User cam mcleman - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T08:56:09Z http://mathoverflow.net/feeds/user/3456 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39430/algebraic-attacks-on-the-odd-perfect-number-problem Algebraic Attacks on the Odd Perfect Number Problem Cam McLeman 2010-09-20T20:27:15Z 2012-08-23T12:04:48Z <p>The <a href="http://en.wikipedia.org/wiki/Perfect_number#Odd_perfect_numbers" rel="nofollow">odd perfect number</a> problem likely needs no introduction. Recent progress (where by recent I mean roughly the last two centuries) seems to have focused on providing restrictions on an odd perfect number which are increasingly difficult for it to satisfy (for example, congruence conditions, or bounding by below the number of distinct prime divisors it must have). By reducing the search space in this manner, and probably due to other algorithmic improvements (factoring, parallelizing, etc.), there has also been significant process improving lower bounds for the size of such a number. A link off of <a href="http://oddperfect.org" rel="nofollow">oddperfect.org</a> claims to have completed the search up to $10^{1250}$. </p> <p>But, assuming my admittedly cursory reading of the landscape is correct, none of the current research seems particularly equipped to prove non-existence. The only compelling argument I've seen on this front is "Pomerance's heuristic" (also described on <a href="http://oddperfect.org" rel="nofollow">oddperfect.org</a>). Worse, and maybe this is really the point of this question, it would be a little disappointing if the non-existence proof was an upper bound of $10^{1250}$ (depending on the techniques used to get the bound) combined with the above brute force search. </p> <p>On the other hand, maybe there's some hope that some insight can be gained into the sum-of-divisors function by modern techniques. For example, the values of the arithmetic functions $$\sigma_{k}(n):=\sum_{d\mid n}d^k,$$ for $k\geq 3$ odd, arise as coefficients of normalied Eisenstein modular forms, and the study of said forms gives amazing proofs of amazing identities between them. For $k=1$, the case of interest, the normalized Eisenstein series $E_2$ is only "quasi-modular", but such forms satisfy sufficiently nice transformation properties that I wonder if $E_2$ has anything to say about the problem. </p> <blockquote>Since no doubt many people on this site will be able to immediately address the previous idea (so please do!), my more general question is whether or not there are applications of the modern machinery of modular forms, mock modular forms, diophantine analysis, Galois representations, abc conjecture, etc., that have anything to say about the odd perfect number problem. Does it descend from or relate to any major open problems from modern algebraic/analytic number theory? </blockquote> <p><sub> Aside: I hope this does not come off as dismissive of "elementary" techniques, or of the algorithmic ones mentioned in the first paragraph. Indeed, they have, to my knowledge, been the only source of progress on this problem, and certainly contain interesting mathematics. Rather, this phrasing stems from my desire to find anything in the intersection of "odd perfect number theory" and "things I know anything about," and perhaps a desire to see the odd perfect number problem settled without the use of a beyond-gigantic brute force search. </sub></p> http://mathoverflow.net/questions/94987/commutative-algebra-with-a-view-toward-algebraic-number-theory/95040#95040 Answer by Cam McLeman for Commutative Algebra with a View Toward Algebraic Number Theory Cam McLeman 2012-04-24T16:28:04Z 2012-04-24T16:28:04Z <p>I concur that Neukirch is a good candidate, so instead of starting with a new recommendation (I'll come back to that later), let me instead disagree with Felipe Voloch's contention that algebraic number theory is all about rings of dimension one (though certainly he had a narrower scope of algebraic number theory in mind than I'm about to describe). So a quick run-down of the fundamental, and reasonably beginner grad-level, commutative algebra I've run into doing algebraic number theory, with the caveat that I've never been very good at figuring out where commutative algebra ends and some of these other things begin (in particular, commutativity tends to fade away somewhat silently):</p> <ol> <li><b>Basic stuff</b>: As mentioned above. Dedekind rings, local rings, valuation theory, integral closures, PIDs/UFDs, etc.<p></li> <li><b>Arithmetic Geometry</b>: Okay, okay, this one's cheating given the context of the question. But still, you can't get <i>too</i> far in algebraic number theory before you run into an elliptic curve, and then you'll want to know something about its function field, and so on.<p> <li><b>Homological Algebra</b>: Free and projective resolutions of groups, most poignantly with the goal of getting to Galois cohomology, which is a natural language for much of algebraic number theory. In particular, there's the cohomological version of class field theory, Cornell and Rosen's treatise on getting much of algebraic number theory cohomologically, Tate-Shafarevich groups, local-global obstructions, etc.<p></li> <li><b>Topological Rings/Fields:</b> e.g., rings of adeles. More generally, direct/inverse limit constructions, especially to get cohomology of profinite groups via limits.<p></li> <li><b>Fancier Stuff</b>: Of which there is probably no end. But I'll just mention that, e.g., Wiles's proof of FLT uses universal deformation rings, complete intersection rings, Gorenstein rings, etc. (Though some of this was subsequently tidied up a little.) </ol> <p>My recommendation would be to start with Neukirch's <i>Algebraic Number Theory</i> for roughly the first bullet point, and as the follow-up book, to go to Manin and Panchishkin's <i>Introduction to Modern Number Theory</i> for basically everything else in the list (with a hat tip to Lorenzini's <i>An Introduction to Arithmetic Geometry</i> as mentioned in the comments). For books that then make heavy <i>use</i> of this material, there's Neukirch et al's follow-up book <i>Cohomology of Number Fields</i>, and Georges Gras's <i>Class Field Theory.</i></p> http://mathoverflow.net/questions/88539/sums-of-rational-squares/88549#88549 Answer by Cam McLeman for sums of rational squares Cam McLeman 2012-02-15T20:01:04Z 2012-02-15T20:01:04Z <p>This result is pretty shy of needing the full Hasse-Minkowski Theorem. Indeed, since Fermat already knew which integers were a sum of two integer squares, it would suffice for him to show that those that weren't (i.e., those with an odd power of some prime congruent to 3 mod 4 showing up in its prime factorization) could also not be written as a sum of two rational squares. But this is the <i>easy</i> direction of Hasse-Minkowski: To show that (let's say) a prime $p\equiv 3\pmod{4}$ can't be written as a sum of two rational squares, it suffices to check that it can't be a sum of two $\ell$-adic rational squares for some $\ell$. Of course, Fermat did not have the language of the $\ell$-adics, so this would have had been replaced with mod-$q^k$ conditions for various $k$. </p> <p>Specifically, the modern Hasse-Minkowski proof boils down to the statement that a prime which is 3 mod 4 can't be written as a sum of two rational squares because it can't be done so 2-adically. Indeed, one can just compute the single Hilbert symbol $$p\equiv 3\pmod{4}\Rightarrow (p,-1)_2=(-1)^{(p-1)/2}=-1,$$ showing that $x^2=pz^2-y^2$ has no $2$-adic, and hence no rational, solutions, which afte the substitutions $a=x/z$ and $b=y/z$, implies one cannot write $p=a^2+b^2$ with $a,b\in\mathbb{Q}$. Of course (again), Fermat did not have Hilbert symbols, but this is just a change of language away from Fermat's approach (I imagine). It would not be hard to unwind the above calculation into a single (probably lengthy) mod-8 calculation, since that's all that goes into deciding which elements of $\mathbb{Q}_2$ are squares, which in turn is essentially all that lives behind the Hilbert symbols.