User robin chapman - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T19:14:28Z http://mathoverflow.net/feeds/user/4213 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48809/relationships-between-the-roots-of-an-entire-function-and-the-roots-of-its-deriva/48815#48815 Answer by Robin Chapman for Relationships between the roots of an entire function and the roots of its derivative Robin Chapman 2010-12-09T19:42:46Z 2010-12-09T19:42:46Z <p>I don't know if there are any general results about these, but when $f$ is a polynomial, these must be in essence results about symmetric functions. If $f(z)=z^n+a_{n-1} z^{n-1}+\cdots+a_1z+a_0$ then $Z(1)=-a_1/a_0$, $Z(2)=Z(1)^2-2a_2/a_0$ etc. In this case your results surely specialize to polynomial identities in the $a_j$.</p> http://mathoverflow.net/questions/48789/rank-of-a-free-module-without-the-axiom-of-choice/48791#48791 Answer by Robin Chapman for Rank of a free module without the axiom of choice Robin Chapman 2010-12-09T17:47:14Z 2010-12-09T17:47:14Z <p>Let $A$ and $B$ be infinite sets. Let $M$ be a rank $|B|$ module with basis $e_b$ for $b\in B$. If we take $|A|$ elements $m_a$ of $M$, then each can be expressed in terms of finitely many of the $e_b$. If $B$ is not the union of $|A|$ finite sets, then there is some $e_b$ not expressible in terms of the $m_a$, so $M$ cannot be free on $|A|$ generators.</p> <p>I'm not expert on set theory without AC, but those here who are will surely tell us if there are models of ZF with non-equinumerous infinite sets such that $A$ is the union of $|B|$ finite sets and <em>vice versa</em>.</p> http://mathoverflow.net/questions/48609/proof-that-the-factors-of-sigmape-have-two-forms/48612#48612 Answer by Robin Chapman for Proof that the factors of sigma(p^e) have two forms. Robin Chapman 2010-12-08T07:57:10Z 2010-12-08T12:57:26Z <p>It's well-known that if $a$ is an integer then a prime factor of the number $\Phi_n(a)$ is either a factor of $n$ or congruent to $1$ modulo $n$. Here $\Phi_n$ is the $n$-th cyclotomic polynomial. The reason is that if $p$ divdes $\Phi_n(a)$ but not $n$ then $a$ has order exactly $n$ in the multiplicative group $(\mathbb Z/p\mathbb Z)^*$. By Lagrange's theorem then $n\mid(p-1)$.</p> <p>When $e+1$ is prime, then $\Phi_{e+1}(X)=X^e+X^{e-1}+\cdots+X+1$, so in this case a prime factor of any $\Phi_{e+1}(a)$ is either $e+1$ or congruent to $1$ modulo $e+1$. In general though $X^e+X^{e-1}+\cdots+X+1$ is the product of the $\Phi_m(X)$ over the factors $m>1$ of $e+1$, so a prime divisor of $a^e+a^{e-1}+\cdots+a+1$ is either a divisor of $e+1$ of congruent to $1$ modulo some prime divisor of $e+1$.</p> http://mathoverflow.net/questions/48351/crookedness-of-convex-curves-milnor/48355#48355 Answer by Robin Chapman for crookedness of convex curves (milnor) Robin Chapman 2010-12-05T10:25:16Z 2010-12-05T10:25:16Z <p>If you have a polygon with say a horizontal side, each point is a maxmimum (or minimum) of the projection onto the $y$-axis. So we must admit the possibility of an infinite number of maxima.</p> http://mathoverflow.net/questions/48065/hopf-algebras-rings-a-question-of-terminology/48069#48069 Answer by Robin Chapman for Hopf Algebras/Rings, A Question of Terminology Robin Chapman 2010-12-02T17:11:33Z 2010-12-02T17:11:33Z <p>The group operation corresponds to the multiplication map $\mu:A\otimes A\to A$ and the identity should be the natural map $\iota:k\to A$. Both these should be coalgebra maps. The inverse should correspond to a map $S:A\to A$ with $\mu\circ(\rm{id}\otimes S)\circ\Delta=\iota\circ\epsilon =\mu\circ(S\otimes\rm{id})\circ\Delta$, so $S$ is the antipode.