User jose capco - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T12:36:20Z http://mathoverflow.net/feeds/user/1245 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/6675/periods-and-commas-in-mathematical-writing Periods and commas in mathematical writing Jose Capco 2009-11-24T11:00:37Z 2013-04-29T18:33:13Z <p>I just realized that I am a barbarian when it comes to writing. But I am not entirely sure, so this might be the right place to ask. When typing display-mode formulae do you guys add a period after the formula ends a sentence? </p> <p>Like:</p> <blockquote> <p>This is the formula for a circle <code>$$x^2 + y^2 = r^2.$$</code><br /> Therefore blabla...</p> </blockquote> <p>or </p> <blockquote> <p>This is the formula for a circle <code>$$x^2 + y^2 = r^2$$</code><br /> Therefore blabla...</p> </blockquote> <p>My supervisor has been complaining a lot that I don't use period and commas in my display-mode formulae. But I get uneasy doing that because it doesn't feel natural to me, I took a look at two books at random and both of them so far do the punctuation in their display formulae.. I know this is stupid of me and its amazing I have never noticed that.</p> <p><strong>Edit:</strong> This would be a fantastic opportunity to see what people <em>actually</em> like as opposed to what they think they like. Everyone who has an opinion on what the punctuation should be should provide an illustrative example of such so that by the voting it can be seen what is actually preferred. If you do this, make your answer just the example (so provide any general homilies in another answer) so that the voting truly reflects the community view of the example.</p> http://mathoverflow.net/questions/33193/reduced-rings-idempotents-and-their-prime-spectrum Reduced rings, idempotents and their prime spectrum Jose Capco 2010-07-24T12:36:31Z 2012-12-21T19:11:39Z <p>Let $B$ be a commutative unitary reduced ring and let $A$ be a subring of it. Let $e$ be an idempotent of $B$. Then we have a natural surjective ring homomorphism $A\rightarrow Ae$ defined by $a\mapsto ae$. $Ae$ is just the ring with $e$ as unity and multiplication, addition induced from $B$ (i.e. $ae\cdot be = abe$ and $ae+be = (a+b)e$).</p> <p>The question is.. How much do we know about Spec $Ae$ ? </p> <p>Edit: Well I realize that the prime ideals of $Ae$ should be of the form $\mathfrak{p}e$ for some prime ideal $\mathfrak{p} \in$ Spec $A$. But how much do we know about the topology of Spec $Ae$? </p> <p>Edit: Apparently I won't know much about Spec $Ae$ unless I know more about $B$. In my specific problem, I wouldn't mind $B$ to be a von Neumann regular (i.e. zero dimensional) ring of fractions of $A$ (i.e. for all $b\in B\backslash${0} there is an $a$ $\in A$ such that $ab$ $\in A\backslash${0}) with extremally disconnected spectrum (i.e. closure of any open set in Spec $B$ is open). I just didn't wanted to add this extra condition to avoid confusion. My more specific question was whether the ring $A[e]$ could have an extremally disconnected minimal prime spectrum if the minimal prime spectrum of $A$ werent exd, for that it suffices for me to know this for $Ae$.</p> http://mathoverflow.net/questions/108024/whats-the-name-of-this-subset-of-polynomials whats the name of this subset of polynomials? Jose Capco 2012-09-25T07:24:27Z 2012-09-25T07:24:27Z <p>Consider the set of polynomials of n-variables (over say a field or a commutative ring) such that any polynomial in this set has maximum m-degrees with respect to any of the n-variables. Is there a name for such a polynomial or does it appear in any literature?</p> http://mathoverflow.net/questions/81054/julia-sets-using-other-fields Julia sets using other fields Jose Capco 2011-11-16T09:38:19Z 2012-09-10T15:21:59Z <p>I hope I am forgiven for my noob question. But, does it make sense to think of Julia sets using other fields? More precisely I would like to think of fields in which closed and bounded isn't necessarily a compact set. I am not sure what this will give us, but some results that we know in complex numbers wouldn't hold (e.g. will the Julia sets and the filled Julia sets still remain compact and nonempty?). </p> http://mathoverflow.net/questions/6477/applications-of-the-other-definition-of-sheaves Applications of the "other" definition of Sheaves Jose Capco 2009-11-22T19:22:28Z 2012-01-11T18:47:18Z <p>In most literature, when you try to look for the definition of sheaves you will see the usual definition for presheaves as a functor from a