature here. Also Felipe is right: based on what you write, you will need to learn some basic algebraic number theory in order to make sense of all this. (If you do not have this background, you can feel free to ask very basic questions on the related site math.stackexchange.com.) http://mathoverflow.net/questions/128267/are-there-modular-elliptic-curves-over-a-field-extension-of-mathbbzi/128270#128270 Comment by Pete L. Clark Pete L. Clark 2013-04-23T03:15:15Z 2013-04-23T03:15:15Z Concerning the first sentence: is it completely obvious that one cannot have an elliptic curve $E$ with rational $j$-invariant which splits off as an isogeny factor of some $J_0(N)$ only over some larger number field? I remember this can't happen when $N$ is squarefree, because by a 1975 theorem of Ribet the geometric endomorphism ring of $J_0(N)$ is equal to its $\mathbb{Q}$-rational endomorphism ring. What about the general case? (I suspect I once knew the answer to this but have forgotten...) http://mathoverflow.net/questions/128194/upper-bound-on-order-of-finite-subgroups-of-gl-nz-p Comment by Pete L. Clark Pete L. Clark 2013-04-21T21:32:46Z 2013-04-21T21:32:46Z By the way, where did I learn about this amazing result of Cassels? He has it in his text <i>Local Fields</i>. Concerning that let me say: a few years ago I was planning out a second graduate course on algebraic number theory -- valuations, local fields and adeles -- and I couldn't find a suitable text, so I ended up making the notes I linked to in my answer. Soon after the course ended I pulled Cassels's text off of my shelf, flipped through it for five minutes and realized, &quot;Oh, crap. I should have used this instead.&quot; http://mathoverflow.net/questions/128194/upper-bound-on-order-of-finite-subgroups-of-gl-nz-p Comment by Pete L. Clark Pete L. Clark 2013-04-21T21:30:37Z 2013-04-21T21:30:37Z @Francois: CET says: let K be finitely generated of characteristic 0, and let x_1,..,x_n be nonzero elements of K. Then there are infinitely many primes p such that K can be field-embedded in Q_p with each x_1,..,x_n mapped to a p-adic unit. When K is a # field this is certainly a Cebotarev condition, as I mention in my writeup. When K is transcendental &quot;a miracle occurs&quot;, of which I do not claim any deep understanding; in particular the argument itself doesn't lead to a Cebotarev condition. But nevertheless, sure, I'll bet that set of primes has a natural density. (Maybe ask it on MO!) http://mathoverflow.net/questions/128194/upper-bound-on-order-of-finite-subgroups-of-gl-nz-p/128196#128196 Comment by Pete L. Clark Pete L. Clark 2013-04-21T21:22:17Z 2013-04-21T21:22:17Z @Geoff: Thanks! http://mathoverflow.net/questions/128194/upper-bound-on-order-of-finite-subgroups-of-gl-nz-p Comment by Pete L. Clark Pete L. Clark 2013-04-21T17:51:54Z 2013-04-21T17:51:54Z (And the Cassels Embedding Theorem shows up in the lecture notes cited in my answer.) http://mathoverflow.net/questions/128194/upper-bound-on-order-of-finite-subgroups-of-gl-nz-p Comment by Pete L. Clark Pete L. Clark 2013-04-21T17:49:21Z 2013-04-21T17:49:21Z @Francois: True. But actually the Cassels Embedding Theorem shows more than this: every finite subgroup of $\operatorname{GL}_n(\mathbb{C})$ occurs in $\operatorname{GL}_n(\mathbb{Z}_p)$ for infinitely many primes $p$. http://mathoverflow.net/questions/128194/upper-bound-on-order-of-finite-subgroups-of-gl-nz-p/128201#128201 Comment by Pete L. Clark Pete L. Clark 2013-04-21T17:46:49Z 2013-04-21T17:46:49Z @Geoff: I'm sorry, evidently I haven't been clear enough on this point: the bound <b>does not</b> only depend on $n$ and $p$. It fully depends on $n$, $p$, the ramification index $e$ and the inertial degree $f$: if you fix any three of them and let the fourth go to infinity, the supremum will be infinite. http://mathoverflow.net/questions/128194/upper-bound-on-order-of-finite-subgroups-of-gl-nz-p Comment by Pete L. Clark Pete L. Clark 2013-04-21T14:22:03Z 2013-04-21T14:22:03Z @Jim: okay, but since $\operatorname{GL}_n(K)$ embeds in $\operatorname{SL}_{n+1}(K)$, up to this shift on $n$, restricting to determinant $1$ can't help attain boundedness. (As I'm sure you know!) http://mathoverflow.net/questions/128130/what-does-a-mathematician-expect-from-mathematics-education Comment by Pete L. Clark Pete L. Clark 2013-04-21T02:32:57Z 2013-04-21T02:32:57Z @Joel: Okay, so I used the wrong word. But functionally the site is explicitly modelled after the SE platform, so it amounts to the same. http://mathoverflow.net/questions/128194/upper-bound-on-order-of-finite-subgroups-of-gl-nz-p/128201#128201 Comment by Pete L. Clark Pete L. Clark 2013-04-21T01:11:24Z 2013-04-21T01:11:24Z Also, although I certainly don't claim that any of these bounds are sharp, I believe, in line with the above comments, that easy examples will show that if you fix any two of $e$ (the ramification index of $K/\mathbb{Q}_p$), $f$ (the residual degree of $K/\mathbb{Q}_p$) and $n$ and let the third one grow, then there will be no uniform bound.