User bill dubuque - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T19:16:03Z http://mathoverflow.net/feeds/user/6716 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/25059/examples-of-undergraduate-mathematics-separation-from-what-mathematicians-should/30939#30939 Answer by Bill Dubuque for Examples of undergraduate mathematics separation from what mathematicians should know Bill Dubuque 2010-07-07T19:08:11Z 2012-01-11T06:04:37Z <p>One might argue that (exotic) counterexamples fall into the first category. For example, when I took Munkres' course in topology it was organized around many very carefully chosen (counter)examples showing how various properties were related. This led me me to delve into many exotic spaces listed in Steen and Seebach's <em>Counterexamples in Topology</em>. While I'll probably never make use of any of those exotic counterexamples, it did help me to learn better how to employ the axiomatic method, e.g. to understand how to constuct examples showing that axioms are independent, to construct pertinent examples for theoretical signposts. summits, etc. One isn't necessarily expected to remember the examples but, rather, the methodology (e.g. various ideas of completion, compactification, ...)</p> http://mathoverflow.net/questions/17778/books-you-would-like-to-see-translated-into-english/30324#30324 Answer by Bill Dubuque for Books you would like to see translated into English. Bill Dubuque 2010-07-02T17:01:39Z 2012-01-02T22:51:08Z <p>The following wonderful 54 page survey by O. Neumann on Kronecker's divisor theory could easily be turned into a book and would fill a very large gap in the English literature on such. I'm interested in helping if anyone is game for such (but, alas, my German is weak).</p> <p>Neumann, O.(D-FSU-MI) 2003k:13021 13F05 (01A55 13G05 20M14)<br> Was sollen und was sind Divisoren? (German. German summary)<br> [What are divisors and what can we do with them?]<br> Math. Semesterber. * 48 (2002), no. 2, 139--192.</p> <p>In the first part of this paper a survey is given of the development of Kronecker's theory of divisors. In the second part the author develops a theory of integral domains $R$ having a divisor theory in the following sense: there exists a monoid $D$ (i.e., a commutative semigroup with cancellation and a unit element) with the GCD-property for the associated group $G$ of quotients, and a homomorphism $\mathrm{div}$ of the multiplicative group $K^*$ of the quotient field of $R$ into $G$ with the following two properties: </p> <p>(i) If $a,b \in K^*$ and $b/a \in R$, then $\mathrm{div}(b)/\mathrm{div}(a) \in D$, and </p> <p>(ii) for every element $d \in D$ there exists a set $A \subseteq K^*$<br> such that $d$ is the gcd of <code>$\{\mathrm{div}(a) : a \in A\}$</code>. </p> <p>The author states that a similar theory was presented in the thesis of F. Lucius ["Ringe mit einer Theorie des groessten gemeinsamen Teilers", Ph.D. thesis, Univ. Gottingen, Gottingen, 1996; Zbl 0901.13002]. After developing the fundamental properties of such divisor theory, relations to the approaches of Kronecker, Zolotarev and Dedekind are established.<br> --Reviewed by W. Narkiewicz</p> http://mathoverflow.net/questions/22579/what-are-the-reasons-for-considering-rings-without-identity/60768#60768 Answer by Bill Dubuque for What are the reasons for considering rings without identity? Bill Dubuque 2011-04-06T04:31:08Z 2011-04-06T04:31:08Z <p>Perhaps you will find the following remarks of interest, excerpted from the preface of Gardner and Wiegandt: <em>Radical Theory of Rings</em>, 2004.<br> <img src="http://i.imgur.com/Zrwfd.jpg" alt="alt text"></p> http://mathoverflow.net/questions/13029/rational-exponential-expressions/31461#31461 Answer by Bill Dubuque for Rational exponential expressions Bill Dubuque 2010-07-11T20:51:56Z 2011-03-10T06:33:53Z <p>Yes, there are algorithms to decide asymptotic dominance - in fact for a much wider class of elementary functions. I discovered the first such algorithm circa 1980 while an undergrad member of the MIT Mathlab group researching effective algorithms for computing limits for the Macsyma symbolic computation system. Another different algorithm was discovered independently a handful of years later by John Shackell. You should be able to find references to the literature by googling the more recent buzzword "transseries". Many computer algebra systems have (partial) implementations of these algorithms. Here's an example of my algorithm on (my generalization of) an example Rich Schroeppel proposed to attempt to stump my algorithm (he was convinced no such algorithm existed). It shows that $\rm{\: lim_{x\to\infty}\ d40 = e^a}$ <img src="http://i29.tinypic.com/312761k.jpg" alt="alt text"><br> Don't dare try L'Hopital's rule on that monster!<br> For further discussion (and the text form of the above image)<br> see my post on sci.math, 1996/03/20, L'Hospital's rule question<br> <a href="http://groups.google.com/group/sci.math/msg/05298104ac44efd2" rel="nofollow">http://groups.google.com/group/sci.math/msg/05298104ac44efd2</a><br> <a href="http://groups.google.com/groups?selm=WGD.96Mar20231913%40berne.ai.mit.edu" rel="nofollow">http://groups.google.com/groups?selm=WGD.96Mar20231913%40berne.ai.mit.edu</a> </p> http://mathoverflow.net/questions/49035/explicit-bezout-cofactors/49087#49087 Answer by Bill Dubuque for Explicit Bezout Cofactors Bill Dubuque 2010-12-11T22:14:57Z 2010-12-11T23:39:48Z <p>For the first query: yes, $\rm\: A - V\ (AT+BS) = S\ (AU - BV) - A\ (SU + TV - 1)$ </p> <p>Now $\rm\ AU = BV = C\ \Rightarrow\ C\ (AT+BS) = BV AT + AU BS = AB\ (TV + US) = AB$</p> <p>Hence $\rm\ \ \ C\: =\: lcm(A,B)\ \Rightarrow\ AT+BS\ =\ AB/C\ =\ gcd(A,B)$</p> <p>For the second question: you may find useful the theory of Grobner bases over a $\rm\: PID$.</p> http://mathoverflow.net/questions/48888/what-is-the-divisibility-theory-for-bezout-domains/48962#48962 Answer by Bill Dubuque for What is the divisibility theory for Bezout Domains? Bill Dubuque 2010-12-10T19:24:11Z 2010-12-10T20:40:47Z <p>Based upon your examples, you may find it more natural to work with the slightly more general class of Prüfer domains (vs. Bezout domains) - i.e. finitely generated ideals$\:\ne 0$ are invertible (vs. principal). Prüfer domains are non-Noetherian generalizations of Dedekind domains. Their ubiquity stems from a remarkable confluence of interesting characterizations, e.g. CRT, or Gauss's Lemma for content ideals, or for ideals $\rm\ A\cap (B + C) = A\cap B + A\cap C\:,\:$ or $\rm\ (A + B)\ (A \cap B) = A\ B\$ etc. It's been estimated that there are close to 100 such characterizations known, e.g. see my <a href="http://groups.google.com/group/sci.math/msg/ba5ac69dddfc4a67" rel="nofollow">sci.math post</a> for 30.</p> <p>As a simple example I'll give the natural Prüfer domain proof of a generalization of your example, viz. the ideal-theoretic $\:$ <em>Freshman's Dream</em> $\rm\ \ (A + B)^n = A^n + B^n\:.\:$ This identity is true for both arithmetic of $\:$ GCDs $\:$ and invertible ideals simply because, in both cases, multiplication is cancellative and addition is idempotent, i.e. $\rm\ A + A = A\$ for ideals and $\rm\ (A,A) = A\$ for GCDs. Combining these properties with the associative, commutative, distributive laws of addition and multiplication we obtain an extremely elementary high-school-level proof of the Freshman's Dream - which is best illustrated for $\rm\: n = 2\:,\:$ viz.</p> <p>$\rm\quad\quad (A + B)^4 \ =\ A^4 + A^3 B + A^2 B^2 + AB^3 + B^4$</p> <p>$\rm\quad\quad\phantom{(A + B)^4 }\ =\ A^2\ (A^2 + AB + B^2) + (A^2 + AB + B^2)\ B^2$</p> <p>$\rm\quad\quad\phantom{(A + B)^4 }\ =\ (A^2 + B^2)\ \:(A + B)^2$</p> <p>So $\rm\ {(A + B)^2 }\ =\ \ A^2 + B^2\$ if $\rm\ A+B\$ is cancellative, e.g. if $\rm\ A+B = 1\$ or if it's invertible.</p> <p>The same proof generalizes for all $\rm\:n\:$ since, as above</p> <p>$\rm\quad\quad (A + B)^{2n}\ =\ A^n\ (A^n + \cdots+ B^n) + (A^n +\cdots+ B^n)\ B^n$</p> <p>$\rm\quad\quad\phantom{(A + B)^{2n}}\ =\ (A^n + B^n)\ (A + B)^n$</p> <p>In the GCD case $\rm\ A+B\ := (A,B) = \gcd(A,B)\$ for $\rm\:A,B\:$ in a GCD-domain, i.e. a domain where $\rm\: \gcd(A,B)\:$ exists for all $\rm\:A,B \ne 0,0\:$. Here too the Dream is true since $\rm\:(A,B)\:$ is cancellable, being nonzero in a domain. (Note: one can unify the GCD and ideal cases by employing Divisor Theory).