</p> http://mathoverflow.net/questions/47871/suggestions-for-sonifying-math/47892#47892 Answer by Robin Chapman for Suggestions for sonifying math Robin Chapman 2010-12-01T11:45:44Z 2010-12-01T11:45:44Z <p>There is <a href="http://en.wikipedia.org/wiki/Per_N%25C3%25B8rg%25C3%25A5rd#Music" rel="nofollow">Per Norgard</a>'s "<a href="http://oeis.org/A004718" rel="nofollow">infinity series</a>" which he used in his Symphony no. 2.</p> http://mathoverflow.net/questions/47695/centers-of-semidirect-products/47698#47698 Answer by Robin Chapman for Centers of Semidirect Products Robin Chapman 2010-11-29T16:50:02Z 2010-11-29T16:50:02Z <p>Suppose that $z=xy$ is in the centre where $x\in N$ and $y\in K$. Then for all $u\in K$, $uxy=xyu$. But $uxy=\phi(u)(x)uy$ so that $x=\phi(u)(x)$ (and $uy=yu$). As this is true for all $u\in K$ then by the assumption on Fix($\phi$), $x=1$. Therefore $z=y\in K$.</p> <p>As $y$ commutes with all elements of $N$ then $y$ lies in Ker($\phi$) and is trivial. So $z=1$ and the centre of $G$ is trivial.</p> http://mathoverflow.net/questions/47469/mertens-function-in-the-light-of-divergent-summation-what-summation-method-were/47489#47489 Answer by Robin Chapman for mertens-function in the light of divergent summation - what summation method were best adapted Robin Chapman 2010-11-27T08:13:02Z 2010-11-27T08:13:02Z <p>Well, $$\sum_{n=1}^\infty\frac{\mu(n)}{n^s}=\frac1{\zeta(s)}$$ for $s>1$, so setting $s=0$ should give $$\sum_{n=1}^\infty\mu(n)=\frac1{\zeta(0)}=-2$$ as $\zeta(0)=-1/2$. :-)</p> <p>I should add that this is a trick often used in analytic number theory (for instance in Eisenstein series). More generally given a divergent sum $$S=\sum_{i\in I}a_i$$ then consider, for an appropriate choice of weights $b_i>0$ the series $$f(s)=\sum_{i\in I}\frac{a_i}{b_i^s}.$$ We hope this converges in a suitable half-plane and can be analytically continued to $0$. Then we "define" $S=f(0)$.</p> http://mathoverflow.net/questions/47429/open-but-not-affine-subschemeexample/47431#47431 Answer by Robin Chapman for open but not affine subscheme?example? Robin Chapman 2010-11-26T14:28:43Z 2010-11-26T18:21:58Z <p>The standard example is to let $X$ be the affine plane over a field, and $U=X-\{(0,0)\}$.</p> http://mathoverflow.net/questions/47271/from-chain-complex-to-simplicial-abelian-group/47272#47272 Answer by Robin Chapman for From chain complex to simplicial abelian group Robin Chapman 2010-11-24T21:34:23Z 2010-11-24T21:34:23Z <p>This is the <a href="http://ncatlab.org/nlab/show/Dold-Kan+correspondence" rel="nofollow">Dold-Kan correspondence</a>.</p> http://mathoverflow.net/questions/47103/is-every-field-the-field-of-fractions-of-an-integral-domain/47106#47106 Answer by Robin Chapman for Is every field the field of fractions of an integral domain? Robin Chapman 2010-11-23T15:23:39Z 2010-11-23T15:23:39Z <p>Every field $F$ of characteristic zero or of prime characteristic but not algebraic over its prime field is the field of fractions of a proper subring of $F$. But no algebraic extension of $\mathbb F_p$ is, since its only subrings are fields.</p> <p>If $F$ is not an algebraic extension of some $\mathbb F_p$ then $F$ contains a subring $A$ isomorphic to $\mathbb Z$ or $\mathbb F_p[X]$. Each of these rings $A$ has a nontrivial valuation $v$. The valuation $v$ can be prolonged to $F$. Its valuation ring is a proper subring of $F$ whose quotient field is $F$.