</p> http://mathoverflow.net/questions/87573/relations-between-class-numbers-of-quadratic-extensions/87619#87619 Answer by Cam McLeman for relations between class numbers of quadratic extensions Cam McLeman 2012-02-05T21:16:38Z 2012-02-05T21:16:38Z <p>The short answer to your question is basically no, there's essentially no connection between the prime powers $q^i$ dividing $h_p$ and $h_{-p}$. </p> <p>It's true that there's a general relationship between the 2-ranks of the class groups of $\mathbb{Q}(\sqrt{m})$ and $\mathbb{Q}(\sqrt{-m})$ as per Franz's answer and the result from Washington you cite (which, incidentally, can be pushed further to 4-ranks, 8-ranks, etc., getting pretty close to a full comparison of the 2-parts of the class numbers given knowledge of the fundamental unit of the real quadratic), but for your case this is almost contentless by genus theory. For other primes $q$, or the class number in whole, it's hard to give a conclusive justification for the "no relationship" claim, though two points bear mentioning: </p> <ol> <li>Larges tables of these class numbers are available or easily generated, a quick survey of which is pretty compellingly against any correlation; and</li> <li>There are all sorts of heuristics out there about how the two classes of class numbers should behave, some of which great imply a lack of relationship. For example, it's very unlikely that $q^i$ dividing $h_{-p}$ could tell you anything about $q^i$ dividing $h_p$ since $h_{-p}$ is non-trivial for all but nine examples, whereas (probably) $h_p=1$ infinitely often. </li></ol> http://mathoverflow.net/questions/87160/geometric-interpretation-of-the-lower-central-series-for-the-fundamental-group/87167#87167 Answer by Cam McLeman for Geometric Interpretation of the Lower Central Series for the Fundamental Group? Cam McLeman 2012-01-31T20:07:31Z 2012-01-31T20:07:31Z <p>A special case you might find informative: If $L$ is a link in $S^3$, then Chen-Milnor theory gives you a presentation of the link group $\pi=\pi_1(S^3-L)$ modulo some deeper terms of the the lower central series of $\pi$, and hence some information about some of the early lower central factors that you're asking about. Particularly neat is that this presentation is directly in terms of combinatorial invariants (linking number, Milnor invariants) of the link, and thus gives concrete interpretations to various cohomological invariants that arise from the algebraic topology viewpoint (cup product, Massey products, etc.)</p> <p>Also worth mentioning is the gadget you get by gluing all of these lower central quotients together, namely the associated graded Lie algebra. (And you can repeat for the lower central p-series, and several other relevant series as well). There's a good amount known about these Lie algebras: For example, John Labute's "The Lie Algebra Associated to the Lower Central Series of a Link Group and Murasugi's Conjecture" (and the rest of Labute's paper for that matter. In particular, some amazing ties to "arithmetic topology" and number-theoretically interesting Galois groups).</p> http://mathoverflow.net/questions/85540/realizing-groups-as-commutator-subgroups/85545#85545 Answer by Cam McLeman for Realizing groups as commutator subgroups Cam McLeman 2012-01-13T01:59:57Z 2012-01-13T12:56:53Z <p>A complete answer seems not to be known. Let me give you the following two nearly-contemporaneous references from the mid-70s:</p> <p>Robert Guralnick, <em>On groups with decomposable commutator subgroups</em></p> <p>Michael Miller, <em>Existence of Finite Groups with Classical Commutator Subgroup</em></p> <p>Both Guralnick and Miller call groups which are commutator subgroups $C$-groups (though I don't know who, if either, originated the term) and give partial answers to your general question. For example, Theorem 4 from Miller gives the following:</p> <blockquote> Let $G$ be a subgroup of $\operatorname{GL}_n(K)$ containing $\operatorname{SL}_n(K)$ for $K$ a finite field of characteristic not equal to 2. Then $G$ is the commutator subgroup of some group unless it is of odd index and $n$ is even. </blockquote> <p>The groupprops-wiki calls such groups commutator-realizable, and give a basic result on such groups, but mention that this terminology is not standard (though is probably safer than the overloaded term $C$-group.)</p> <p>Edit: Some googling around led to the following slick argument of Schoof (from his <em>Semistable abelian varieties with good reduction outside 15</em>), which is closely related to your observation in bullet (3), and also serves to eliminate the symmetric groups. I'll quote verbatim except for change of variable names:</p> <blockquote> Let $G$ be a group and let $G'$ be its commutator subgroup. Conjugation gives rise to a homomorphism $G \to \operatorname{Aut}(G')$. On the one hand it maps $G'$ to the commutator subgroup of $\operatorname{Aut}(G')$. On the other hand the image of $G'$ is the group $\operatorname{Inn}(G')$ of inner automorphisms of $G'$. Therefore, if a group $X$ is the commutator subgroup of some group, we must have $\operatorname{Inn}(X)\subset \operatorname{Aut}(X)'$. </blockquote> http://mathoverflow.net/questions/82847/a-metabelian-quotient-of-a-free-group/82849#82849 Answer by Cam McLeman for A metabelian quotient of a free group Cam McLeman 2011-12-07T04:28:22Z 2011-12-07T04:55:12Z <p>Do you mean the augmentation map on the group ring $\mathbb{Z}[F]$ (and in the pro-$p$ case, the completed group ring $\mathbb{Z}_p[[F]]$?) I ask only because this augmentation map comes up frequently and significantly in the study of large number-theoretic Galois groups.</p> <p>Assuming this is the case (and apologies for misinterpreting if not -- hopefully the answer will still be of some use to you), there is a tremendous amount of machinery set up for dealing exactly with questions of this sort -- probably the best starting place is the phrase "pro-p Fox Differential Calculus." (And so, indeed, your intuition that solving the discrete problem turns out to provide the correct pro-$p$ analog is correct. It was Iwasawa who carefully established the fundamental analogy here. In fact, thanks to the topology of $\mathbb{Z}_p$, in some ways the pro-p Fox calculus is nicer than the discrete version.) In particular, if you filter the group ring $\mathbb{Z}_p[[F]]$ by powers of the augmentation ideal (the group-ring version of your $A$), you land upon the sequence of "dimension subgroups" of F.</p> <p>These subgroups have shown up repeatedly in the analysis of pro-$p$-groups arising in the study of large Galois groups arising from restricted ramification questions (as appears to be the case for you). A couple of the highlights of the theory are the work of Vogel and Morishita interpreting number-theoretic analogs of the <i>a priori</i> knot-theoretic notion of Milnor invariants, refined versions of Golod-Shafarevich-type inequalities, and perhaps most relevant for your question, work of Arrigoni (e.g., "On Schur $\sigma$-groups") which I think explicitly answers questions of your type. For a more fundamental reference, see Koch's "Galois theory of $p$-extensions.")</p> <p>Sorry to be mostly hand-wavey -- I'm away from good references at the moment.</p> http://mathoverflow.net/questions/76616/unramified-extensions-of-number-fields/76617#76617 Answer by Cam McLeman for Unramified extensions of number fields Cam McLeman 2011-09-28T09:01:23Z 2011-09-28T09:01:23Z <p>Two things: </p> <p>1) Yes, certainly. By class field theory and the finiteness of the class group, the maximal <em>abelian</em> unramified extension of <em>any</em> number field is of finite degree. Thus any infinite unramified extension is non-abelian -- in particular, any infinite class field tower. The Golod-Shafarevich examples and oodles of refinements since then all give examples.</p> <p>2) Perhaps you are actually interested in non-<em>solvable</em> infinite extensions, i.e., unramified extensions which are not built up as an infinite series of unramified abelian extensions. In this case, the answer is also yes. In fact, this can even be done over number fields of class number 1. For examples, see Maire's "On Infinite Unramified Extensions."