</p> <p>In fact this yields yet another characterization: a domain is Prufer iff it satisfies the Freshman's Dream for all finitely generated ideals. See said sci.math post for further discussion. </p> http://mathoverflow.net/questions/30220/abstract-thought-vs-calculation/30313#30313 Answer by Bill Dubuque for Abstract Thought vs Calculation Bill Dubuque 2010-07-02T14:53:41Z 2010-12-10T18:07:15Z <p>Some of the prettiest examples of Dedekind's structuralism arise from revisiting proofs in elementary number theory from a highbrow viewpoint, e.g. by reformulating them after noticing hidden structure (ideals, modules, etc). A striking example of such is the generalization and unification of elementary irrationality proofs of n'th roots by way of Dedekind's notion of <strong>conductor ideal</strong>. This gem seems to be little-known (even to some number theorists, e.g. Estermann and Niven). Since I've already explained this at length elsewhere I'll simply <a href="http://groups.google.com/group/sci.math/msg/b547bca171fc24be" rel="nofollow">link [1]</a> to it. </p> <p>At first glance the various "elementary" proofs seem to be magically pulled out of a hat since the crucial structure of the conductor ideal is obfuscated by the descent "calculations" of various lemmas (that have all been inlined vs. abstracted out). However, once one abstracts out the hidden innate structure the proof becomes a striking one-liner: simply remark that in a PID a conductor ideal is principal so cancelable, thus PIDs are integrally closed. Here, the complexity of the calculations verifying the descent (induction) etc are abstracted out and tidily encapsulated once-and-for-all in the lemma that Euclidean domains are PIDs. Following Dedekind's ground-breaking insight, we recognize in many number-theoretical contexts the innate structure of an ideal, and we exploit that structure whenever possible. For much further detail and discussion see <em>all</em> of my posts in the thread <a href="http://groups.google.com/group/sci.math/msg/b547bca171fc24be" rel="nofollow">1</a> (click on the thread's title/subject at the top of the frame to see a threaded view in the Google Groups usenet web interface)</p> <p>When I teach such topics I emphasize that one should always look for "hidden ideals" and other obfuscated innate structure. Alas, too many students cannot resist the urge to dive in and "calculate" before pursuing conceptual investigations. It was such methodological principles that led Dedekind to discover most all of the fundamental algebraic structures. Nowadays we often take for granted such structural abstractions and methodology. But it was certainly a nontrivial task to discover these in the rarefied mathematical atmosphere of Dedekind's day (and it remains so even nowadays for students when first learning such topics). Emmy Noether wasn't joking when she said "it's all already in Dedekind". It deserves emphasis that this remark also remains true for methodological principles.</p> <p><a href="http://groups.google.com/group/sci.math/msg/b547bca171fc24be" rel="nofollow">1</a> sci.math, 20 May 2009, Irrationality of sqrt (n^2 - 1)<br> <a href="http://groups.google.com/groups?selm=y8z3ab08eh3.fsf%40nestle.csail.mit.edu" rel="nofollow">http://groups.google.com/groups?selm=y8z3ab08eh3.fsf%40nestle.csail.mit.edu</a></p> http://mathoverflow.net/questions/35524/how-fast-can-the-base-bumping-function-in-goodsteins-theorem-grow/35546#35546 Answer by Bill Dubuque for How fast can the base-bumping function in Goodstein's theorem grow? Bill Dubuque 2010-08-13T22:05:31Z 2010-12-01T02:52:43Z <p>Goodstein actually employed arbitrary increasing base-bumping functions. He showed that the convergence of all such is equivalent to transfinite induction below $\epsilon_0$. This is illustrated somewhat more graphically by the Hercules vs. Hydra game. See the references from my old <a href="http://groups.google.com/groups?selm=WGD.95Dec11023450@martigny.ai.mit.edu" rel="nofollow">post [1]</a> of 1995 which helped serve to popularize these topics on (use)net. Curiously that post received far more feedback than any of my other posts - from popular science writers to researchers, teachers and students. </p> <p>[1] Bill Dubuque, sci.math, Dec 11, 1995. Goedel's theorem: about anything in real world?<br> <a href="http://groups.google.com/groups?selm=WGD.95Dec11023450@martigny.ai.mit.edu" rel="nofollow">http://groups.google.com/groups?selm=WGD.95Dec11023450@martigny.ai.mit.edu</a></p> http://mathoverflow.net/questions/35127/why-does-the-algebraic-condition-of-flatness-on-the-structure-sheaves-give-a-goo Why does the algebraic condition of flatness on the structure sheaves give a good definition of family? Bill Dubuque 2010-08-10T15:30:01Z 2010-08-26T20:07:28Z <p>Hartshorne remarks that is is something of a mystery as to why the algebraic condition of flatness on the structure sheaves gives a good definition of a family (see below). Are there any known enlightening explanations that help serve to unravel this mystery? Below is Hartshorne's introductory motivation to flat families containing said remark:</p> <blockquote> <p>For many reasons it is important to have a good notion of an algebraic family of varieties or schemes. The most naive definition would be just to take the fibres of a morphism. To get a good notion, however, we should require that certain numerical invariants remain constant in a family, such as the dimension of the fibres. It turns out that if we are dealing with non- singular (or even normal) varieties over a field, then the naive definition is already a good one. Evidence for this is the theorem (9.13) that in such a family, the arithmetic genus is constant. </p> <p>On the other hand, if we deal with nonnormal varieties, or more general schemes, the naive definition will not do. So we consider a flat family of schemes, which means the fibres of a flat morphism, and this is a very good notion. Why the algebraic condition of flatness on the structure sheaves should give a good definition of a family is something of a mystery. But at least we will justify this choice by showing that flat families have many good properties, and by giving necessary and sufficient conditions for flatness in some special cases. In particular, we will show that a family of closed subschemes of projective space (over an integral scheme) is flat if and only if the Hilbert polynomials of the fibres are the same. -- Hartshorne, Algebraic Geometry, 1977, III.9.5, p. 256</p> </blockquote> http://mathoverflow.net/questions/35140/interesting-applications-in-pure-mathematics-of-first-year-calculus/35147#35147 Answer by Bill Dubuque for Interesting applications (in pure mathematics) of first-year calculus Bill Dubuque 2010-08-10T17:12:38Z 2010-08-10T17:55:49Z <p>An interesting application of calculus is the elementary polynomial case of Mason's ABC theorem. This yields, for instance, a completely trivial proof of the polynomial case of FLT (Fermat's Last Theorem). That this works so effectively for polynomials (functions) vs. numbers is due to the fact that for functions we have available the derivative, which implies that we can exploit Wronskians as a measure of algebraic independence. Such Wronskian estimates serve as fundamental tools in diophantine approximation. See my <a href="http://google.com/groups?selm=WGD.96Jul17041312@berne.ai.mit.edu" rel="nofollow">post [1]</a> for further details and references. </p> <p>[1] sci.math.research, 1996/07/17<br> poly FLT, abc theorem, Wronskian formalism [was: Entire solutions of f^2+g^2=1]<br> <a href="http://groups.google.com/group/sci.math/msg/4a53c1e94f1705ed" rel="nofollow">http://groups.google.com/group/sci.math/msg/4a53c1e94f1705ed</a><br> <a href="http://google.com/groups?selm=WGD.96Jul17041312@berne.ai.mit.edu" rel="nofollow">http://google.com/groups?selm=WGD.96Jul17041312@berne.ai.mit.edu</a> </p> http://mathoverflow.net/questions/34544/size-of-a-groebner-basis/35013#35013 Answer by Bill Dubuque for Size of a Groebner basis Bill Dubuque 2010-08-09T15:19:49Z 2010-08-09T15:19:49Z <p>There is a long strand of research attempting to show that the worst-case doubly-exponential bounds don't apply to "interesting" problems - dating back to Bayer and Stillman's work on Castelnuovo-Mumford regularity in the eighties to very recent work of Mayr on sharpened bounds for low-dimensional ideals. The obvious keyword searches, e.g. Grobner exponential, should quickly locate the pertinent literature.</p> http://mathoverflow.net/questions/34806/what-was-the-relative-importance-of-flt-vs-higher-reciprocity-laws-in-kummers-i/34857#34857 Answer by Bill Dubuque for What was the relative importance of FLT vs. higher reciprocity laws in Kummer's invention of algebraic number theory? Bill Dubuque 2010-08-07T19:08:52Z 2010-08-07T19:08:52Z <p>I too defer to Franz's deep knowledge on such topics. Everything that I have read in both the primary and secondary literature completely agrees with what Franz has written here and elsewhere. Besides the link I gave to the discussion in his interesting <a href="http://www.rzuser.uni-heidelberg.de/~hb3/ideal3.pdf" rel="nofollow">paper</a> on Jacobi and Kummer's ideal numbers, one may also find helpful the following passage from p. 15 of his beautiful <a href="http://www.rzuser.uni-heidelberg.de/~hb3/rec.html" rel="nofollow">book</a> on reciprocity laws. It provides a concise summary of what we currently know about such matters. I quote it below since some readers may not have convenient access to the book.