</p> http://mathoverflow.net/questions/46475/infinite-direct-product-of-the-integers-not-a-free-module-over-the-integers/46481#46481 Answer by Robin Chapman for Infinite direct product of the integers not a free module over the integers Robin Chapman 2010-11-18T13:33:05Z 2010-11-18T19:00:27Z <p>Another reference to a proof of Specker's theorem is <a href="http://www-groups.dcs.st-and.ac.uk/~john/Zagier/Problems.html" rel="nofollow">Zagier's St Andrews problems</a>.</p> <p><strong>Added</strong> Also rings such as $\mathbb{Z}$ with this property are called <a href="http://projecteuclid.org/DPubS?verb=Display&amp;version=1.0&amp;service=UI&amp;handle=euclid.pjm/1102945104&amp;page=record" rel="nofollow">slender rings</a>.</p> http://mathoverflow.net/questions/46485/parametrization-of-o3/46487#46487 Answer by Robin Chapman for Parametrization of O(3) Robin Chapman 2010-11-18T15:04:58Z 2010-11-18T15:04:58Z <p>The general element is $\pm\exp(A)$ where $A$ is skew-symmetric. (This gives each element infinitely often). This trick essentially works for all compact Lie groups.</p> <p>There is also the Cayley parameterization: $(I+A)(I-A)^{-1}$ for skew-symmetric $A$ is the general element of $SO(3)$ which lacks an eigenvalue $-1$ (so isn't a half-turn.) This parameterizes all such matrices once each.</p> http://mathoverflow.net/questions/46350/between-mu-and-primitive-recursion/46351#46351 Answer by Robin Chapman for Between mu- and primitive recursion Robin Chapman 2010-11-17T13:47:36Z 2010-11-17T13:47:36Z <p>You might look up <a href="http://en.wikipedia.org/wiki/Fast-growing_hierarchy" rel="nofollow">fast-growing hierarchies</a>.</p> http://mathoverflow.net/questions/46258/generalizations-of-belyis-theorem/46261#46261 Answer by Robin Chapman for Generalizations of Belyi's theorem Robin Chapman 2010-11-16T17:08:50Z 2010-11-16T17:08:50Z <p>The compactification is the usual one coming up in the theory of modular forms, with the cusps being orbits of $\Gamma$ on $\mathbb{Q}\cup\{\infty\}$.</p> <p>As for the proof, I like <a href="http://uk.arxiv.org/abs/math/0108222" rel="nofollow">this paper</a> by Bernhard Koeck.</p> http://mathoverflow.net/questions/44801/a-5-extension-of-number-fields-unramified-everywhere/44809#44809 Answer by Robin Chapman for $A_5$-extension of number fields unramified everywhere Robin Chapman 2010-11-04T11:23:45Z 2010-11-04T11:37:47Z <p>Here's the standard example. I found it in Lang's <em>Algebraic Number Theory</em> where he attributes it to Artin. Let $K$ be the splitting field of $X^5-X+1$ over $\mathbb{Q}$. Then $K$ has Galois group $S_5$ over $\mathbb{Q}$ and $A_5$ over $L=\mathbb{Q}(\sqrt{2869})$. Also $K$ is unramified over $L$.</p> http://mathoverflow.net/questions/44666/does-the-hausdorff-dimension-depend-on-the-lp-norm/44668#44668 Answer by Robin Chapman for Does the Hausdorff dimension depend on the L^p-norm? Robin Chapman 2010-11-03T11:34:43Z 2010-11-03T11:34:43Z <p>Let $B_p$ denote the 1-ball with centre 0 with respect to the $l^p$ norm. For any $p$ and $q$ there is a number $N$ such that $B_p$ is covered by $N$ translates of $B_q$. Then any $\epsilon$-ball in the $l^p$ norm is covered by $N$ $\epsilon$-balls in the $l^q$ norm. Thus within a constant factor, the number of $\epsilon$-balls required to cover a set in the $l^p$ and $l^q$ norms is the same. This constant factor won't affect the asymptotic power in the number of $\epsilon$-balls required to cover a given set, so the Hausdorff dimension in both cases is the same.</p> http://mathoverflow.net/questions/44115/analytic-continuation-via-square-of-absolute-value/44126#44126 Answer by Robin Chapman for Analytic continuation via square of absolute value Robin Chapman 2010-10-29T13:00:22Z 2010-10-29T13:00:22Z <p>Rather obviously not: if $f(z)=\sqrt{z}$ on $U$, the plane slit along the negative real axis, then $|f(z)|^2=|z|$ is real analytic on $V$ the plane with the origin removed but $f$ does not analytically continue from $U$ to $V$.