</p> <p>Hope that helps.</p> http://mathoverflow.net/questions/74373/lower-bound-on-the-class-group-of-the-p-hilbert-class-field-of-an-imaginary-quadr/74377#74377 Answer by Cam McLeman for Lower bound on the class group of the p-Hilbert class field of an imaginary quadr. field Cam McLeman 2011-09-02T16:44:48Z 2011-09-02T16:44:48Z <p>If $G$ is the Galois group of the $p$-class field tower over $K$, then $A(H(K))=G'/G''$ is a quotient of $G_2/G_4$, where $G_i$ denotes the lower central series. By Arrigoni's calculation that $G_2/G_4$ has $p$-rank exactly $\frac{d(d-1)(2d+5)}{6}$, this serves as a lower bound for the $p$-rank of $A(H(K))$. When $d=2$, you get the bound of $3$ you mention in the question. Note that the calculation is actually much more precise: The size of $A(H(K))$ depends not only on the rank, but on the orders of the generators of the $p$-class group. This will give you a better bound for the class number than simply raising $p$ to the rank bound given above.</p> http://mathoverflow.net/questions/22910/exactness-of-2nd-order-differential-equations-via-differential-forms Exactness of 2nd-Order Differential Equations via Differential Forms Cam McLeman 2010-04-28T23:37:11Z 2011-08-13T22:15:57Z <p>This (probably very elementary) question came up the last time I taught differential equations, and I've been toying with it for a while with no success:</p> <p>A 1st-order differential equation $M(x,y)dx+N(x,y)dy=0$ is exact if $$M(x,y)dx+N(x,y)dy=f_x(x,y)dx+f_y(x,y)dy$$ for some differentiable function $f(x,y)$ defined on the domain of $\omega$. In this case, we easily arrive at an implicitly-defined solution to the differential equation. Importantly, there is a nice test for exactness stemming from Clairaut's theorem -- for everywhere smooth $M$ and $N$ (for simplicity/laziness...obvious generalizations abound), the differential equation is exact iff $N_y=M_x$. Of course, this procedure is easily re-interpreted as saying that by the triviality of $H^1(\mathbb{R}^2)$, a one-form is closed if and only if it is exact.</p> <p>Now let's move one degree higher. Boyce and Di Prima define a 2nd-order differential equation $P(x)y''+Q(x)y'+R(x)y=0$ to be exact if there exists a differentiable function $f(x,y)$ such that the differential equation can be written</p> <p>$$P(x)y''+Q(x)y'+R(x)y=[P(x)y']'+[f(x)y]'=0.$$</p> <p>The analogous expression to Clairaut's theorem seems to be that (again, for sufficiently smooth inputs) an equation of that form is exact iff $P''(x)-Q'(x)+R(x)=0.$ Of importance is that such forms can be integrated once to leave us with a 1st-order differential equation. So we've successfully lowered the degree of our problem.</p> <p>This feels to me very much like an analogous $H^2$ calculation. We have a condition on some coefficients that very much looks like an alternating sum coming from a $d$ map on forms, and lets us conclude that the equation "comes from" a one-degree-smaller differential equation. </p> <p>But! (and here's the question) I can't seem to fit any 2-forms into this picture that would explain this analogy. Presumably there's some big story here linking the two notions of exactness about which I'd love to be enlightened.</p> <p>Side remark: I once received a partial response that there might be a link with Cartan tableau, which I've been unsuccessful in pursuing, if that helps spark an idea.</p> http://mathoverflow.net/questions/69617/on-what-kind-of-objects-do-the-galois-groups-act/69636#69636 Answer by Cam McLeman for On what kind of objects do the Galois groups act? Cam McLeman 2011-07-06T14:14:09Z 2011-07-06T14:14:09Z <p>This is not exactly an incarnation of the question you asked, in the sense that is not so much an action of a Galois group but rather an action whose existence is governed by a Galois group of number-theoretic origin, but it seems likely to be of interest.</p> <p>Let $K$ be a number field, and let $K^{(1)}$ be the maximal unramified abelian extension of $K$. The Galois group of $K^{(1)}/K$ is a subquotient of Gal$(\overline{\mathbb{Q}}/\mathbb{Q})$) which is isomorphic to the class group of $K$. Note that by Minhyong Kim's answer <a href="http://mathoverflow.net/questions/49960/are-class-numbers-encoded-in-the-absolute-galois-group-of-mathbb-q/49967#49967" rel="nofollow">here</a>, we can characterize this subquotient purely Galois-theoretically. Several authors have discovered surprising links between the arithmetic of number fields and actions of groups on spheres. In particular, when $K$ is the real cyclotomic field $K_m=\mathbb{Q}(\zeta_m+\zeta_m^{-1})$, the class group appears to govern the free actions of binary dihedral groups on spheres $S^n$ with $n\equiv 3\pmod{4}$. Let me loosely quote/paraphrase from Lang's "Units and Class Groups in Number Theory and Algebraic Geometry" (bolding mine):</p> <blockquote> C. T. C. Wall has already shown to depend in part on the 2-primary component of the ideal class group in real cyclotomic fields $K_m^+$ for suitable $m$...Using the algebraic background of a paper of Wall, applied to the surgery exact sequence, Thomas gives examples for the binary dihedral group $D_{4p}$ of order $4p$ operating freely on $S^{4k-1}$ with $k\geq 2$, <b>when the order of</b> $[(K_p^+)^{( 1)}:K_p^+]$ <b>is odd.</b> <br> <p>... <br> Furthermore, according to Thomas, there exist free actions by $D_{4p}$ which can be topologically distinguished <b>only</b> by an invariant in the 2-primary part of the ideal class group of $K_p^+$. </blockquote> <p>Perhaps needless to say, the study of these degrees $[(K_p^+)^{( 1)}:K_p^+]$, even their 2-part, is of tremendous interest in algebraic number theory (Vandiver's conjecture, etc.), so the link to actions on spheres is surprising.</p> http://mathoverflow.net/questions/64795/partitions-into-0-1-and-2-with-a-partial-sum-condition Partitions into 0,1, and 2 with a partial sum condition. Cam McLeman 2011-05-12T14:58:34Z 2011-05-12T16:04:44Z <p>On a tangent to a problem I've been working on, I've run into a combinatorial/partition-theoretic problem that I wondered if anyone had run into before. </p> <p>Let $N$ be a positive integer, and <em>ad-hoc</em>-ly call an (ordered) non-negative partition of $N$ into exactly $N$ parts $$N=n_1+n_2+\cdots+n_N$$ <em>valid</em> if</p> <p></p> <ul> <li>$0\leq n_i\leq 2$ for all $i$; and </li> </ul> <p><ul> <li><code>$\sum\limits_{k=1}^i n_k&lt;i$</code> for all <code>$i&lt;N$</code>.</li></ul></p> <p>So these are something like partitions where the running total is always bounded by the number of terms added thus far. (So the running average of the elements of the partition is less than 1.) In particular, this forces $n_1=0$ and $n_N=2$.</p> <p>I'm more interested in whether this notion of a "valid" partition has arisen previously in the literature than an explicit count of how many of them there are for a given $N$ (probably a reasonably straight-forward linear recurrence or something), so any such references would be appreciated.</p> http://mathoverflow.net/questions/63879/conceptualizing-weil-pairing-for-elliptic-curves-and-number-fields/63883#63883 Answer by Cam McLeman for Conceptualizing Weil Pairing for elliptic curves ( and number fields) Cam McLeman 2011-05-04T06:06:50Z 2011-05-04T18:29:50Z <p>The unifying picture you're looking for is probably most transparent the other way around -- by re-writing the Weil pairing on elliptic curves (in fact, this works more generally for Jacobians) to make it look like Hilbert symbols. Indeed, once you view the Weil pairing as a class-field-theoretic construction and pass it through the standard function-field-to-number-field analogy, you get exactly the Hilbert symbols. This is made very explicit in, for example, Everett Howe's "The Weil Pairing and the Hilbert Symbol." With notation in the paper, compare the Weil pairing formula </p> <p>\begin{equation*} e_m([X],[Y])=\prod_{p}(-1)^{m(\text{ord}_P(D))(\text{ord}_P(E))}\frac{g^{\text{ord}_P(D)}}{f^{\text{ord}_P(E)}}(P) \end{equation*}</p> <p>(here, $X$ and $Y$ are $m$-torsion divisors on the Jacobian of a curve with $mX=div(f)$ and $mY=div(g)$, with $P$ running over geometric points of the curve) with Schmidt's formula for the Hilbert symbol, reveals a striking similarity.