</p> <blockquote> <p>The role of Fermat's Last Theorem in the development of algebraic number theory is often overrated, probably due to Hensel's (false) story claiming Kummer's first manuscript to have been an incorrect proof of this problem. The crème de la crème of French mathematicians - Lame, Legendre, Liouville, and Cauchy - tried their luck but didn't really advance algebraic number theory during their work on FLT. Gauss did not value it very highly, but admits that it made him take up his investigations in number theory again: in a letter to Olbers from March 21, 1816, he writes</p> <blockquote> <p>I admit, that Fermat's Theorem, as an isolated result, has little interest for me, since I can easily make a lot of such claims that can be neither proved nor disproved. Nevertheless it made me take up again some old ideas about a large extension of the higher arithmetic. [...] Yet I am convinced, if luck should do more than I may expect and if I succeed in making some of the main steps in that theory, then Fermat's Theorem will appear as one of the less interesting corollaries.</p> </blockquote> <p>Gauss's last remark clearly indicates that he was at least thinking about the arithmetic in cyclotomic number fields ${\mathbb Q}(\zeta_p)$, even when, in a letter to Bessel a few months later, he reveals that the investigations in question had to do with the part that he eventually would publish, namely the theory of biquadratic residues. Parts of his research were published in 1828 [272] and 1832 [273], and the last paper contains the statement that cubic reciprocity is best described in ${\mathbb Z}[\rho]$, where $\rho^2 + \rho + 1 = 0$, and that, more generally, the study of higher reciprocity laws should be done after adjoining higher roots of unity.</p> <p>Even Kummer, who is responsible for the greatest advance towards of Fermat's Last Theorem before the recent developments, got the main motivation for studying cyclotomic fields from his desire to find a general reciprocity law (which he called his "main enemy" in [Ku, Feb 25, 1848]). In almost every letter to Kronecker written between 1842 and 1848, Kummer mentions results related to reciprocity; the Fermat equation is mentioned for the first time in [Ku, Apr. 02, 1847]. In [Ku, Sept. 17, 1849] he informs Kronecker about the prize of 3000 Francs that the French Academy had offered to pay for a solution of Fermat's Last Theorem, and in [Ku, Jan. 14, 1850] he writes</p> <blockquote> <p>Once I will have fathomed whether this is so or not, then I will drop the avaricious plans and work again only for the science, especially for the reciprocity laws for which I have already envisaged some ideas.</p> </blockquote> <p>It is therefore safe to say that it was the quest for higher reciprocity laws that made Kummer study abelian extensions of $\mathbb Q$, Eisenstein those of ${\mathbb Q}(i)$ and Hilbert those of general number fields. Ironically, it was Hilbert himself who started the rumour that it was FLT that led Kummer to his ideal numbers: in his famous address at the ICM in Paris 1900, he wrote</p> <blockquote> <p>stimulated by Fermat's problem, Kummer arrived at the introduction of his ideal numbers and discovered the theorem of unique factorization of the integers of cyclotomic fields into ideal prime factors.</p> </blockquote> <p>Already in 1910, Hensel talked about "incontestable evidence" (actually it was something that Gundelfinger had heard from H.G. Grassmann) for the existence of a manuscript in which Kummer had claimed to have solved Fermat's Last Theorem, a rumour eventually dismissed by Edwards [Ed1, Ed2] and Neumann [Neu]. </p> </blockquote> http://mathoverflow.net/questions/33697/assistance-with-understanding-parent-child-relationships-in-pythagorean-triples/33726#33726 Answer by Bill Dubuque for Assistance with understanding parent/child relationships in Pythagorean Triples Bill Dubuque 2010-07-28T22:35:09Z 2010-07-31T02:53:43Z <p>A little-known chatoyant gem of elementary number theory is that the tree of Pythagorean triples has a beautiful geometric genesis in terms of reflections. This viewpoint should clarify the points that you raise. Below is a brief sketch excerpted from some emails I sent to John Conway and R. K. Guy, after noticing that they mention this topic (too) briefly in their "Book of Numbers". Namely, on p. 172 they write: <img src="http://i30.tinypic.com/156fm2g.jpg" alt="alt text"> <img src="http://i31.tinypic.com/25ptso3.jpg" alt="alt text"> . </p> <p>Below I explain briefly how to view this in terms of reflections and I mention some generalizations and closely related topics. I plan to discuss this at greater length in a future MO post when time permits.</p> <p>Consider the quadratic space $Z$ of the form $Q(x,y,z) = x^2 + y^2 - z^2$. It has Lorentzian inner product $(Q(x+y)-Q(x)-Q(y))/2$ given by $\; v \cdot u = v_1 u_1 + v_2 u_2 - v_3 u_3$. Recall that here one defines the </p> <p>$\quad$ <strong>reflection</strong> of $v$ in $u$</p> <p>$\quad\quad v \mapsto v - 2 \dfrac{v \cdot u}{u \cdot u} u \quad\quad$ Reflectivity is clear: $\; u \mapsto -u$, and $\; v \mapsto v$ if $\; v\perp u, \;$ i.e. $v\cdot u = 0$.</p> <p>With $\; v = (x,y,z)$ and $\; u = (1,1,1)$ of norm 1</p> <p>$\quad\quad (x,y,z)\; \mapsto (x,y,z) - 2 \dfrac{(x,y,z)\cdot(1,1,1)}{(1,1,1)\cdot(1,1,1)} (1,1,1)$</p> <p>$\quad\quad\quad\quad\quad\quad = (x,y,z) - 2 \; (x+y-z) \; (1,1,1)$</p> <p>$\quad\quad\quad\quad\quad\quad = (-x-2y+2z, \; -2x-y+2z, \; -2x-2y+3z)$</p> <p>This is the nontrivial reflection that effects the descent in the triples tree. Said simpler: if $x^2 + y^2 = z^2$ then $(x/z, y/z)$ is a rational point $P$ on the unit circle $C$. A simple calculation shows that the line through $P$ and $(1,1)$ intersects $C$ in a <em>smaller</em> rational point, given projectively via the above reflection, e.g. </p> <p>$\quad\quad (5,12,13) \mapsto (5,12,13) - 2 \; (5+12-13) \; (1,1,1) = (-3,4,5)$ </p> <p><img src="http://i27.tinypic.com/ohimfs.jpg" alt="alt text"><br> We ascend the tree by inverting this reflection, combined with trivial sign-changing reflections:</p> <p>$\quad\quad (-3,+4,5) \mapsto (-3,+4,5) - 2 \; (-3+4-5) \; (1,1,1) = ( 5,12,13)$</p> <p>$\quad\quad (-3,-4,5) \mapsto (-3,-4,5) - 2 \; (-3-4-5) \; (1,1,1) = (21,20,29)$</p> <p>$\quad\quad (+3,-4,5) \mapsto (+3,-4,5) - 2 \; (+3-4-5) \; (1,1,1) = (15,8,17)$</p> <p>Continuing in this manner one may reflectively generate the entire tree of primitive Pythagorean triples, e.g. the topmost edge of the triples tree corresponds to the ascending $C$-inscribed zigzag line $(-1,0), (3/5,4/5), (-3/5,4/5), (5/12,12/13), (-5/12,12/13), (7/25,24/25), (-7/25,24/25) \ldots$</p> <p>This technique easily generalizes to the form $x_1^2 + x_2^2 + \cdots + x_{n-1}^2 = x_n^2$ for $4 \le n \le 9$, but for $n \ge 10$ the Pythagorean n-tuples fall into at least $[(n+6)/8]$ distinct orbits under the automorphism group of the form - see <a href="http://www.jstor.org/stable/2048355" rel="nofollow">Cass &amp; Arpaia (1990) [1]</a> </p> <p>There are also generalizations to different shape forms that were first used by L. Aubry (Sphinx-Oedipe 7 (1912), 81-84) to give elementary proofs of the 3 &amp; 4 square theorem (see Appendix 3.2 p. 292 of Weil's: Number Theory an Approach Through History). These results show that if an integer is represented by a form rationally, then it must also be so integrally. In particular, the following class of forms is included $x^2+y^2, x^2 \pm 2y^2, x^2 \pm 3y^2, x^2+y^2+2z^2, x^2+y^2+z^2+t^2,\ldots$ More precisely, essentially the same proof as for Pythagorean triples shows</p> <p>THEOREM Suppose that the $n$-ary quadratic form $F(x)$ has integral coefficients and has no nontrivial zero in ${\mathbb Z}^n$, and suppose further that for any $x \in {\mathbb Q}^n$ there exists $y \in {\mathbb Z}^n$ such that $\; |F(x-y)| &lt; 1$. Then $F$ represents $m$ over $\mathbb Q$ $\iff$ $F$ represents $m$ over $\mathbb Z$, for all nonzero integers $m$.</p> <p>The condition $|F(x-y)| &lt; 1$ is closely connected to the Euclidean algorithm. In fact there is a function-field analog that employs the Euclidean algorithm which was independently rediscovered by Cassels in 1963. Namely, a polynomial is a sum of $n$ squares in $k(x)$ iff the same holds true in $k[x]$. Pfister immediately applied this to obtain a complete solution of the level problem for fields. Shortly thereafter he generalized Cassels result to arbitrary quadratic forms, founding the modern algebraic theory of quadratic forms ("Pfister forms").</p> <p>Aubry's results are, in fact, very special cases of general results of Wall, Vinberg, Scharlau et al. on <strong>reflective lattices</strong>, i.e. arithmetic groups of isometries generated by reflections in hyperplanes. Generally reflections generate the orthogonal group of Lorentzian quadratic forms in dim &lt; 10.</p> <p><a href="http://www.jstor.org/stable/2048355" rel="nofollow">1</a> Daniel Cass; Pasquale J. Arpaia<br> Matrix Generation of Pythagorean n-Tuples.<br> Proc. Amer. Math. Soc. 109, 1, 1990, 1-7.<br> <a href="http://www.jstor.org/stable/2048355" rel="nofollow">http://www.jstor.org/stable/2048355</a></p> http://mathoverflow.net/questions/33687/ideals-in-the-ring-of-single-variable-laurent-polynomials-with-integer-coefficien/33707#33707 Answer by Bill Dubuque for Ideals in the ring of single-variable Laurent polynomials with integer coefficients Bill Dubuque 2010-07-28T20:27:27Z 2010-07-28T20:41:20Z <p>It is fairly straighttforward to adapt standard Grobner basis techniques to such algebras, e.g. see the <a href="http://www.springerlink.com/content/qgbwymag351atn71/fulltext.pdf" rel="nofollow">paper [1]</a>. See also the <a href="http://www.springerlink.com/content/m704326653727425/fulltext.pdf" rel="nofollow">paper [0]</a> which applies such algorithms to the problem at hand.