</p> http://mathoverflow.net/questions/43650/which-elements-in-sl2q-are-conjugated-to-an-element-in-sl2z/43651#43651 Answer by Robin Chapman for Which elements in SL2(Q) are conjugated to an element in SL2(Z) Robin Chapman 2010-10-26T10:01:36Z 2010-10-26T10:01:36Z <p>You can do this if and only if the trace of $M$ is an integer. By the theory of the rational canonical form if matrices $A$ and $B$ over $\mathbb{Q}$ have the same characteristic polynomial and neither has a repeated eigenvalue they are conjugate by a matrix over $\mathbb{Q}$. This almost does it, save for some fiddling about when the eignvalue of $M$ is repeated.</p> http://mathoverflow.net/questions/43489/analysis-of-a-quadratic-diophantine-equation/43490#43490 Answer by Robin Chapman for Analysis of a quadratic diophantine equation Robin Chapman 2010-10-25T06:59:51Z 2010-10-25T15:17:32Z <p>One thing to do is to try to express these in terms of squares. Note that $$12x(3x-1)=36x^2-12x=(6x-1)^2-1$$ so that your equations become $$a_1^2+b_1^2=c_1^2+1$$ and $$a_1^2-b_1^2=d_1^2-1$$ where $a_1=6a-1$ etc. Then the variables $a_1$ etc are constrained to be congruent to $5$ modulo $6$.</p> <p>Homogenizing these gives $$X^2+Y^2=Z^2+T^2$$ and $$X^2-Y^2=Z^2-T^2.$$ Searching for rational solutions of your equation is essentially looking for rational points on the intersection of these two quadrics in $\mathbf{P}^3$. In general the intersection of two quadrics in $\mathbf{P}^3$ is an elliptic curve, so it looks like your problem will boil down to something like finding the integer points on an elliptic curve.</p> <p><strong>Added</strong> There's a blunder in the above: I must thank Fedor for noticing that the second equation should be $$X^2-Y^2=W^2-T^2.$$ So the variety is the intersection of two quadrics in $\mathbf{P}^4$. Hartshorne mentions in passing that in general this construction gives a del Pezzo surface. Del Pezzo surfaces are rational so there should be a birational parametrizion (in terms of two affine parameters) of the <strong>rational</strong> solutions to the original pair of equations.</p> http://mathoverflow.net/questions/43180/integration-problem-int-pi-pi-log-1-exp-i-nu-mathr/43185#43185 Answer by Robin Chapman for Integration problem: $\int_{-\pi}^{\pi} | \log( | 1 + \exp(- I \nu ) | ) | \mathrm{d}\nu < \infty$ Robin Chapman 2010-10-22T14:35:46Z 2010-10-22T14:35:46Z <p>You want to show that $$\int_{-\pi}^\pi|\log|1+e^{-it}||dt$$ is finite. Now $$|1+e^{-it}|=|e^{it/2}+e^{-it/2}|=2\cos(t/2)$$ so your integral is $$\int_{-\pi}^\pi|\log|2\cos(t/2)||dt =2\int_0^\pi|\log|2\cos(t/2)||dt.$$ Replacing $t$ by $\pi-2$ in the last integral gives $$2\int_0^\pi|\log|2\sin(t/2)||dt.$$ The integrand is nicely continuous away from $0$. Near $0$, $\sin (t/2)=tf(t)$ where $f(t)\to1/2$ as $t\to0$. Then the integrand is $|\log t+g(t)|$ where $g$ is continuous at $0$ and now finiteness follows since $$\int_0^1|\log t|dt$$ is finite (integration by parts).</p> http://mathoverflow.net/questions/42809/how-many-hecke-operators-span-the-level-1-hecke-algebra/42811#42811 Answer by Robin Chapman for How many Hecke operators span the level 1 Hecke algebra? Robin Chapman 2010-10-19T17:45:55Z 2010-10-19T17:45:55Z <p>The answer is yes when $k$ is a multiple of $4$. There is a unique form of weight $k$ of the form $f_k=1+a_dq^d+\cdots$. When $k$ is a multiple of $4$ this is the theta series for a putative extremal even unimodular lattice of rank $2k$. Theorem 20 in chapter 7 of Conway and Sloane's <em>Sphere Packings, Lattices and Groups</em> asserts that $a_d>0$. They give several references for the proof, including a 1969 paper of Siegel.