</p> <p>I'm not sure if I have anything coherent to say about an improved conceptual explanation other than that the class-field-theoretic approach makes the Weil pairing appear as a natural and canonical construction, whereas the standard divisor construction feels rather <i>ad hoc</i> at first.</p> http://mathoverflow.net/questions/59020/where-can-i-find-a-modern-write-up-of-heegners-solution-of-gauss-class-number-1/59030#59030 Answer by Cam McLeman for Where can I find a modern write-up of Heegner's solution of Gauss' class number 1 problem? Cam McLeman 2011-03-21T02:40:08Z 2011-03-21T02:40:08Z <p>In his article <i>On the "gap'' in a theorem of Heegner</i>, Stark does a pretty thorough job of explaining where people thought the purported gap came from, to what extent it actually was a gap, and what you would need to fix such a thing if it existed. I'm paraphrasing, but he basically argues that the confusion stemmed from some errors (typos?) in some analytic results of Weber that Heegner had heavily used. So in a literal sense, Heegner had not proved it because he had cited faulty results, but Stark shows that he deserved credit for the theorem since using Heegner's argument with the correct versions of Weber results (which were indeed known to Weber), the job gets done.</p> <p>Here's the mathscinet review of the article:</p> <p><a href="http://www.ams.org/mathscinet-getitem?mr=241384" rel="nofollow">http://www.ams.org/mathscinet-getitem?mr=241384</a></p> http://mathoverflow.net/questions/58047/explicitly-describable-maximal-unramified-extension-of-a-number-field/58075#58075 Answer by Cam McLeman for Explicitly describable maximal unramified extension of a number field Cam McLeman 2011-03-10T14:13:00Z 2011-03-11T19:08:07Z <p>No, I'm pretty sure not.</p> <p>In general, the theory is much more developed for the maximal <i>pro-p-quotient</i> of the groups you're asking about, and even in this more explored setting, not a single explicit presentation of an infinite such group is known (to me, for sure, but I think to anyone -- <b>Edit:</b> Nigel Boston appears to concur in his survey paper <i>Galois $p$-groups unramified at $p$</i>). In fact, this is true even if we generalize to ask about the Galois groups $G_{S,p}(K)$ of maximal $p$-extensions unramified outside of a finite set of primes $S$ (with some tameness conditions on $S$ -- obviously taking $K=\mathbf{Q}(\zeta_p)$ and $S$ as the set of primes above $p$ gives a counterexample to my claims.) </p> <p>To elaborate, what we <i>do</i> have are certain approximations to presentations for such groups. If you consider a pro-$p$ presentation of $G_{S,p}(K)$: \begin{align*} 1\to R\to F\to G_{S,p}(K)\to 1, \end{align*} where $F$ is a free pro-$p$-group, then in some cases (for example, $K=\mathbb{Q}$) you can find an approximation to a minimal generating set for the relation module $R$ in the sense that you can give an explicit description of these generators in some quotient of $F$, e.g., modulo the third step in the lower central series of $F$. (Actually, it's the "Zassenhaus filtration" for which the results are sharpest.) A lot of authors have written on this idea, which roughly originated with Koch -- I'd recommend NSW's "Cohomology of Number Fields" for an overview, and then work of Morishita, Vogel, or possibly myself for more details. In brief, these approximations are determined by the arithmetic of $K$ and $S$ (e.g., $p$-th power residue symbols or other class-field-theoretic symbols evaluated at the primes in $S$). This was a fairly resounding triumph of the theory -- it would be outright revolutionary to lift these congruences of relations mod $F_3$ to literal equalities in $F$.</p> <p>Let me finish by bringing the general case back toward your original unramified setting. The relevance of the more general case is as follows: Say, for example, that you have a quadratic extension $K$ of $\mathbb{Q}$ and would like to know its maximal unramified 2-extension. If we let $S$ be the set of primes dividing the discriminant of $K$, then the maximal unramified 2-extension of $K$ corresponds to an index-2 subgroup of $G_{S,2}(\mathbb{Q})$, which is, as above, difficult to get ones hands on (if infinite). Finally, I should also mention work of Boston conjecturing explicit (non-approximate) presentations for $G_{S,p}(K)$, though even here, the unramified case is less concrete. </p> http://mathoverflow.net/questions/48448/william-rowan-hamilton-and-algebra-as-time William Rowan Hamilton and Algebra as Time Cam McLeman 2010-12-06T15:10:11Z 2011-02-26T22:31:03Z <p>This question ended up longer than I intended (though most of the bulk is interesting remarks by Hamilton), so I thought it might be good to include my question at the beginning before the admittedly-lengthy background:</p> <p><b>Question:</b> Why did Hamilton view the scalar part of a quaternion as representing time? Does the modern viewpoint of quaternions in physics admit an interpretation that involves time but does not require relativity and related thoughts as a prerequisite?</p> <p><b>Background:</b> It strikes me as remarkably ahead-of-his-time that Hamilton preferred to think, or perhaps insisted on thinking, of algebra as the study of a time variable. In fact, while I'm certainly no math-historian, by my reading he is actually quite uncomfortable with the relatively newfound spread of abstraction in algebra, particularly in terms of imaginary numbers. He laments on the chasm between this abstraction and the firm footing of science:</p> <blockquote>Yet a natural regret might be felt, if such were the destiny of Algebra; if a study, which is continually engaging mathematicians more and more, and has almost superseded the Study of Geometrical Science, were found at last to be not, in any strict or proper sense, the Study of a Science at all.... </blockquote> <p>...and later...</p> <blockquote> The author acknowledges with pleasure that he agrees with M. Cauchy, in considering every (so-called) Imaginary Equation as a symbolic representation of two separate Real Equations: but he differs from that excellent mathematician in his method generally, and especially in not introducing the sign $\sqrt{-1}$ until he has provided for it, by his Theory of Couples, a possible and real meaning, as a symbol of the couple (0, 1). </blockquote> <p>As a solution to his quandry, Hamilton postulates that the interpretation of algebra as the study of time is the way to base algebra with imaginary numbers on a scientific footing, writing:</p> <p><p></p> <blockquote> It is the genius of Algebra to consider what it reasons on as flowing, as it was the genius of Geometry to consider what it reasoned on as fixed. </blockquote> <p>In his treatise on the subject: "Theory of Conjugate Functions, or Algebraic Couples; with a Preliminary and Elementary Essay on Algebra as The Science of Pure Time," he develops a tremendous amount of basic algebra (from addition and ordering to indeterminate forms and exponentiation) through this lens. The sticky part is that he does not seem (to me, at least) to resolve the issue at hand; that of providing an intuitive formulation of algebra in which one can relate time and imaginary numbers, at least beyond that of Cauchy's theory of couples referenced above. And yet he himself, however, declares victory on the matter, writing that this "Theory of Couples is published to make manifest that hidden meaning." He is so taken by this point of view that he later interprets quaternions as a "scalar plus vector" as a "time plus space" element of spacetime:</p> <blockquote>Time is said to have only one dimension, and space to have three dimensions. […] The mathematical quaternion partakes of both these elements; in technical language it may be said to be "time plus space", or "space plus time": and in this sense it has, or at least involves a reference to, four dimensions.