topological space (or from a Grothendieck topology) to some category and then sheaves would require this category to be complete and you have some exactness/equalizer condition. </p> <p>But then for some categories there is another equivalent definition. You are defined a "protosheaf" (there are various names for these creatures), a sheaf space, a base space, a local homeomorphism between the sheaf space and the base space, you are even already defined a stalk.. but this definition seems not to be very abstract in the category-theoretical point of view as I only see this kind of definition for very specific categories (for instance in the category of groups or rings, you want the addition operation defined on the fiber product of the sheaf space over the base space to be continuous). What is the equivalent category theoretical way of defining a sheaf using this method? In which cases does this definition give us a more psychological advantage than the aforementioned one? I have personally found the former definition more advantageous in my practice, but there are some mathematical practices by which the latter definition might be more useful.</p> http://mathoverflow.net/questions/82361/how-would-you-define-a-hyper-cylinder How would you define a hyper-cylinder? Jose Capco 2011-12-01T09:53:38Z 2011-12-02T04:23:48Z <p>I'd like to describe a "bottleneck" in n-dimensional topological vector space $\mathbb R^n$. For that I should probably try to define an "n-dimensional" cylinder.</p> <p>I'd like to define something similar to a "hollow" cylinder but in say $\mathbb R ^n$ for n >= 2. I would probably define it as something homeomorphic to $S^{n-2} \times L$ where $L$ is a line crossing 0 (for a cylinder with finite volume I'd then assume this line to have two endpoints). Is there a name for this in Geometry or a reference where such a thing is discussed? Would this be the right way of defining a cylinder in n-dimensions? Any other suggestions?</p> http://mathoverflow.net/questions/36638/effective-algorithm-to-test-positivity/81166#81166 Answer by Jose Capco for Effective algorithm to test positivity Jose Capco 2011-11-17T11:42:53Z 2011-11-17T11:42:53Z <p>If you could find an upper bound for degrees of polynomials used in Positivstellensatz I guess you can create an algorithm for this.</p> <p>You would like to take a look at Positivstellensatz, for instance in the book of Bochnak-Coste-Roy "Real Algebraic Geometry" under Corollary 4.4.3 (this corollary combines Positivstellnsatz, Negativstellensatz and Nullstellensatz for real closed fields). A more abstract form can be found in a German book "Einführung in die Reelle Algebra" by Knebusch and Scheiderer (the book is now freely available for download from the university of Regensburg library: <a href="http://epub.uni-regensburg.de/12857/1/ubr06693_ocr.pdf" rel="nofollow">here</a>). You would also like to see the works by Schmüdgen and Putinar on various forms of Positivstellensätze.</p> http://mathoverflow.net/questions/70209/fractal-questions-weierstrass-mandelbrot Fractal questions: Weierstraß-Mandelbrot Jose Capco 2011-07-13T10:29:41Z 2011-07-14T14:47:38Z <p>Hi,</p> <p>Coming from a specific field in algebraic geometry I am a total noob in Fractal Theory and I'd like to learn it a bit. I hope I am tolerated for my maybe-trivial questions. I just read about the Weierstrass-Mandelbrot fractal (it's also simply called Weierstrass fractal using the Weierstrass function.. but there are dozens of Weierstrass functions so I'd rather call it "Weierstrass-Mandelbrot" function). The definition of this fractal is found in <a href="http://en.wikipedia.org/wiki/Weierstrass_function" rel="nofollow">wikipedia</a>. I got easily impressed by it.</p> <p>My question is whether there are nowhere differentiable continuous functions (between real numbers) whose graph are not fractals? Is the WM function the easiest example of a nowhere differentialbe continuous function? </p> <p>The other question is quite basic (for experts probably). I have seen the definition of fractal in wikipedia. This definition uses self-similarity. But in a reference of mine (from a lecture note) I get a definition that makes use of an inequality with Hausdorff-dimension and inductive dimension. Are these definitions equivalent or are the precise definition still under debate (my reference suggests that the suggested definition was former definition by Mandelbrot and then this definition was changed as Mandelbrot fractals don't follow this definition). A little