</p> http://mathoverflow.net/questions/42016/algorithms-for-finding-rational-points-on-an-elliptic-curve/42021#42021 Answer by Robin Chapman for Algorithms for finding rational points on an elliptic curve? Robin Chapman 2010-10-13T14:04:52Z 2010-10-13T14:04:52Z <p>There is a whole industry devoted to this. The basic method is by <em>descent</em>, which is a formalized version of the infinite descent proofs of Fermat and Euler. It helps if there are rational 2-torsion points but it's not essential. Chapter X in Silverman's <em>The Arithmetic of Elliptic Curves</em> is called "Computing the Mordell-Weil group". It has lots of good information, but maybe isn't so easy for a beginner due to its heavy use of group cohomology.</p> http://mathoverflow.net/questions/41757/eigenvalues-of-sum-of-two-anti-commuting-matrices/41761#41761 Answer by Robin Chapman for Eigenvalues of sum of two anti-commuting matrices Robin Chapman 2010-10-11T08:23:42Z 2010-10-11T18:35:00Z <p>Suppose for simplicity's sak that $A$ and $B$ are diagonalizable over $\mathbb{R}$ and are non-singular.</p> <p>Let $V_a$ be the $a$-eigenspace of $A$. Then by anti-commutativity, we find $BV_a\subseteq V_{-a}$ etc. As $A$ and $B^2$ commute then there is an eigenvector $v\in V_a$ with $B^2v=b^2 v$ for some $v$. If we let $w=bv+Bv$ then $Bw=bw$ so $b$ is real (assuming $B$ has real eigenvectors). On the space $W$ spanned by $v$ and $w=b^{-1}Bv$ the linear transformation $A+B$ has matrix $$\left(\begin{array}{rr} a&amp;b\\ b&amp;-a\\ \end{array}\right)$$ which has an eigenvectors with eigenvalues $\pm\sqrt{a^2+b^2}$.</p> http://mathoverflow.net/questions/41606/convex-sets-and-projections/41608#41608 Answer by Robin Chapman for Convex sets and projections Robin Chapman 2010-10-09T16:13:34Z 2010-10-09T16:13:34Z <p>I presume what you want to prove is the following. Let $S$ be a nonempty closed subset of $\mathbb{R}^n$. Then if there is a point $y\in\mathbb{R}^n$ and there are at least two points $p$ and $q$ in $S$ with Euclidean distance $d$ from $y$ (where $d$ is the distance of $y$ from $S$), then $S$ is not convex. To see this, note that the midpoint $r$ of the line segment $pq$ is closer to $y$ than $p$ of $q$ is, and so cannot lie in $S$. Hence $S$ isn't convex.</p> http://mathoverflow.net/questions/41345/how-to-resolve-an-issue-with-pranesachar-et-al-s-formula-for-the-number-of-four/41371#41371 Answer by Robin Chapman for How to resolve an issue with Pranesachar et al.'s formula for the number of four-line Latin rectangles? Robin Chapman 2010-10-07T06:40:14Z 2010-10-07T06:40:14Z <p>Obviously, as you know, writing down something like $(-3)!$ is absurd and meaningless. But I would contend that absurd and meaningless as an expression like $$\frac{(-3)!}{(-6)!}$$ is, that it still equals $(-3)(-4)(-5)=-60$. If you have these negative factorials paired up into numerators and denominators, you can calculate like this. Essentially this is interpolating the factorial in the obvious way via the gamma function and noting that sometimes poles cancel to give removable singularities.</p> <p>I don't know if using a convention like this will successfully resolve your particular problem though.