</blockquote> <p>Ever since Einstein and Minkowski (and others), it is quite commonplace to think in terms of spacetime (and indeed the concept apparently dates back to d'Alembert in 1754), but without relativity/Lorenz metrics/etc. at one's disposal, it is striking how dedicated Hamilton was to the point of view of relating time and imaginary quantities. </p> <p><b>Question (redux):</b> It is really Hamilton's strikingly-modern interpretation of the scalar part of a quaternion as representing time that is the basis for this question. Why did he do this? Does the modern viewpoint of quaternions in physics admit an interpretation that involves time but does not require relativity and related thoughts as a prerequisite?</p> http://mathoverflow.net/questions/47620/does-pi-1spec-mathbbz1-p-depend-on-p/47628#47628 Answer by Cam McLeman for Does $\pi_1(Spec(\mathbb{Z}[1/p]))$ depend on p? Cam McLeman 2010-11-29T02:18:04Z 2010-11-29T14:10:18Z <p>To add on to Pete's answer, let me comment that the differences are even more pronounced if we look at the maximal pro-$p$ quotient $\pi_1(\operatorname{Spec}(\mathbb{Z}\left[\frac{1}{p_1p_2\cdots p_r}\right])^{(p)}$ of this etale fundamental group. For example, if $r\geq 4$, then this group is infinite, in fact non-$p$-adic-analytic, if each $p_i\equiv 1\pmod{p}$ and is trivial if each $p_i\not\equiv1\pmod{p}$. The latter is basically for stupid reasons (only primes which are 1 mod p can ramify in a $p$-extension). But even ignoring stupid cases, there's a lot of fantastic arithmetic going on here. For example, $\pi_1(\operatorname{Spec}(\mathbb{Z}\left[\frac{1}{19\cdot 103}\right])^{(2)}$ is finite whereas $\pi_1(\operatorname{Spec}(\mathbb{Z}\left[\frac{1}{17\cdot 103}\right])^{(2)}$ is infinite, results which stem from simple quadratic residue calculations. Figuring out to generate these kinds of results more generally is an active and difficult area of research.</p> http://mathoverflow.net/questions/41219/number-fields-with-no-unramified-extensions/41227#41227 Answer by Cam McLeman for number fields with no unramified extensions? Cam McLeman 2010-10-06T02:40:17Z 2010-10-06T11:22:09Z <p>The question itself is certainly still open. Mostly as an exercise for myself, I'll coalesce my comments above into an answer, and add in some details about where various pieces of the philosophy come from. </p> <p>The starting point is the following philosophy:</p> <blockquote> The ring of integers in any number field with sufficiently small root discriminant admits no non-trivial unramified extensions. </blockquote> <p>This philosophy can occasionally be made precise. For example, Yamamura uses tables of root discriminant bounds from Diaz y Diaz to conclude that for a quadratic imaginary number field $K$ of discriminant $|d|\leq 499$ (or $|d|\leq 2003$ under GRH), the maximal unramified extension of $K$ is a finite extension. This is particularly relevant since each of these maximal unramified extensions clearly has the property that you ask about, that they themselves admit no unramified extensions.</p> <p>The bad news is that the set of number fields with sufficiently small root discriminant to apply these results (at least, without a tremendous of extra effort analyzing carefully constructed extensions) is finite. In particular, results of Odlyzko imply that that there are only finitely many number fields with root discriminant less than $4\pi e^\gamma\approx 22.3$, where $\gamma$ is the Euler-Mascheroni constant (yeah, that Euler-Mascheroni constant!). Under GRH, this remains true for the larger bound $8\pi e^\gamma$. In fact, as a nice concrete factoid to hold on to, Jones and Roberts have shown that there are exactly 7063 <em>abelian</em> number fields with root discriminant under $8\pi e^\gamma$, and sort these according to their Galois group.</p> <p>Back to good news: So we now ask ourselves whether or not these numbers $4\pi e^\gamma$ and $8\pi e^\gamma$ can be improved. The answer is a definite yes. If we partition number fields based on their proportion of real and complex embeddings, we can get improvements on fields with increased proportion of complex embeddings, up to an improvement factor of $e$ for totally complex number fields. Further, since Odlyzko's argument stems from work of Stark estimating values of $L$-functions, it seems plausible to believe there are analytic improvements to be made as well. So maybe we can keep pushing these bounds higher and higher, enough so that we find infinitely many number fields with smaller root discriminant.</p> <p>More bad news: There's an inherent limit to how good we can make these bounds, coming from the study of class field towers. (Okay, so this is actually really good news for those of us who like to study class field towers, but I digress...) Namely, since root discriminants are unchanged when moving up an unramified extension, fields with an infinite class field tower provide a stopping point for any claim of the form "there are finitely many number fields with root discriminant less than such-and-such bound." This is also something that can be partitioned by proportion of real and complex embeddings, and it's been a hot topic recently to see how limited these Odlyzko-type bounds can get. Recently, Hajir and Maire have further refined this line of thought by considering towers of number fields with tame ramification.</p> <p>So, long story short, from this point of view, the big unknown is whether, once we know optimal bounds on root discriminants, whether or not there will be infinitely many number fields with root discriminants less than that bound. Of course, there's also the possibility that there are other techniques for proving that a number field has no unramified extensions that do not go through root discriminants -- perhaps a form of non-abelian class field theory can come to the rescue, as abelian class field theory can address only the weaker (but still open and fantastically interesting) question of fields with no <em>abelian</em> unramified extensions. </p> http://mathoverflow.net/questions/41187/a-coverage-question/41230#41230 Answer by Cam McLeman for A coverage question Cam McLeman 2010-10-06T03:14:35Z 2010-10-06T03:14:35Z <p>It is very likely that <em>every</em> (positive) odd number is covered by a sum of this type.</p> <p>As Robin Chapman points out, this is equivalent to asking, for a given odd number $h$, whether there exists an odd prime $q\equiv3\pmod{4}$ such that $\mathbb{Q}(\sqrt{-q})$ has class number $h$. Let $N(h)$ be the number of quadratic imaginary number fields with class number $h$ (for odd $h$) -- note that such a field is already necessarily of the form $\mathbb{Q}(\sqrt{-q})$ for $q\equiv3\pmod{4}$ by genus theory. This rephrases your question as "Is $N(h)>0$ for all odd $h$"?</p> <p>By the calculation of mark Watkins mentioned in Stoppie's answer, it is known that $N(h)>0$ for all $h\leq 100$. More importantly, $N(h)$ gets rather large -- $N(1)=9$ is Heegner-Stark, and this is the smallest value of $N(h)$ in this range. For instance, $N(h)>100$ for all $h>37$, and $N(99)=289$. So we'd like an assurance of some sort of a rough upward trend to guarantee that $N(h)$ does not hit zero at some point down the road. Enter Soundararajan's article "The number of imaginary quadratic number fields with a given class number", which studies asymptotics of $N(h)$. Soundararajan remarks that $N(h)$ should be on the order of $h$, and conjectures more precisely that $$\frac{h}{\log h}\ll N(h)\ll h\log h.$$ This is accompanied by a probabilistic argument as to why this is a reasonable conjecture. I have not worked through the analysis, but it seems likely to me that if one's only goal was to prove that $N(h)>0$ for all $h$ (a lower bound was not a goal of the paper), this heuristic argument could be made sufficiently rigorous, especially when combined with the data for $h&lt;100$ above, to rule out pathologies for small inputs.