</p> <p><a href="http://www.springerlink.com/content/m704326653727425/fulltext.pdf" rel="nofollow">0</a> Jesus Gago-Vargas; Isabel Hartillo-Hermoso; Jose Marya Ucha-Enryquez<br> Algorithmic Invariants for Alexander Modules. LNCS 4194, 149-154<br> <a href="http://www.springerlink.com/content/m704326653727425/fulltext.pdf" rel="nofollow">http://www.springerlink.com/content/m704326653727425/fulltext.pdf</a></p> <p>Abstract. Let G be a group given by generators and relations. It is possible to compute a presentation matrix of a module over a ring through Fox's differential calculus. We show how to use Grobner bases as an algorithmic tool to compare the chains of elementary ideals defined by the matrix. We apply this technique to classical examples of groups and to compute the elementary ideals of Alexander matrix of knots up to 11 crossings with the same Alexander polynomial.</p> <p><a href="http://www.springerlink.com/content/qgbwymag351atn71/fulltext.pdf" rel="nofollow">1</a> Franz Pauer, Andreas Unterkircher.<br> Grobner Bases for Ideals in Laurent Polynomial Rings and their Application to Systems of Difference Equations.<br> AAECC 9, 271-291 (1999)<br> <a href="http://www.springerlink.com/content/qgbwymag351atn71/fulltext.pdf" rel="nofollow">http://www.springerlink.com/content/qgbwymag351atn71/fulltext.pdf</a></p> <p><em>Abstract.</em> We develop a basic theory of Grobner bases for ideals in the algebra of Laurent polynomials (and, more generally, in its monomial subalgebras). For this we have to generalize the notion of term order. The theory is applied to systems of linear partial difference equations (with constant coefficients) on ${\mathbb Z}^n$. Furthermore, we present a method to compute the intersection of an ideal in the algebra of Laurent polynomials with the subalgebra of all polynomials.</p> http://mathoverflow.net/questions/33526/deformations-and-the-dual-numbers/33531#33531 Answer by Bill Dubuque for Deformations and the dual numbers Bill Dubuque 2010-07-27T15:28:16Z 2010-07-28T15:18:37Z <p>Such first-order infinitesimal deformations allow one to compute the Zariski tangent space when good moduli space exists. You can find some nice motivational remarks in the first chapter of Hartshorne's <em>Deformation Theory</em>. See esp. the two paragraphs preceding the Exercises 1.1.</p> <p>You may find it helpful to first understand analogous applications of dual numbers in simpler contexts. Below is an excerpt from one of my old sci.math posts which may prove useful in this regard.</p> <blockquote> <p>What is the factor ring R[x]/(x^2) ?</p> </blockquote> <p>It is known as the algebra of dual numbers over R, for R a commutative ring. It and its higher order analogs R[x]/(x^n) prove useful when studying derivations. E.g. they permit easy transfer of properties of homomorphisms to derivations -- see section 8.15 in Jacobson, Basic Algebra II. They yield algebraic models of tangent spaces. </p> <p>They've been applied in many contexts, e.g. <a href="http://people.maths.ox.ac.uk/szendroi/defth.pdf" rel="nofollow">deformation theory [2]</a>, <a href="http://bt.pa.msu.edu/cgi-bin/display.pl?name=dahap" rel="nofollow">numerical analysis [3]</a> (along with <a href="http://bt.pa.msu.edu/pub/papers" rel="nofollow">Levi-Civita fields</a>), where they're viewed simply as truncated Taylor / power series, and in Synthetic Differential Geometry <a href="http://www.jstor.org/stable/20116534" rel="nofollow">(SDG) [1]</a>, another rigorization of inifinitesimals based on work of Lawvere and Kock. SDG employs nilpotent infinitesimals, unlike Abe Robinson's nonstandard analysis which has invertible infinitesimals, hence infinities. </p> <p><a href="http://www.jstor.org/stable/20116534" rel="nofollow">1</a> Bell, J. L. Infinitesimals. Synthese 75 (1988) #3, 285--315.<br> <a href="http://www.jstor.org/stable/20116534" rel="nofollow">http://www.jstor.org/stable/20116534</a></p> <p><a href="http://people.maths.ox.ac.uk/szendroi/defth.pdf" rel="nofollow">2</a> Szendroi, B. The unbearable lightness of deformation theory, a tutorial introduction.<br> <a href="http://people.maths.ox.ac.uk/szendroi/defth.pdf" rel="nofollow">http://people.maths.ox.ac.uk/szendroi/defth.pdf</a></p> <p><a href="http://bt.pa.msu.edu/cgi-bin/display.pl?name=dahap" rel="nofollow">3</a> M. Berz, Differential Algebraic Techniques,<br> in "Handbook of Accelerator Physics and Engineering, M. Tigner, A.Chao (Eds.)" (World Scientific, 1998)<br> <a href="http://bt.pa.msu.edu/cgi-bin/display.pl?name=dahape" rel="nofollow">http://bt.pa.msu.edu/cgi-bin/display.pl?name=dahape</a><br> <a href="http://bt.pa.msu.edu/NA/" rel="nofollow">http://bt.pa.msu.edu/NA/</a><br> <a href="http://bt.pa.msu.edu/pub/papers/" rel="nofollow">http://bt.pa.msu.edu/pub/papers/</a> </p> http://mathoverflow.net/questions/31007/what-are-the-units-in-the-ring-of-laurent-polynomials/33550#33550 Answer by Bill Dubuque for What are the units in the ring of Laurent polynomials? Bill Dubuque 2010-07-27T17:40:32Z 2010-07-27T17:40:32Z <p>You can find a more general result in the <a href="http://dx.doi.org/10.1007/s10440-008-9370-8" rel="nofollow">paper [1]</a>, which determines the units and nilpotents in arbitrary group rings $\rm R[G]$ where $\rm G$ is a unique-product group - which includes ordered groups. As the author remarks, his note was prompted by an earlier <a href="http://homepage.uibk.ac.at/~c70202/jordan/archive/remarks/remarks.pdf" rel="nofollow">paper [2]</a> which explicitly treats the Laurent case.</p> <p><a href="http://dx.doi.org/10.1007/s10440-008-9370-8" rel="nofollow">1</a> Erhard Neher. Invertible and Nilpotent Elements in the Group Algebra of a Unique Product Group<br> Acta Appl Math (2009) 108: 135-139<br> <a href="http://dx.doi.org/10.1007/s10440-008-9370-8" rel="nofollow">http://dx.doi.org/10.1007/s10440-008-9370-8</a><br> <a href="http://homepage.uibk.ac.at/~c70202/jordan/archive/note/note.pdf" rel="nofollow">http://homepage.uibk.ac.at/~c70202/jordan/archive/note/note.pdf</a></p> <p><a href="http://homepage.uibk.ac.at/~c70202/jordan/archive/remarks/remarks.pdf" rel="nofollow">2</a> Ottmar Loos. Remarks on Holger P. Petersson's "Idempotent 2-by-2 matrices" <a href="http://homepage.uibk.ac.at/~c70202/jordan/archive/remarks/remarks.pdf" rel="nofollow">http://homepage.uibk.ac.at/~c70202/jordan/archive/remarks/remarks.pdf</a></p> http://mathoverflow.net/questions/32968/slick-ways-to-make-annoying-verifications/33435#33435 Answer by Bill Dubuque for Slick ways to make annoying verifications Bill Dubuque 2010-07-26T19:35:36Z 2010-07-26T19:35:36Z <p>Proofs exploiting universality often provide nice examples of slick ways to avoid annoying special cases. For example, the matrix identities below have trivial algebraic proofs by proceeding "generically", i.e. let the matrix entries $a_{ij}, b_{ij}$ be indeterminates and perform the proof over the polynomial ring $\mathbb Z[a_{ij}, b_{ij}]$. </p> <p>$\rm\quad\; det(I-AB) = det(I-BA)\;\:$ by taking $\;\rm det\;$ of $\;\;\rm (I-AB)\;A = A\;(I-BA)\;$ then canceling $\;\rm det \:A$</p> <p>$\rm\quad\quad det(adj \:A) = (det \:A)^{n-1}\quad$ by taking $\;\rm det\;$ of $\;\rm\quad A\;(adj\: A) = (det\: A) \;I\quad\;\;$ then canceling $\;\rm det \:A$</p> <p>Contrast these absolutely trivial <em>algebraic</em> proofs with the more complex and more frequently presented <em>topological</em> proofs via density arguments. See my <a href="http://mathoverflow.net/questions/32133/expressing-adja-as-a-polynomial-in-a/32343#32343" rel="nofollow">post [1]</a> and its comments for some further discussion.</p> http://mathoverflow.net/questions/32133/expressing-adja-as-a-polynomial-in-a/32343#32343 Answer by Bill Dubuque for Expressing adj(A) as a polynomial in A? Bill Dubuque 2010-07-18T06:42:02Z 2010-07-26T19:34:42Z <p><strong>HINT</strong> $\;$ Work "<strong>generically</strong>", i.e. let the entries $\;\rm a_{i,j}$ of $\rm A\;$ be <strong>indeterminates</strong> and work in the matrix ring $\rm M = M_n(R)\;$ over $\;\rm R = {\mathbb Z}[a_{i,j}\:]. \;$ We wish to prove $\rm B = C$ from $\rm d\: B = d\: C$ for $\rm d = det\: A \in R, \;\; B,C \in M.$ But this is equivalent to $\rm d\: b_{i,j} = d\: c_{i,j}$ in the <em>domain</em> $\rm R = {\mathbb Z}[a_{i,j}\:]$ where $\;\rm d = det\: A \ne 0$, so $\rm d$ is cancelable, yielding $\;\rm b_{i,j} = c_{i,j}\;$ hence $\rm B = C$. This identity remains true over every commutative ring $\rm S$ since, by the <strong>universality</strong> of polynomial rings, there exists an eval homomorphism that evaluates $\;\rm a_{i,j}\;$ at any $\;\rm s_{i,j}\in S$. </p> <p>Notice that the crucial insight is that $\;\rm b_{i,j}\:, \; c_{i,j}\:,\; d\;$ have <em>polynomial form</em> in $\;\rm a_{i,j}\:$, i.e. they are elts of the polynomial ring $\;\rm R = {\mathbb Z}[a_{i,j}\:] = {\mathbb Z}[a_{1,1},\cdots,a_{n,n}\:]$ which, being a <em>domain</em>, enjoys cancelation of elts $\ne 0$. Working generically allows us to cancel $\rm d$ and deduce the identity <em>before</em> any evaluation where $\rm d\mapsto 0.