</p> http://mathoverflow.net/questions/41310/any-sum-of-2-dice-with-equal-probability/41311#41311 Answer by Robin Chapman for Any sum of 2 dice with equal probability Robin Chapman 2010-10-06T18:39:55Z 2010-10-06T18:39:55Z <p>You can write this as a single polynomial equation $$p(x)q(x)=\frac1{11}(x^2+x^3+\cdots+x^{12})$$ where $p(x)=p_1x+p_2x^2+\cdots+p_6x^6$ and similarly for $q(x)$. So this reduces to the question of factorizing $(x^2+\cdots+x^{12})/11$ where the factors satisfy some extra conditions (coefficients positive, $p(1)=1$ etc.).</p> <p>This is a standard method (generating functions).</p> http://mathoverflow.net/questions/41245/proof-that-bases-etc-exist-in-early-linear-algebra-course/41248#41248 Answer by Robin Chapman for Proof that bases etc. exist in early linear algebra course? Robin Chapman 2010-10-06T09:59:08Z 2010-10-06T09:59:08Z <p>If a vector space had bases of two different finite sizes $m &lt; n$ say, then expressing one in terms of the other gives $m$ by $n$ and $n$ by $m$ matrices $A$ and $B$ such that $BA=I_n$. Now use Gaussian elimination (quote their first-year course) to "prove" (with an appropriate amount of hand-waving) that $Av=0$ has a nonzero solution, and thus derive a contradiction. Surely less than a lecture :-)</p> http://mathoverflow.net/questions/41187/a-coverage-question/41191#41191 Answer by Robin Chapman for A coverage question Robin Chapman 2010-10-05T18:14:22Z 2010-10-05T18:14:22Z <p>These odd multiples are (save for $q=3$) the class numbers $h_{-q}$ of the fields $\mathbb{Q}(\sqrt{-q})$. (This follows from the analytic class number formula,)</p> <p>By the <a href="http://en.wikipedia.org/wiki/Brauer%25E2%2580%2593Siegel_theorem" rel="nofollow">Brauer-Siegel theorem</a>, $\log h_{-q}\sim\log\sqrt{q}$. From this it looks plausible that all possible odd $h$ could occur. I'm not aware of any proof that this is the case (nor of any potential counterexample).</p> http://mathoverflow.net/questions/40816/fibonacci-series-mod-a-number/40818#40818 Answer by Robin Chapman for fibonacci series mod a number Robin Chapman 2010-10-02T06:51:18Z 2010-10-02T06:51:18Z <p>This is really just an expansion of Gerhard's comment. One has the matrix formula $$\begin{pmatrix} 1&amp;1\\ 1&amp;0 \end{pmatrix}^n= \begin{pmatrix} F_{n+1}&amp;F_n\\ F_n&amp;F_{n-1} \end{pmatrix}$$ so the problem reduces to computing $A^n$ modulo $k$ where $$A=\begin{pmatrix} 1&amp;1\\ 1&amp;0 \end{pmatrix}.$$ This can be done by the <a href="http://en.wikipedia.org/wiki/Exponentiation_by_squaring" rel="nofollow">repeated squaring</a> method often used in <a href="http://en.wikipedia.org/wiki/Modular_exponentiation" rel="nofollow">modular exponentiation</a>. The idea is to compute $A^n$ recursively either as $(A^m)^2$ or $A(A^m)^2$ according to whether $n=2m$ or $n=2m+1$.</p> http://mathoverflow.net/questions/49729/free-research-web-pages Comment by Robin Chapman Robin Chapman 2010-12-17T14:05:22Z 2010-12-17T14:05:22Z What is &quot;free/independent&quot;? http://mathoverflow.net/questions/20664/why-is-complex-projective-space-triangulable/35241#35241 Comment by Robin Chapman Robin Chapman 2010-12-16T14:10:13Z 2010-12-16T14:10:13Z According to the authors of <a href="http://uk.arxiv.org/abs/1012.3235" rel="nofollow">uk.arxiv.org/abs/1012.3235</a> &quot;no explicit triangulation of $CP^3$ was known so far&quot;. http://mathoverflow.net/questions/49485/modular-lambda-function-as-a-cross-ratio Comment by Robin Chapman Robin Chapman 2010-12-15T08:07:26Z 2010-12-15T08:07:26Z An elliptic curve has (affine) equation $y^2=f(x)$ where $f$ is a cubic or a quartic. In either cases there is a map from $E$ to $P^1$ given by $(x,y)\mapsto x$. This is a double cover, ramified at the zeros of $f$, and also at $\infty$ when $f$ is cubic. http://mathoverflow.net/questions/49398/sieve-numbers-conjecture