</p> http://mathoverflow.net/questions/41103/congruences-mod-primes-in-galois-extensions/41105#41105 Answer by Cam McLeman for Congruences mod primes in Galois extensions Cam McLeman 2010-10-05T04:14:16Z 2010-10-05T16:22:58Z <p>Sure. $a\equiv b\pmod{\mathfrak{P}}$ just means $a-b\in\mathfrak{P}$. Taking norms to any subfield $K$ of $\mathbb{Q}(\zeta_n)$ (e.g., $\mathbb{Q}$ or $\mathbb{Q}(\zeta_m)$) gives you $N_{\mathbb{Q}(\zeta_n)/K}(a-b)\in N_{\mathbb{Q(\zeta_n)}/K}\mathfrak{P}.$ </p> <p>For $K=\mathbb{Q}$, the latter norm is just $p^f$ where $f$ is the order of $p\pmod{n}$.</p> <p>For $K=\mathbb{Q}(\zeta_m)$, the former norm is $(a-b)^{\phi(n)/\phi(m)}$ and the latter is $\mathfrak{p}^{f'}$, where $f'$ is the easily-calculated relative residue degree. </p> <p>This doesn't give you an explicit congruence between $a$ and $b$, but given Gerry's answer, that might have been too much to ask for anyway. On the other hand, if $\phi(n)/\phi(m)$ is small or (as in Alex's answer) if $p$ has few factors in $\mathbb{Q}(\zeta_m)$, you get something at least slightly non-stupid out.</p> http://mathoverflow.net/questions/40728/algebraic-integers-on-the-unit-circle/40744#40744 Answer by Cam McLeman for Algebraic integers on the unit circle Cam McLeman 2010-10-01T13:35:04Z 2010-10-01T14:17:43Z <p>"Do these objects have a name"</p> <p>Probably not. "Multiplicative subgroup of $S^1$ generated by algebraic integers" is pretty descriptive, and not of such fundamental importance that it's worth shortening.</p> <p>"If the generating set is the set of roots of an irreducible polynomial, what kind of information would they contain?"</p> <p>Almost none. Assuming you're still talking about algebraic integers, if all of the roots of a monic irreducible polynomial have absolute value 1, then the polynomial is cylcotomic and the roots are roots of unity. Your multiplicative group is then finite and well-understood.</p> <p>I'm not sure I completely understand your third question, but it looks like Scott Carnahan's first comment points you in the right direction. To elaborate very slightly, note that if $\alpha$ is an algebraic integer on the unit circle, then so is $\alpha^r$ for any $r\in\mathbb{Q}$, so you get a copy of $\mathbb{Q}$ in your multiplicative subgroup for each "independent" such $\alpha$. Of course, if you're inside a <em>fixed</em> number field (which re-reading seems to be the focus of this question), you at least get the subgroup $\alpha^n$ for $n\in\mathbb{Z}$.</p> http://mathoverflow.net/questions/40528/spencer-browns-claimed-proof-of-the-four-color-theorem/40536#40536 Answer by Cam McLeman for Spencer-Brown's claimed proof of the four color theorem Cam McLeman 2010-09-29T20:29:42Z 2010-09-29T20:29:42Z <p>I spent some time with this a couple of years ago out of curiosity, but did not make it any farther than you're like to be able to find online. The water is definitely murky.</p> <p><a href="http://arxiv.org/PS_cache/math/pdf/0112/0112266v2.pdf" rel="nofollow">Here</a> is a discussion of the work which makes it clear that there are substantial ideas in Spencer-Brown's work, though Kauffman makes it clear that he is not evaluating the work <em>per se</em>. Kauffman reports that Spencer-Brown definitely gives (with proof) a reformulation of the 4-Color Theorem into something called the Primacy Principle, that "A minimal planar (non-empty) uncolorable trail is prime." (You'll have to see the references for the terminology). Kauffman points out that Spencer-Brown has set up his logical foundations to have basically unintentionally axiomatized this Principle (hence the line on Wikipedia that Spencer-Brown's work "straddles the boundary between mathematics and of philosophy"), which is the source of Spencer-Brown's claimed proof. This seems to be far as Kauffman is willing to go in giving caution as to the validity of the proof. </p> <p>The Wikipedia article on <a href="http://en.wikipedia.org/wiki/Laws_of_Form" rel="nofollow">Laws of Form</a> is decidedly less charitable.</p> http://mathoverflow.net/questions/33538/what-is-the-ring-of-integers-of-the-pythagorean-field What is the ring of integers of the Pythagorean field? Cam McLeman 2010-07-27T16:07:57Z 2010-09-29T18:16:39Z <p>Following Hilbert, we call the complex numbers constructible via compass and straight-edge the field of <i>Euclidean numbers</i>, and the totally real such numbers the field of <i>Pythagorean numbers.</i> (Among other possible definitions, an algebraic number is totally real if its minimal polynomial has all real roots). For a reference, Richard Alperin gives a description of these and related fields from a constructibility viewpoint in his paper "Trisections and Totally Real Origami." </p> <p>There is a remarkably nice characterization of the Pythagorean numbers -- the Pythagorean field is the smallest field containing the rationals and closed under the operation $x\rightarrow \sqrt{1+x^2}$. Or, from an only slightly different viewpoint, it is the <i>Pythagorean closure</i> of $\mathbb{Q}$, in the sense of</p> <p><a href="http://planetmath.org/encyclopedia/PythagoreanField.html" rel="nofollow">http://planetmath.org/encyclopedia/PythagoreanField.html</a></p> <p>Because it's a nice "hands-on" intro to this field, let me include in the question Daniel Litt's comment below that since $\sqrt{2}=\sqrt{1+1^2}$, and $\sqrt{3}=\sqrt{1+\sqrt{2}^2}$, and so on, the Pythagorean field contains $\sqrt{n}$ for all $n\geq 0$, and hence contains the compositum of all real quadratic fields.</p> <p><b>My Question:</b></p> <pre> What is the ring of integers of the Pythagorean field? </pre> <p>Note that the most naive guess of it being the smallest subring of algebraic integers closed under the operation $x\rightarrow \sqrt{1+x^2}$ is incorrect -- this ring does not include $\frac{1+\sqrt{5}}{2}$, which is certainly a totally real Euclidean algebraic integer. I suspect/hope (though this may just be the second most naive guess) that there's some description of the form "smallest subring of the algebraic integers closed under $x\rightarrow \sqrt{1+x^2}$ and division by 2 when certain conditions are met." I've done a little bit of a literature search on rings of integers of totally real multiquadratic extensions of $\mathbb{Q}$, but haven't found anything even remotely inspiring something of this form.</p> <p>I don't have much to offer in terms of motivation, except that I have come across a variety of rings of integers in my research, and I'm trying to decide if any are exactly the ring of Pythagorean integers. It would be nice to be able to compare them to the Pythagorean integers just by seeing whether or not one of these rings satisfies certain closure operations.</p> http://mathoverflow.net/questions/38341/class-numbers-and-163 Class Numbers and 163 Cam McLeman 2010-09-10T19:16:22Z 2010-09-17T23:17:16Z <p>This is a bit fluffier of a question than I usually aim for, so apologies in advance if this doesn't pass the smell test for suitability.</p> <p>Likely my favorite fun fact in all of number theory is the juxtaposition of two "extremal and opposite" properties about the prime 163 in relation to class numbers:</p> <ul> <li> $p=163$ is the <i>largest</i> value of $p$ for which the quadratic imaginary number field $\mathbb{Q}(\sqrt{-p})$ has class number <i>equal</i> to one. (Baker-Heegner-Stark) </li> <p> <li> $p=163$ is the <i>smallest</i> value of $p$ for which the real cyclotomic field $\mathbb{Q}(\zeta_p+\zeta_p^{-1})$ has class number <i>greater</i> than one. (Schoof) </ul> <p>Of the various ways I know of "pushing up" and "pushing down" class numbers (class field theory, Herglotz-type formulas, Scholz-type reflection theorems), none seem to give any indication that these two class numbers should be related, let alone inversely so. Of course, since the smaller Heegner discriminants don't correspond to analogous real cyclotomic fields with positive class number, this is not surprising.