$ </p> <p>Such proofs by way of <strong>universal polynomial identities</strong> emphasize the power of the abstraction of a <em>formal</em> polynomial (vs. polynomial <em>function</em>). Alas, many algebra textbooks fail to explicitly emphasize this universal viewpoint. As a result, many students cannot easily resist the obvious topological temptations and instead derive hairier proofs employing density arguments (e.g see elswhere in this thread).</p> <p>Analogously, the same <em>generic</em> method of proof works for many other polynomial identities, e.g.</p> <p>$\rm\quad\; det(I-AB) = det(I-BA)\;\:$ by taking $\;\rm det\;$ of $\;\;\rm (I-AB)\;A = A\;(I-BA)\;$ then canceling $\;\rm det \:A$</p> <p>$\rm\quad\quad det(adj \:A) = (det \:A)^{n-1}\quad$ by taking $\;\rm det\;$ of $\;\rm\quad A\;(adj\: A) = (det\: A) \;I\quad\;\;$ then canceling $\;\rm det \:A$</p> <p>Now, for our pièce de <em>résistance of topology</em>, we derive the polynomial derivative purely formally.</p> <p>For $\rm f(x) \in R[x]$ define $\rm D f(x) = f_0(x,x)$ where $\rm f_0(x,y) = \frac{f(x)-f(y)}{x-y}.$ Note that the existence and uniqueness of this derivative follows from the Factor Theorem, i.e. $\;\rm x-y \; | \; f(x)-f(y)\;$ in $\;\rm R[x,y],\;$ and, from the cancelation law $\;\rm (x-y) g = (x-y) h \implies g = h$ for $\rm g,h \in R[x,y].$ It's clear this agrees on polynomials with the analytic derivative definition since it is linear and it takes the same value on the basis monomials $\rm x^n$. <em>Resisting</em> limits again, we get the product rule rule for derivatives from the trivial difference product rule </p> <p>$$\rm f(x)g(x) - f(y)g(y)\; = \;(f(x)-f(y)) g(x) + f(y) (g(x)-g(y))$$</p> <p>$\quad\quad\quad\quad\rm\quad\quad\quad \Longrightarrow \quad\quad\quad\quad\quad\; D(fg)\quad = \quad (Df) \; g \; + \; f \; (Dg)$</p> <p>by canceling $\rm x-y$ in the first equation, then evaluating at $\rm y = x$, i.e. specialize the difference "quotient" from the product rule for differences. Here the <em>formal</em> cancelation of the factor $\;\rm x-y\;$ <em>before</em> evaluation at $\;\rm y = x\;$ is precisely analogous to the <em>formal</em> cancelation of $\;\rm det \:A\;$ in all of the examples given above.</p> http://mathoverflow.net/questions/33265/how-do-we-know-that-fermat-wrote-his-famous-note-in-1637/33328#33328 Answer by Bill Dubuque for How do we know that Fermat wrote his famous note in 1637? Bill Dubuque 2010-07-25T21:27:19Z 2010-07-25T21:27:19Z <p>Below is Andre Weil's opinion on this general topic, from his historical treatise <em>Number Theory</em>, p.104. </p> <blockquote> <p>As we have observed in Chap. I, S.X, the most significant problems in Diophantus are concerned with curves of genus 0 or 1. With Fermat this turns into an almost exclusive concentration on such curves. Only on one ill-fated occasion did Fermat ever mention a curve of higher genus, and there can hardly remain any doubt that this was due to some misapprehension on his part, even though, by a curious twist of fate, his reputation in the eyes of the ignorant came to rest chiefly upon it. By this we refer of course to the incautious words "et generaliter nullam in infinitum potestatem" in his statement of "Fermat's last theorem" as it came to be vulgarly called: "No cube can be split into two cubes, nor any biquadrate into two biquadrates, nor generally any power beyond the second into two of the same kind" is what he wrote into the margin of an early section of his Diophantus (Fe.I.291, Obs.II), adding that he had discovered a truly remarkable proof for this "which this margin is too narrow to hold". How could he have guessed that he was writing for eternity? We know his proof for biquadrates (cf. above, S.X); he may well have constructed a proof for cubes, similar to the one which Euler discovered in 1753 (cf. infra, S.XVI); he frequently repeated those two statements (e.g. Fe.II.65,376,433), but never the more general one. For a brief moment perhaps, and perhaps in his younger days (cf. above, S.III), he must have deluded himself into thinking that he had the principle of a general proof; what he had in mind on that day can never be known.</p> </blockquote> http://mathoverflow.net/questions/33304/mareys-problem-generating-all-prime-numbers-in-n-1-n-2/33324#33324 Answer by Bill Dubuque for Marey's problem: Generating all prime numbers in $[n_1,n_2]$ Bill Dubuque 2010-07-25T21:12:45Z 2010-07-25T21:12:45Z <p>Unless you have very specialized needs, you can probably apply standard sieving techniques. Here are a couple references. First, hot off the press is Kjell Wooding's Calgary <a href="http://math.ucalgary.ca/~hwilliam/files/wooding10thesis.pdf" rel="nofollow">thesis [1]</a>, which includes a good overview of both software and hardware sieving techniques (e.g. FPGA-based sieves, Calgary's scalable sieve, etc). Its bibliography should prove useful as an entry point into the literature. Also, <a href="http://www.ams.org/journals/mcom/1953-07-041/S0025-5718-1953-0052876-7/S0025-5718-1953-0052876-7.pdf" rel="nofollow">here [2]</a> is a charming classic paper by D.H. Lehmer - one of the pioneers of computational number theory. It gives a nice succinct introduction to sieving in general. Such methods have a long venerable history that stimulated much development in both number theory and computer science, e.g. google "Lehmer sieve" and you'll discover many ingenious machines devised to carry out number theoretical computations long before the dawn of the modern digital computer, e.g. via bicycles chains, photoelectric devices, etc.</p> <p><a href="http://math.ucalgary.ca/~hwilliam/files/wooding10thesis.pdf" rel="nofollow">1</a> Wooding, Kjell. The Sieve Problem in One- and Two-Dimensions.<br> PhD Thesis. Calgary, Alberta. April, 2010</p> <p><a href="http://www.ams.org/journals/mcom/1953-07-041/S0025-5718-1953-0052876-7/S0025-5718-1953-0052876-7.pdf" rel="nofollow">2</a> Lehmer, D.H. The sieve problem for all-purpose computers, MTAC, v. 7 1953, p. 6-14</p> http://mathoverflow.net/questions/29590/a-condition-that-implies-commutativity/33234#33234 Answer by Bill Dubuque for A condition that implies commutativity Bill Dubuque 2010-07-24T21:38:30Z 2010-07-24T22:09:43Z <p>For fixed $n \in \mathbb{N}$, Birkhoff's completeness theorem implies that such a proof must exist in the first-order equational theory of rings - as I mentioned here in a recent <a href="http://mathoverflow.net/questions/30220/abstract-thought-vs-calculation/30273#30273" rel="nofollow">post</a>. Many years ago Stan Burris told me that John Lawrence discovered such an equational proof that works uniformly for all $n$ (possibly also for Jacobson's form $x^{n(x)} = x$). I don't know if the proof is published yet, but some clues as to how it may proceed might be gleaned from their earlier joint <a href="http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PAPERS/fields3.pdf" rel="nofollow">work [1]</a></p> <p><a href="http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PAPERS/fields3.pdf" rel="nofollow">1</a> S. Burris and J. Lawrence, Term rewrite rules for finite fields.<br> International J. Algebra and Computation 1 (1991), 353-369. <a href="http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PAPERS/fields3.pdf" rel="nofollow">http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PAPERS/fields3.pdf</a> </p> http://mathoverflow.net/questions/32262/representation-of-rings/32305#32305 Answer by Bill Dubuque for Representation of rings Bill Dubuque 2010-07-17T18:58:21Z 2010-07-17T22:18:19Z <p>Generally it's difficult to characterize rings isomorphic to an endomorphism ring of an abelian group. Interest in such problems was sparked by a problem given by Fuchs in his widely-read monograph <em>Abelian Groups</em>, cf. the excerpt below from the introduction to the <a href="http://dx.doi.org/10.1016/j.jalgebra.2006.01.036" rel="nofollow">paper [1]</a></p> <blockquote> <p>The notion of an E-ring goes back to a seminal paper of Schultz [20] written in response to Problem 45 in the well-known book Abelian Groups' by Laszlo Fuchs [11]. In this paper Schultz distinguished between two possibly different approaches, the first we will continue to call an E-ring, while the second we shall refer to as a generalized E-ring. Thus a ring R is said to be an E-ring if R is isomorphic to the endomorphism ring of its underlying additive group, R+, via the mapping sending an element r $\in$ R to the endomorphism given by left multiplication by r, whilst R is a generalized E-ring if some isomorphism, not necessarily left multiplication, exists between R and its endomorphism ring End(R+). Since right multiplication is always an endomorphism, it is not difficult to see that E-rings are necessarily commutative. The existence of a non-commutative generalized E-ring has recently been established [15], and so it follows that the class of generalized E-rings is strictly larger than the class of E-rings.