Comment by Robin Chapman Robin Chapman 2010-12-14T16:06:13Z 2010-12-14T16:06:13Z What's the question? http://mathoverflow.net/questions/48809/relationships-between-the-roots-of-an-entire-function-and-the-roots-of-its-deriva/48815#48815 Comment by Robin Chapman Robin Chapman 2010-12-09T19:59:19Z 2010-12-09T19:59:19Z And also in Macdonald's <i>Symmetric Functions and Hall Polynomials</i>. I don't know off-hand if they have results like this though. http://mathoverflow.net/questions/48771/proofs-that-require-fundamentally-new-ways-of-thinking/48808#48808 Comment by Robin Chapman Robin Chapman 2010-12-09T19:44:39Z 2010-12-09T19:44:39Z Ans others think it's the usual proof in disguise :-) http://mathoverflow.net/questions/48789/rank-of-a-free-module-without-the-axiom-of-choice/48791#48791 Comment by Robin Chapman Robin Chapman 2010-12-09T18:06:42Z 2010-12-09T18:06:42Z For finitely generated free modules over commutative rings, the result is elementary. For finitely generated free modules over non-commutative rings, the assertion can fail. http://mathoverflow.net/questions/32011/direct-proof-of-irrationality/48780#48780 Comment by Robin Chapman Robin Chapman 2010-12-09T16:50:26Z 2010-12-09T16:50:26Z So why does $p/q$ being in lowest terms entail that $p^2/q^2$ is? If &quot;in lowest terms&quot; means having no common factors save units, then this implication doesn't hold in all integral domains. http://mathoverflow.net/questions/32967/have-any-long-suspected-irrational-numbers-turned-out-to-be-rational/48663#48663 Comment by Robin Chapman Robin Chapman 2010-12-09T16:28:01Z 2010-12-09T16:28:01Z Version 2 of the first paper has now appeared on ArXiv. In it, &quot;Lemma 2&quot; has been downgraded to &quot;Conjecture 1&quot;, but a footnote claims that a cited reference has proved it. :-) http://mathoverflow.net/questions/48623/the-numbers-of-zeta-zeros-how-did-riemann-find-it Comment by Robin Chapman Robin Chapman 2010-12-08T10:40:54Z 2010-12-08T10:40:54Z No, Riemann gives an aysmptotic formula not for the number of zeroes on the critical line, but rather in the critical strip. A translation of Riemann's paper appears in an appendix to Harold Edwards's book <i>Riemann's Zeta Function</i> now avaliable as a Dover paperback. Edwards proves all the results in Riemann's paper and a lot more. http://mathoverflow.net/questions/48426/detect-if-directed-cycle-is-clockwise-or-counterclockwise-in-3d Comment by Robin Chapman Robin Chapman 2010-12-06T10:22:42Z 2010-12-06T10:22:42Z Clockwise and anticlockwise don't mean anything in three dimensions. http://mathoverflow.net/questions/48350/list-of-inclusions Comment by Robin Chapman Robin Chapman 2010-12-05T10:26:37Z 2010-12-05T10:26:37Z <a href="http://www.amazon.co.uk/Introduction-Differential-Inclusions-Graduate-Mathematics/dp/0821829777/ref=sr_1_10?ie=UTF8&amp;qid=1291544765&amp;sr=8-10" rel="nofollow">amazon.co.uk/&hellip;</a> http://mathoverflow.net/questions/48351/crookedness-of-convex-curves-milnor Comment by Robin Chapman Robin Chapman 2010-12-05T09:52:19Z 2010-12-05T09:52:19Z Could you please remind those of us without immediate access to Milnor's paper of what $\mu(P,b)$ denotes. http://mathoverflow.net/questions/48268/find-a-linearly-independent-set Comment by Robin Chapman Robin Chapman 2010-12-04T11:17:53Z 2010-12-04T11:17:53Z This question is very difficult to read :-( http://mathoverflow.net/questions/48065/hopf-algebras-rings-a-question-of-terminology/48069#48069 Comment by Robin Chapman Robin Chapman 2010-12-02T18:35:51Z 2010-12-02T18:35:51Z Thanks Todd, but doesn't one also have group objects in monoidal categories?