</p> <p>This leads me to wonder if there's an analytic link between these two quantities -- for example, by relating their zeta-functions and looking at the corresponding class number formulas. My initial, admittedly naive, attempts to extract anything from the relationship between the zeta-functions for $\mathbb{Q}(\zeta_p)$ and $\mathbb{Q}(\zeta_p+\zeta_p^{-1})$ have come up empty. So my question is:</p> <blockquote> Are there fancier analytic (or other) techniques that might shed some light on the "miracle" above? </blockquote> <p>Of course, I'm also aware that this juxtaposition might be purely coincidental, a mildly large example of the law of small numbers at work. I might even prefer it that way. </p> <p><b>Edit to incorporate some computations and comments from below.</b></p> <p>For primes congruent to 7 mod 12, the real cyclotomic field of conductor p contains a unique cyclic cubic subfield. By class field theory, there is a surjection of class groups from the real cyclotomic to the cubic. Since for 163, the cyclic cubic has class number 4, the non-triviality of the class number for the real cyclotomic can be said to "come from" the cubic. In a sense, the coincidence thus reduces to the fact that the first cyclic cubic field ("first" with respect to ordering by conductor) of prime conductor 7 mod 12 with non-trivial class group is the one of conductor 163. The fact that 163 is only the 11th prime in this congruence class may modify (in which direction I'm not sure) your opinion of whether or not this is a coincidence. </p> <p>Barring any insight as to why the class number of this quadratic and cubic would be related, which may be unlikely given Franz Lemmermeyer's answer, it would be interesting to know if one could devise a clever probabilistic test for evaluating how surprised one should be to see ten class-number-one cubics in a row. I imagine that it's not very unlikely -- I just ran a computation, and class number 1 seems to very prevalent for cyclic cubics of small conductor ($p&lt;5000$), and some heuristic (sorry, Andrew) evidence in the literature seems to agree.</p> http://mathoverflow.net/questions/38680/can-an-algebraic-number-on-the-unit-circle-have-a-conjugate-with-absolute-value-d/38683#38683 Answer by Cam McLeman for Can an algebraic number on the unit circle have a conjugate with absolute value different from 1? Cam McLeman 2010-09-14T13:09:24Z 2010-09-14T13:09:24Z <p>Yes. Take $$\alpha=\sqrt{2-\sqrt{2}}+i\sqrt{\sqrt{2}-1}.$$ Neither of the conjugates $$\sqrt{2+\sqrt{2}}\pm \sqrt{\sqrt{2}+1}$$ have absolute value 1. </p> <p>It is impossible, however, if $\mathbb{Q}(\alpha)/\mathbb{Q}$ is abelian, since then all automorphisms commute with complex conjugation.</p> <p>This was all stolen from Washington's Cyclotomic Fields book.</p> http://mathoverflow.net/questions/37880/kummer-generator-for-the-ribet-extension/37893#37893 Answer by Cam McLeman for Kummer generator for the Ribet extension Cam McLeman 2010-09-06T14:56:03Z 2010-09-07T12:57:56Z <p>Here is an explicit construction$^*$.</p> <p>Since there exists such an unramified $p$-extension, by class field theory the $p$-part of the class group of $\mathbb{Q}(\mu_p)$ is non-trivial. Further, specifying the $\Delta$-action gives more; namely, that the $\omega$-eigenspace of the $p$-part of the class group is non-trivial (for $\omega=\chi^{1-k}$). By Herbrand-Ribet, the $\omega$-eigenspace of the class group has the same order as the $\omega$-component of the $p$-part of $($units mod cyclotomic units$)$ in $\mathbb{Q}(\mu_p)$., so this quotient too is non-trivial. Now, following the proof of Theorem 15.8 in Washington's Cyclotomic Fields (roughly), we choose a unit $u$ whose $\omega$-projection $\varepsilon_\omega u$ in this quotient group is:</p> <ul> <li> Congruent to 1 modulo the prime above $p$ in $\mathbb{Z}[\zeta_p]^+$ </li><li> Not a $p$-th power of such a unit. </li><li> *Is* a $p$-th power of an element of the topological closure of the group of these units. </li></ul> <p>Such a thing exists by the converse to Herbrand-Ribet$^{**}$. Then $\mathbb{Q}(\zeta_p,u^{1/p})/\mathbb{Q}(\zeta_p)$ is everywhere unramified and carries the proper action of $\Delta$, so this is the unit you're looking for.</p> <p>$^*$: "Construction" may be a bit of an exaggeration. Following the proof of Theorem 15.8, however, I'm not immediately clear on what would be difficult to do explicitly. I think SAGE could handle local units well enough to carry out the construction in the proof. Unless someone comes and shoots down this answer, I might see if I can't get SAGE to do this explicitly. <b>Edit</b>: Chris Wuthrich makes a good point below -- even if there's no theoretical obstruction to doing everything explicitly, at a practical level computations would quickly become infeasible.</p> <p>$^{**}$: Actually, there's one more case to consider, which amounts to doing a similar construction in a different ("reflected") eigenspace, but I think this is good enough to get the gist of the argument.</p> http://mathoverflow.net/questions/13428/please-check-my-6-line-proof-of-fermats-last-theorem Please check my 6-line proof of Fermat's Last Theorem. Cam McLeman 2010-01-29T22:02:26Z 2010-09-02T19:59:49Z <p>Kidding, kidding. But I <i>do</i> have a question about an $n$-line outline of a proof of the first case of FLT, with $n$ relatively small.</p> <p>Here's a result of Eichler (remark after Theorem 6.23 in Washington's Cyclotomic Fields): If $p$ is prime and the $p$-rank of the class group of $\mathbb{Q}(\zeta_p)$ satisfies $d_p&lt;\sqrt{p}-2$, then the first case of FLT has no non-trivial solutions. Once you know Herbrand-Ribet and related stuff, the proof of this result is even rather elementary.</p> <p>The condition that $d_p&lt;\sqrt{p}-2$ seems reminiscent of rank bounds used with Golod-Shafarevich to prove class field towers infinite. More specifically, a possibly slightly off (and definitely improvable) napkin calculation gives me that for $d_p>2+2\sqrt{(p-1)/2}$, the $p$-th cyclotomic field $\mathbb{Q}(\zeta_p)$ has an infinite $p$-class field tower. It's probably worth emphasizing at this point that by the recent calculation of Buhler and Harvey, the largest index of irregularity for primes less than 163 million is a paltry 7. </p> <p>So it seems natural to me to conjecture, or at least wonder about, a relationship between the unsolvability of the first case of FLT and the finiteness of the $p$-class field tower over $\mathbb{Q}(\zeta_p)$. Particularly compelling for me is the observation that regular primes (i.e., primes for which $d_p=0$) are precisely the primes for which this tower has length 0, and have obvious historical significance in the solution of this problem. In fact, the mechanics of the proof would probably/hopefully be to lift the arithmetic to the top of the (assumed finite) p-Hilbert class field tower, and then use that its class number is prime to $p$ to make arguments completely analogously to the regular prime case. </p> <p>I haven't seen this approach anywhere. Does anyone know if it's been tried and what the major obstacles are, or demonstrated why it's likely to fail? Or maybe it works, and I just don't know about it?</p> <p><b>Edit</b>: Franz's answer indicates that even for a relatively simple Diophantine equation (and relatively simple class field tower), moving to the top of the tower introduces as many problems as it rectifies. This seems pretty compelling. But if anyone has any more information, I'd still like to know if anyone knows or can come up with an example of a Diophantine equation which <i>does</i> benefit from this approach.