</p> <p>Since Schultz's original paper there has been a great deal of interest in E-rings and some natural generalizations, see e.g. [1,2,4,6,8-10,17,19,21]. A notable feature of much of this recent work has been the use of so-called realization theorems, whereby a cotorsion-free ring is realized, using combinatorial ideas derived from Shelah's Black Box - see e.g. [7] for details of this technique - as the endomorphism ring of an Abelian group. This present work arose from an observation of the second author in response to a question from the first about the existence of generalized E-algebras over the ring $J_p$ of p-adic integers; see [16] for further details. A natural question which arises, is to what extent is it necessary for a ring to be cotorsion-free in order to be a generalized E-ring and the principal objective of this work is to characterize generalized E-rings modulo cotorsion-free groups.' The characterization is quite elementary but seems to have been overlooked heretofore. It should be noted that Bowshell and Schultz showed in [2] that a reduced cotorsion E-ring has the form $\prod_{p \in U} {\mathbb Z}(p^{k_p}) \oplus \prod_{p\in V} J_p$ where $U,V$ are disjoint sets of primes.</p> </blockquote> <p><a href="http://dx.doi.org/10.1016/j.jalgebra.2006.01.036" rel="nofollow">1</a> R. Gobel, B. Goldsmith.<br> Classifying E-algebras over Dedekind domains<br> Jnl. Algebra, Vol. 306, 2006, 566-575</p> http://mathoverflow.net/questions/32011/direct-proof-of-irrationality/32017#32017 Answer by Bill Dubuque for Direct proof of irrationality? Bill Dubuque 2010-07-15T15:34:53Z 2010-07-16T17:22:27Z <p>Below is a simple direct proof that I found as a teenager:</p> <p><strong>THEOREM</strong> $\;\rm r = \sqrt{n}\;$ is integral if rational, for $\;\rm n\in\mathbb{N}$. </p> <p>Proof: $\;\rm r = a/b,\;\; {\text gcd}(a,b) = 1 \implies ad-bc = 1\;$ for some $\rm c,d \in \mathbb{Z}$, by Bezout </p> <p>so: $\;\rm 0 = (a-br) (c+dr) = ac-bdn + r \implies r \in \mathbb{Z} \quad\square$ </p> <p>Nowadays my favorite proof is the 1-line gem using Dedekind's <strong>conductor ideal</strong> - which, as I explained at length <a href="http://mathoverflow.net/questions/30220/abstract-thought-vs-calculation/30313#30313" rel="nofollow">elsewhere</a>, beautifully encapsulates the descent in ad-hoc "elementary" irrationality proofs.</p> http://mathoverflow.net/questions/32009/factoring-and-solving-trinomials/32049#32049 Answer by Bill Dubuque for Factoring and solving trinomials Bill Dubuque 2010-07-15T18:24:25Z 2010-07-16T14:21:11Z <p>One can of course apply general algorithms for irreducibility testing and factorization, so I presume you are asking if there is something more efficient or more explicit that can be said in the case of trinomials. Except for special cases I don't believe that is the case. </p> <p>While it is known that every binomial in $\mathbb{Q}(x)$ must have an irreducible factor that is either binomial or trinomial, no analogous bound is known in the trinomial case. It is at least 8 terms due to the <a href="http://journals.cambridge.org/action/displayFulltext?type=1&amp;fid=5054152&amp;jid=GMJ&amp;volumeId=34&amp;issueId=01&amp;aid=5054144" rel="nofollow">known [1]</a> example</p> <p>$$f(x)f(-x) = - x^{14} - 27180501562500 x^2 + 1244325625000000$$</p> <p>for $f(x) = x^7 + 20 x^6 + 200 x^5 + 2450 x^4 + 29000 x^3 + 545000 x^2 + 8101250 x + 35275000$</p> <p><a href="http://journals.cambridge.org/action/displayFulltext?type=1&amp;fid=5054152&amp;jid=GMJ&amp;volumeId=34&amp;issueId=01&amp;aid=5054144" rel="nofollow">1</a> Choudhry and A. Schinzel (1992).<br> On the number of terms in the irreducible factors of a polynomial over $\mathbb Q$.<br> Glasgow Mathematical Journal (1992), 34, 11-15.</p> <p>To dig deeper I suggest starting with the work of Schinzel - who has studied these and related factorization problems intensively for almost half a century, e.g. see</p> <blockquote> <p>MR1254093 (95d:11146) 11R09 (12E05 12E10)<br> Schinzel, Andrzej. On reducible trinomials.<br> Dissertationes Math. (Rozprawy Mat.) 329 (1993), 83 pp. </p> <p>Let $K$ be a field. It is well known that a binomial $x^n+a\in K[x]$ is reducible iff it has the form $x^{pk}-b^p$ ($p$ prime) or $x^{4k}+4b^4$. In this treatise the reducibility of trinomials $x^n+ax^m+b$ $(a,b\neq 0)$ is investigated. It turns out that the situation is very complicated. A satisfactory answer is obtained if $K$ is a rational function field. For algebraic function fields in one variable and for algebraic number fields, less complete results are proved. It is assumed throughout that the characteristic of $K$ does not divide $mn(n-m)$. </p> <p>It is easy to find trinomials with linear or quadratic factors. Table 1 of this paper provides additional families of reducible trinomials if $(n,m)$ belongs to a list of 12 pairs, the largest being $(15,5)$. Perhaps the simplest example is $$x^6+4(v+1)x^2-v^2=(x^3+2x^2+2x-v)(x^3-2x^2+2x+v).$$ Every reducible trinomial $f(x)=x^n+ax^m+b$ gives rise to additional examples by considering $u^nf(x^l/u)$ (with $u\in K^\times$ and $l\geq 1$) or $x^nf(1/x)/b$. Theorem 1 essentially states that every reducible trinomial arises in this manner from the examples indicated before if $K$ is a rational function field. (More precisely, it is assumed that $a^{-n}b^{n-m}$ is not a constant.) Table 2 lists $7$ families of reducible trinomials $x^n+A(v,w)x^m+B(v,w)$ with $(v,w)\in E(K)$, where $A$, $B$ are polynomials over $\mathbb Z$ and $E$ is an elliptic curve defined by an equation $z^2=C(w)$, where $C$ is a monic polynomial over $\mathbb Z$. The polynomials $A,B$ and the corresponding factorizations of the trinomials are too complicated to be included in this review. (For the largest pair $(n,m)=(21,7)$ the corresponding $A$ fills 10 lines in the paper.) In Theorem 2 it is assumed that $K$ is a finite extension of a rational function field $F(t)$ such that $\overline FK$ has genus $g>0$ and $a^{-n}b^{n-m}\notin \overline{F}$. If $g=1$ then there are no additional examples of reducible trinomials. If $g>1$ then essentially new examples with $n&lt;24g$ may exist. Theorem 3 reduces the case where $K$ is a finite separable extension of $F(t)$ and $a^{-n}b^{n-m}\in\overline F$ to studying reducibility over $K\cap\overline F$. If $K$ is an algebraic number field then for fixed $n$, $m$ a finite number of essentially new examples of reducible trinomials $x^n+ax^m+b$ may exist (Theorem 6). The author conjectures that for every $K$ there is only a finite number of these sporadic trinomials''. If the conjecture holds then there exists a constant $c(K)$ such that every trinomial over $K$ has an irreducible factor with at most $c(K)$ nonzero coefficients (Consequence 2). Table 5 contains all 52 sporadic trinomials over $\mathbb Q$ known to the author. Their degrees lie in the range from $8$ to $52$. The rest of the paper is devoted to studying the reducibility of $ax^n+bx^m+c\in\mathbb Z[x]$. Theorem 9 (refining a result of Nagell) derives necessary conditions, which in the case $(m,n)=1$ yield an explicit bound for $b$ in terms of $a,c,m,n$. For every positive integer $d$ there exist only finitely many $n,m,b$ with $n/(m,n)>d$ and $|b|>2$ such that $x^n+bx^m\pm 1$ has a factor of degree $d$; and these can be effectively computed. Theorem 10 derives necessary conditions from the existence of a factor (of $ax^n+bx^m+c)$ of given degree $d$. These imply that there exists $n_0(d)$ such that $x^n+bx^m+1$ is irreducible if $n\geq n_0(d)$, $n\neq 2m$, $|b|>2$. By Theorem 8, for every $n$ there exist only finitely many reducible trinomials $x^n+bx^m+1$ with $n\neq 2m$. </p> <p>The proof of Theorem 10 does not depend on the other results of the paper. The same applies to Theorem 9. All other theorems except for Theorem 3 are based on lower estimates for the genus of certain function fields. These estimates show that the existence of a factor of degree $k$ of $x^n+ax^m+b\in K[x]$ imposes severe restrictions on $k,m,n,a,b$ provided $K$ is a function field. The remaining cases are treated in a long series of lemmas applying to every field $K$ whose characteristic does not divide $mn(n-m)$. In several cases the proofs require extensive manipulations (with polynomials in several variables) which were performed by means of computer algebra systems. Faltings' theorem (solving Mordell's conjecture) is invoked in the proof of Theorem 6 (dealing with number fields). Theorems 7 and 8 (concerning $ax^n+bx^m+c\in\mathbb Z[x])$ are proved by using the corresponding theorems for rational function fields together with a lemma which may be viewed as a refinement of Hilbert's irreducibility theorem. The proof of this lemma is based on Siegel's theorem (on integral points of curves of positive genus) and on a result of Maillet (1919) dealing with rational functions over $\mathbb Q$ taking infinitely many integral values at rational points.</p> <p>{Reviewer's remarks: In Theorem 2 the term $u^{\nu-\mu}$ in the expression for $B$ has to be replaced by $u^\nu$. The proof of Lemma 27 employs Lemma 2(c) although this lemma only applies to separable extensions. In order to prove Lemma 49 one has to know that every finite separable extension $L$ of $K(t)$ with $L\subseteq \overline K(t)$ is contained in $K'(t)$ for some separable extension $K'$ of $K$. (One can in fact prove that $L=K'(t)$ for suitable $K'$. This need not be true for inseparable $L$.) The proof of Theorem 6 is apparently based on the incorrect assumption that a divisor $P$ of a function field $L=K(t,y)$ has degree $1$ or is ramified with respect to $K(t)$ if $t$ and $y$ are congruent to elements of $K\bmod P$.} </p> <p>REVISED (1995)</p> <p>Reviewed by G. Turnwald</p> </blockquote> http://mathoverflow.net/questions/30220/abstract-thought-vs-calculation/30273#30273 Answer by Bill Dubuque for Abstract Thought vs Calculation Bill Dubuque 2010-07-02T05:54:24Z 2010-07-15T17:11:59Z <p>One striking example that comes to mind is Nathan Jacobson's proof that rings satisfying the identity $X^m = X$ are commutative. This is model-theoretic and proceeds by a certain type of factorization which reduces the problem to the (subdirectly) irreducible factors of the variety. These turn out to be certain finite fields, which are commutative, as desired. By (Birkhoff) completeness there must also exist a purely equational proof (in the language of rings) but even for small $m$ this is notoriously difficult, e.g. $m = 3$ is often posed as a difficult <a href="http://groups-beta.google.com/group/sci.math/msg/9b884af731351f10" rel="nofollow">exercise</a>. It's only recently that such a general non-model-theoretic equational proof was discovered by John Lawrence (as Stan Burris informed me). I don't know if it has been published yet, but see their earlier <a href="http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PAPERS/fields3.pdf" rel="nofollow">work [1]</a></p> <p>So here, by "higher-order" conceptual structural reasoning, one is able to escape the confines of first-order equational logic and give a more conceptual proof than the brute-force equational proofs - arguments so devoid of intuition that they can been discovered by an automatic theorem prover.</p> <p><a href="http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PAPERS/fields3.pdf" rel="nofollow">1</a> S. Burris and J. Lawrence, Term rewrite rules for finite fields.<br> International J. Algebra and Computation 1 (1991), 353-369. <a href="http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PAPERS/fields3.pdf" rel="nofollow">http://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PAPERS/fields3.pdf</a> </p> http://mathoverflow.net/questions/30156/demystifying-complex-numbers/30322#30322 Answer by Bill Dubuque for Demystifying complex numbers Bill Dubuque 2010-07-02T16:28:15Z 2010-07-15T04:06:50Z <p>One cannot over-emphasize that passing to complex numbers often permits a great simplification by <em>linearizing</em> what would otherwise be more complex <em>nonlinear</em> phenomena. One example familiar to any calculus student is the fact that integration of rational functions is much simpler over $\mathbb C$ (vs. $\mathbb R$) since partial fraction decompositions involve at most linear (vs quadratic) polynomials in the denominator. Similarly one reduces higher-order constant coefficient differential and difference equations to linear (first-order) equations by factoring the linear operators over $\mathbb C$. More generally one might argue that such <strong>simplification by linearization</strong> was at the heart of the development of abstract algebra. Namely, Dedekind, by abstracting out the essential linear structures (ideals and modules) in number theory, greatly simplified the prior nonlinear theory based on quadratic forms. This enabled him to exploit to the hilt the power of linear algebra. Examples abound of the revolutionary power that this brought to number theory and algebra - e.g. for one little-known gem see my recent <a href="http://mathoverflow.net/questions/30220/abstract-thought-vs-calculation/30313#30313" rel="nofollow">post</a> explaining how Dedekind's notion of <em>conductor ideal</em> beautifully encapsulates the essence of elementary irrationality proofs of n'th roots.</p> http://mathoverflow.net/questions/15703/newton-and-newton-polygon/31894#31894 Answer by Bill Dubuque for Newton and Newton polygon Bill Dubuque 2010-07-14T19:16:23Z 2010-07-15T03:37:22Z <p>If memory serves correct the history of Newton's polygon and Puiseaux series has some subtleties, so be a bit wary of secondary historical sources. Histories of mathematics are bursting at the seams with romanticized legends, so it is always best to consult primary sources if you wish to know the real history. The following note from Chrystal's <em>Algebra</em> may serve as a helpful entry into the primary literature.</p> <blockquote> <p><strong>Historical Note</strong>. - As has already been remarked, the fundamental idea of the reversion of series, and of the expansion of the roots of algebraical or other equations in power-series originated with Newton. His famous" Parallelogram" is first mentioned in the second letter to Oldenburg; but is more fully explained in the <em>Geometria Analytica</em> (see Horsley's edition of Newton's Works, t. i., p. 398). The method was well understood by Newton's followers, Stirling and Taylor; but seems to have been lost sight of in England after their time. It was much used (in a modified form of De Gua's) by Cramer in his well-known <em>Analyse dea Lignes Courbea Algebriques</em> (1750). Lagrange gave a complete analytical form to Newton's method in his "Memoire sur l'Usage des Fractions Continues," <em>Nouv. Mem. d. l'Ac. roy. d. Sciences d. Berlin</em> (1776). (See <em>OEuvres de Lagrange</em>, t. iv.) </p> <p>Notwithstanding its great utility, the method was everywhere all but forgotten in the early part of this century, as has been pointed out by De Morgan in an interesting account of it given in the <em>Cambridge Philosophical Transactions</em>, vol.ix. (1855). </p> <p>The idea of demonstrating, a priori, the possibility of expansions such as the reversion-formulae of S.18 originated with Cauchy; and to him, in effect, are due the methods employed in SS.18 and 19. See his memoirs on the Integration of Partial Differential Equations, on the Calculus of Limits, and on the Nature and Properties of the Roots of an Equation which contains a Variable Parameter, <em>Exercices d'Analyse et de Physique Mathematique</em>, t. i. (1840), p. 327; t. ii. (1841), pp. 41, 109. The form of the demonstrations given in SS. 18, 19 has been borrowed partly from Thomae, <em>El. Theorie der Analytischen Functionen einer Complexen Veranderlichen</em> (Halle, 1880), p. 107; partly from Stolz, <em>Allgemeine Arithmetik</em>, I. Th. (Leipzig, 1885), p. 296. </p> <p>The Parallelogram of Newton was used for the theoretical purpose of establishing the expansibility of the branches of an algebraic function by Puiseaux in his Classical Memoir on the Algebraic Functions (<em>Liouv. Math. Jour</em>., 1850). Puiseaux and Briot and Bouquet (<em>Theorie des Fonctions Elliptiques</em> (1875), p. 19) use Cauchy's Theorem regarding the number of the roots of an algebraic equation in a given contour; and thus infer the continuity of the roots. The demonstration given in S.21 depends upon the proof, a priori, of the possibility of an expansion in a power-series; and in this respect follows the original idea of Newton. </p> <p>The reader who desires to pursue the subject further may consult Durege, <em>Elemente der Theorie der Functionen einer Complexen Veranderlichen Grosse</em>, for a good introduction to this great branch of modern function-theory. </p> <p>The applications are very numerous, for example, to the finding of curvatures and curves of closest contact, and to curve-tracing generally. A number of beautiful examples will be found in that much-to.be-recommended text-book, Frost's <em>Curve Tracing</em>. -- G. Chrystal: <em>Algebra, Part II, p.370</em></p> </blockquote> http://mathoverflow.net/questions/31595/how-would-you-solve-this-tantalizing-halmos-problem How would you solve this tantalizing Halmos problem? Bill Dubuque 2010-07-12T18:25:37Z 2010-07-14T20:03:00Z <p>1-ab invertible => 1-ba invertible has a slick power series "proof" as below, where Halmos asks for an explanation of why this tantalizing derivation succeeds. Do you know one?</p> <hr> <p><em>Geometric series.</em> In a not necessarily commutative ring with unit (e.g., in the set of all 3 x 3 square matrices with real entries), if $1 - ab$ is invertible, then $1 - ba$ is invertible. However plausible this may seem, few people can see their way to a proof immediately; the most revealing approach belongs to a different and distant subject.</p> <p>Every student knows that $1 - x^2 = (1 + x) (1 - x),$ and some even know that $1 - x^3 =(1+x +x^2) (1 - x).$ The generalization $1 - x^{n+1} = (1 + x + \cdots + x^n) (1 - x)$ is not far away. Divide by $1 - x$ and let $n$ tend to infinity; if |x| &lt; 1, then $x^{n+1}$ tends to $0$, and the conclusion is that $\frac{1}{1 - x} = 1 + x + x^2 + \cdots$. This simple classical argument begins with easy algebra, but the meat of the matter is analysis: numbers, absolute values, inequalities, and convergence are needed not only for the proof but even for the final equation to make sense.</p> <p>In the general ring theory question there are no numbers, no absolute values, no inequalities, and no limits - those concepts are totally inappropriate and cannot be brought to bear. Nevertheless an impressive-sounding classical phrase, "the principle of permanence of functional form", comes to the rescue and yields an analytically inspired proof in pure algebra. The idea is to pretend that $\frac{1}{1 - ba}$ can be expanded in a geometric series (which is utter nonsense), so that $(1 - ba)^{-1} = 1 + ba + baba + bababa + \cdots$ It follows (it doesn't really, but it's fun to keep pretending) that $(1 - ba)^{-1} = 1 + b (1 + ab + abab + ababab + \cdots) a.