</p> http://mathoverflow.net/questions/37214/why-arent-there-more-classifying-spaces-in-number-theory Why aren't there more classifying spaces in number theory? Cam McLeman 2010-08-31T04:11:09Z 2010-09-01T12:23:08Z <p>Much of modern algebraic number theory can be phrased in the framework of group cohomology. (Okay, this is a bit of a stretch -- much of the part of algebraic number theory that I'm interested in...). As examples, Cornell and Rosen develop basically all of genus theory from cohomological point of view, a significant chunk of class field theory is encoded as a very elegant statement about a cup product in the Tate cohomology of the formation module, and Neukirch-Schmidt-Wingberg's fantastic tome "Cohomology of Number Fields" convincingly shows that cohomology is the principal beacon we have to shine light on prescribed-ramification Galois groups. </p> <p>Of course, we also know that group cohomology can be studied via topological methods via the (topological) group's classifying space. My question is:</p> <blockquote> Question: Why doesn't this actually happen? </blockquote> <p>More elaborately: I'm fairly well-acquainted with the "Galois cohomology for number theory" literature, and not once have I come across an argument that passes to the classifying space to use a slick topological trick for a cohomological argument or computation (though I'd love to be enlightened). On the other hand, for example, are things like Tyler's answer to my question </p> <p><a href="http://mathoverflow.net/questions/15375/coboundary-representations-for-trivial-cup-products" rel="nofollow">http://mathoverflow.net/questions/15375/coboundary-representations-for-trivial-cup-products</a></p> <p>which strikes me as saying that there may be plenty of opportunities to carry over interesting constructions and/or lines of reasoning from the topological side to the number-theoretic one. </p> <p>Maybe the classifying spaces for gigantic profinite groups are too hideous to think about? (Though there's plenty of interesting Galois cohomology going on for finite Galois groups...). Or maybe I'm just ignorant to the history, and that indeed the topological viewpoint guided the development of group cohomology and was so fantastically successful at setting up a good theory (definition of differentials, cup/Massey products, spectral sequences, etc.) that the setup and proofs could be recast entirely without reference to the original topological arguments? </p> <p>(<b>Edit</b>: This apparently is indeed the case. In a comment, Richard Borcherds gives the link <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.bams/1183537593" rel="nofollow">http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.bams/1183537593</a> and JS Milne suggests MacLane 1978 (Origins of the cohomology of groups. Enseign. Math. (2) 24 (1978), no. 1-2, 1--29. MR0497280)., both of which look like good reads.)</p> http://mathoverflow.net/questions/33162/how-divisible-is-the-average-integer/33164#33164 Answer by Cam McLeman for How divisible is the average integer? Cam McLeman 2010-07-24T02:43:29Z 2010-07-25T00:41:03Z <p>Hopefully I've read all your notation correctly. If so, by playing (very) fast and loose with heuristics, I think your friend is right that the answer is 0.</p> <p>Your function $Log(n)$ is the additive function $\Omega(n)$. According to the mathworld entry</p> <p><a href="http://mathworld.wolfram.com/PrimeFactor.html" rel="nofollow">http://mathworld.wolfram.com/PrimeFactor.html</a>,</p> <p>$\Omega(n)$ has been dubbed the "multiprimality of $n$" by Conway, and satisfies </p> <p>$$\Omega(n)\sim \ln\ln(n)+\text{mess},$$ so (very roughly), $$D(n)\sim \frac{\ln\ln(n)}{\ln(n)},$$ and $$E(p)\sim \frac{1}{p}\int_e^p \frac{\ln\ln n}{\ln n}dn.$$ This goes to 0 (very very slowly) as $p\rightarrow\infty$.</p> http://mathoverflow.net/questions/32967/have-any-long-suspected-irrational-numbers-turned-out-to-be-rational/33041#33041 Answer by Cam McLeman for Have any long-suspected irrational numbers turned out to be rational? Cam McLeman 2010-07-23T04:39:08Z 2010-07-23T04:39:08Z <p>This certainly doesn't answer the question, but I can't help but mention Conway's constant:</p> <p><a href="http://mathworld.wolfram.com/ConwaysConstant.html" rel="nofollow">http://mathworld.wolfram.com/ConwaysConstant.html</a></p> <p>It relates to Pete's comment about "bumping" it up a notch, in that it gives an example of a number that I think any reasonable person would conjecture to be transcendental, but turns out to be algebraic (of degree 71, of all things). And algebraic numbers are sort of finitely far from being rational, so...</p> http://mathoverflow.net/questions/127717/how-to-visualize-the-class-number-of-an-imaginary-quadratic-field Comment by Cam McLeman Cam McLeman 2013-04-16T17:57:24Z 2013-04-16T17:57:24Z This doesn't quite &quot;visualize the class number&quot;, but you can at least visualize when the division algorithm fails when the fundamental rectangle is too large. http://mathoverflow.net/questions/125877/computing-certain-class-numbers-modulo-4 Comment by Cam McLeman Cam McLeman 2013-03-29T12:48:55Z 2013-03-29T12:48:55Z On a side note, I think it's better to think of this as a statement about the 2-divisibility of the class number than it is to think about it as the value of the class number mod 4. http://mathoverflow.net/questions/92652/are-cantor-type-numbers-algebraic Comment by Cam McLeman Cam McLeman 2013-03-14T15:16:42Z 2013-03-14T15:16:42Z Unknown (google) seems to be of two different minds on the matter. http://mathoverflow.net/questions/122035/where-does-the-name-euler-system-come-from Comment by Cam McLeman Cam McLeman 2013-02-17T01:18:08Z 2013-02-17T01:18:08Z I'm pretty sure it comes from &quot;Euler factors.&quot; http://mathoverflow.net/questions/119329/the-riemann-hypothesis-and-the-langlands-program/119532#119532 Comment by Cam McLeman Cam McLeman 2013-01-22T03:35:55Z 2013-01-22T03:35:55Z Do you have a reference for the NAS report? http://mathoverflow.net/questions/13349/tate-module-of-cm-elliptic-curves/90056#90056 Comment by Cam McLeman Cam McLeman 2012-03-02T19:14:22Z 2012-03-02T19:14:22Z @oxeimon: It would be best to ask this as a separate question, including a link back to this one. http://mathoverflow.net/questions/77519/optic-fibers-after-joseph-orourke Comment by Cam McLeman Cam McLeman 2012-03-01T14:33:59Z 2012-03-01T14:33:59Z The title of this question always makes me think this would make for a fantastic sci-fi movie. http://mathoverflow.net/questions/89389/difference-between-2nd-and-5th-corrected-printing-of-a-classical-introduction-to Comment by Cam McLeman Cam McLeman 2012-02-24T13:22:48Z 2012-02-24T13:22:48Z If they're different printings rather than different editions, there may be no change. http://mathoverflow.net/questions/89269/on-odd-perfect-numbers Comment by Cam McLeman Cam McLeman 2012-02-23T15:15:24Z 2012-02-23T15:15:24Z I think the votes to close are premature. Arnie: This would be a better question if you focused in on a particular case. Forget about cases 1 and 3, and maybe even 2b. Give it a little bit of motivation as to where your question comes from, and then ask if anyone knows an improvement of a specific bound. http://mathoverflow.net/questions/18271/what-out-of-print-books-would-you-like-to-see-re-printed/18272#18272 Comment by Cam McLeman Cam McLeman 2012-02-19T16:42:04Z 2012-02-19T16:42:04Z Neat. Thanks!! http://mathoverflow.net/questions/88539/sums-of-rational-squares/88549#88549 Comment by Cam McLeman Cam McLeman 2012-02-15T20:23:32Z 2012-02-15T20:23:32Z Ah, the ol' &quot;clearing the denominator trick.&quot; Great! http://mathoverflow.net/questions/88048/on-the-class-number Comment by Cam McLeman Cam McLeman 2012-02-10T02:30:53Z 2012-02-10T02:30:53Z @Frank: I concur entirely. Still, it seems very likely to be the true. http://mathoverflow.net/questions/88048/on-the-class-number Comment by Cam McLeman Cam McLeman 2012-02-10T00:27:01Z 2012-02-10T00:27:01Z I answered a similar question to your surjectivity question here: <a href="http://mathoverflow.net/questions/41187/a-coverage-question" rel="nofollow" title="a coverage question">mathoverflow.net/questions/41187/&hellip;</a> http://mathoverflow.net/questions/86990/effective-lower-bound-for-class-numbers-of-cyclotomic-fields Comment by Cam McLeman Cam McLeman 2012-02-08T13:55:43Z 2012-02-08T13:55:43Z @Bernikov: There's a lot to be said on that front. A good starting point is Goldfeld's &quot;THE GAUSS CLASS NUMBER PROBLEM FOR IMAGINARY QUADRATIC FIELDS&quot;. http://mathoverflow.net/questions/86422/example-of-unramified-abelian-extension Comment by Cam McLeman Cam McLeman 2012-01-23T05:00:27Z 2012-01-23T05:00:27Z For other examples, I'd probably start by looking up the Hilbert class field.