$ and, after one more application of the geometric series pretense, this yields $(1 -ba)^{-1} = 1 + b (1 - ab)^{-1} a.$</p> <p>Now stop the pretense and verify that, despite its unlawful derivation, the formula works. If, that is, $c = (1 - ab)^{-1}$, so that $(1 - ab)c = c(1 - ab) = 1,$ then $1 + bca$ is the inverse of $1 - ba.$ Once the statement is put this way, its proof becomes a matter of (perfectly legal) mechanical computation.</p> <p>Why does it all this work? What goes on here? Why does it seem that the formula for the sum of an infinite geometric series is true even for an abstract ring in which convergence is meaningless? What general truth does the formula embody? I don't know the answer, but I note that the formula is applicable in other situations where it ought not to be, and I wonder whether it deserves to be called one of the (computational) elements of mathematics. -- P. R. Halmos [1]</p> <p>[1] Halmos, P.R. Does mathematics have elements?<br> Math. Intelligencer 3 (1980/81), no. 4, 147-153<br> <a href="http://dx.doi.org/10.1007/BF03022973" rel="nofollow">http://dx.doi.org/10.1007/BF03022973</a></p> http://mathoverflow.net/questions/31681/euclidean-function-at-0/31704#31704 Answer by Bill Dubuque for Euclidean Function at 0 Bill Dubuque 2010-07-13T14:03:57Z 2010-07-13T14:09:43Z <p>You'll find your answer and much more in the little-known paper [1] which surveys all of the dozen known ways of axiomatizing Euclidean rings (including those of Nagata and Samuel), and explores in-depth all of their logical interrelations. It's a convenient reference to have at hand when you're comparing texts which use (seemingly) different definitions of Euclidean rings / domains.</p> <p>[1] Euclidean Rings. A. G. Agargun, C. R. Fletcher<br> Tr. J. of Mathematics, 19, 1995, 291 - 299.<br> <a href="http://journals.tubitak.gov.tr/math/issues/mat-95-19-3/pp-291-299.pdf" rel="nofollow">http://journals.tubitak.gov.tr/math/issues/mat-95-19-3/pp-291-299.pdf</a> </p> http://mathoverflow.net/questions/31635/theorems-which-say-such-and-such-method-cannot-possibly-prove-flt/31651#31651 Answer by Bill Dubuque for Theorems which say "such and such method cannot possibly prove FLT" Bill Dubuque 2010-07-13T02:30:07Z 2010-07-13T02:30:07Z <p>You may find it of interest to peruse the very readable introduction in the following paper. It explains briefly what's known about obstructions to a local-global principle for the generalized Fermat equation.</p> <p>H. Darmon, A. Granville,<br> On the equations $z^m = F(x, y)$ and $A x^p + B y^q = C z^r$.<br> Bull. London Math. Soc. 27 (1995), 513–543.<br> <a href="http://blms.oxfordjournals.org/cgi/content/short/27/6/513" rel="nofollow">http://blms.oxfordjournals.org/cgi/content/short/27/6/513</a></p> http://mathoverflow.net/questions/26549/is-there-much-difference-between-kroneckers-and-dedekinds-methods-in-algebraic/26717#26717 Comment by Bill Dubuque Bill Dubuque 2012-01-03T06:52:23Z 2012-01-03T06:52:23Z @KConrad An English summary of Lucius' thesis is [Rings with a theory of greatest common divisors,](<a href="http://www.digizeitschriften.de/dms/img/?PPN=GDZPPN002239620" rel="nofollow">digizeitschriften.de/dms/img/?PPN=GDZPPN002239620</a>) Manuscripta Math. 95, 117-136 (1998). This is the best English introduction as far as I know. http://mathoverflow.net/questions/32968/slick-ways-to-make-annoying-verifications/33435#33435 Comment by Bill Dubuque Bill Dubuque 2010-12-12T07:54:21Z 2010-12-12T07:54:21Z @Michael: It can be extended to nonsquare matrices by appropriately padding them to make them square. http://mathoverflow.net/questions/35524/how-fast-can-the-base-bumping-function-in-goodsteins-theorem-grow/35546#35546 Comment by Bill Dubuque Bill Dubuque 2010-12-01T02:54:47Z 2010-12-01T02:54:47Z @Scott: Thanks. Such links used to work ok before recent changes. http://mathoverflow.net/questions/32133/expressing-adja-as-a-polynomial-in-a/32609#32609 Comment by Bill Dubuque Bill Dubuque 2010-10-01T23:01:26Z 2010-10-01T23:01:26Z @Pierre-Yves Gaillard: +1, thanks for the interesting post. http://mathoverflow.net/questions/35127/why-does-the-algebraic-condition-of-flatness-on-the-structure-sheaves-give-a-goo Comment by Bill Dubuque Bill Dubuque 2010-08-17T18:54:02Z 2010-08-17T18:54:02Z Thanks to all for the many interesting remarks, here and below. http://mathoverflow.net/questions/34806/what-was-the-relative-importance-of-flt-vs-higher-reciprocity-laws-in-kummers-i/34857#34857 Comment by Bill Dubuque Bill Dubuque 2010-08-07T19:22:56Z 2010-08-07T19:22:56Z @Robin: thanks for posting the link. Alas, such links don't always work due to various factors (viewing limits, country restrictions, etc). In fact that page is currently blocked for me, and I doubt I've reached viewing limits since I have my own copy of the book. That's why I posted the excerpt. http://mathoverflow.net/questions/33865/indeterminate-x-in-abstract-algebra-ring-theory Comment by Bill Dubuque Bill Dubuque 2010-08-03T21:36:30Z 2010-08-03T21:36:30Z @VP Per Scott's request, 3 days ago I created a meta thread to discuss such matters, see <a href="http://meta.mathoverflow.net/discussion/568" rel="nofollow">meta.mathoverflow.net/discussion/568</a> There is much further discussion there. My posts there elaborate on my views on such matters. If you wish to continue this discussion please do so there. http://mathoverflow.net/questions/33865/indeterminate-x-in-abstract-algebra-ring-theory Comment by Bill Dubuque Bill Dubuque 2010-08-01T16:42:35Z 2010-08-01T16:42:35Z @VP: I see the opposite. 33252 appears in almost <i>every</i> algebra textbook, whereas the question here (esp. as I reformulated it in the meta thread) is rarely, if ever, discussed in textbooks. So if any author lacks &quot;due diligence&quot; it is the author of 33252. I don't see any &quot;chaotic discussion&quot; or &quot;soapboxing&quot; in the current replies below. Nor can I possibly imagine this topic ever degenerating into such (unless MO was quickly invaded by a marauding band of cranks). Could you please stop making unfounded claims such as your insinuation that I &quot;view MO as a platform to cure all the worlds sins&quot;. http://mathoverflow.net/questions/33865/indeterminate-x-in-abstract-algebra-ring-theory Comment by Bill Dubuque Bill Dubuque 2010-07-31T22:09:55Z 2010-07-31T22:09:55Z @VP I don't agree at all because 33252 can easily be answered by a textbook reference to almost any algebra textbook but this question cannot. Indeed, most algebra textbooks do a very poor job (if any) of motivating the <i>real</i> reason why algebraists work with formal rather than functional polynomial rings. http://mathoverflow.net/questions/33865/indeterminate-x-in-abstract-algebra-ring-theory Comment by Bill Dubuque Bill Dubuque 2010-07-31T14:50:25Z 2010-07-31T14:50:25Z I too think it should be reopened. This is a common question on other math newsgroups and it would be nice to be able to point people at good expositions here. http://mathoverflow.net/questions/33697/assistance-with-understanding-parent-child-relationships-in-pythagorean-triples/33726#33726 Comment by Bill Dubuque Bill Dubuque 2010-07-31T02:58:05Z 2010-07-31T02:58:05Z @Gerry: Yes, pun intended - sharp eye! http://mathoverflow.net/questions/33526/deformations-and-the-dual-numbers/33531#33531 Comment by Bill Dubuque Bill Dubuque 2010-07-28T15:22:51Z 2010-07-28T15:22:51Z @altgr Thanks for pointing out the link rot - I've corrected it. ALas, I don't always have the time to check for link rot when excerpting old posts. http://mathoverflow.net/questions/11964/strong-induction-without-a-base-case/12061#12061 Comment by Bill Dubuque Bill Dubuque 2010-07-22T22:13:48Z 2010-07-22T22:13:48Z @Joel Yes, without a rigorous definition of what it means to &quot;not treat the base case separately&quot; the question cannot be answered. I've seen many dozens of similar questions in various forums and a consensus is never reached due to the lack of such precision. http://mathoverflow.net/questions/32133/expressing-adja-as-a-polynomial-in-a/32383#32383 Comment by Bill Dubuque Bill Dubuque 2010-07-19T01:38:59Z 2010-07-19T01:38:59Z @Q The equation in Victor's comment is indeed the equation that I started with, viz. $\rm d B = d C$ for $\rm d = det A$. http://mathoverflow.net/questions/32133/expressing-adja-as-a-polynomial-in-a/32343#32343 Comment by Bill Dubuque Bill Dubuque 2010-07-18T22:36:17Z 2010-07-18T22:36:17Z @Victor: to ensure there is no confusion, I remark that my use of &quot;generic&quot; above is not intended to denote anything topological or geometrical. Rather, it is meant to be understood as exploiting the universality of a free objects. The proof I gave does not require any knowledge of topology or (algebraic) geometry. I'm not saying that imposing other such viewpoints isn't interesting or useful - just that such is not required for problems of this sort. Further doing so adds complexity to what is - at the heart - trivial (yet elegant) algebra.