enlightening would help :)</p> http://mathoverflow.net/questions/63423/checkmate-in-omega-moves/63507#63507 Answer by Jose Capco for Checkmate in $\omega$ moves? Jose Capco 2011-04-30T07:10:42Z 2011-04-30T07:29:17Z <p>Yes there is if the finite number of pieces are large enough. But to answer your question I want black to win (because I hate it when problems most often require white to win). Let's consider the case for $N\times N$ boards and extrapolate it and then I will show how this is done for infinite boards. For a usual $8\times 8$ board there is no argument on the fact that we have mates in $1,2,\dots,7$ (and Im sure even more). I claim that for an $N\times N$ board we can have mates in $1,2,\dots,N-1$, and then I show how this is done for an infinite board.</p> <p>Consider this classical mate in 7 (white to move, black mates in 6 plies, white moves maximum 7 plies) position (actually its mate in 8 if a queenside castling was allowed and we had black rook in a8 and king in e8, its funny position I always show people who never considered castling when solving such problems):</p> <p><img src="http://i.harepix.com/i/389837748.jpg" alt="alt text"></p> <p>Now you can reproduce the same position for an $N\times N$ board and get a mate by $N-1$ for any $N>8$. To make this work for an infinite board just surround an infinite board by black's pawn to "create" a bounded $N\times N$ board. So to make an $8\times 8$ board like the one in the diagram below. Just surround the $10\times 10$ area by Black's pawn. White's king cannot move away from the "boundary" because of the pawns. So in this way we see that we get mate in $7,8,9,\dots$ for an infinite board. For $1,2,\dots,6$ moves to mate, we do the same by only creating an $8\times 8$ board but hopefully in such a way that the "boundary" cannot be taken or moved by white (I don't think its difficult to provide the particular examples here). </p> <p>My instincts tell me that this particular example can be done for any ordinal as well.</p> http://mathoverflow.net/questions/61155/normal-domains-with-algebraically-closed-quotient-field normal domains with algebraically closed quotient field Jose Capco 2011-04-09T17:38:05Z 2011-04-12T18:40:50Z <p>I am looking for an integral domain $A$ with the following properties:</p> <ol> <li>$A$ is not integrally closed</li> <li>$A$ has a quotient field $K$ that is algebraically closed and that has characteristic 0</li> <li>There is an integral element $x\in K$ (<em>over</em> $A$) such that $A[x]$ is integrally closed.</li> </ol> <p>Can someone help to tell me if the above is even possible? </p> <p><strong>Edit</strong>: Lubin easily gave me an example. Now I want to consider the case when I replace the condition 2. by:</p> <p>2'. $A$ has a quotient field $K$ that is real closed.</p> http://mathoverflow.net/questions/59263/ring-of-idempotents-of-the-integral-extension-of-a-ring ring of idempotents of the integral extension of a ring Jose Capco 2011-03-23T05:15:08Z 2011-03-24T11:52:43Z <p>For any commutative ring $A$, the set of idempotents of $A$ will be denoted as $E(A)$. This set has a (canonical) ring structure. With addition defined by: $$e+'f=e(1−f)+f(1−e)$$ where $+$ and $−$ are operation in the ring itself. The multiplication operation is the same as the ring itself.</p> <p>Suppose now I have a commutative unital ring $A$ and let $B$ be another commutative ring that is an integral extension of $A$. Then clearly $E(B)$ is an over-ring of $E(A)$, but is it clear that $E(B)$ is an integral extension of $E(A)$. Are there easy counter examples for this? Would it help if I assumed that $B$ is a finite integral extension of $A$?</p> <p><strong>Edit:</strong> The question above had a trivial answer as Todd pointed out. Now Im curious what happens if I really want $E(B)$ to be a finite integral extension of $E(A)$. Does $B$ being a finite integral extension of $A$ gauarantee that? </p> http://mathoverflow.net/questions/2770/can-any-topological-space-be-the-result-of-a-scheme Can any topological space be the result of a scheme? Jose Capco 2009-10-27T04:52:37Z 2010-07-31T15:26:03Z <p>Maybe this is trivial but lets give it a try anyways.. </p> <p>Obviously there is a forgetful functor from schemes to topological space.. but is it surjective on objects? i.e. I ask whether any topological space is a result of using the forgetful functor on a certain scheme? Certainly this is not true if we consider ONLY affine schemes (which are spectral spaces i.e. Kolmogorov, Compact, compactness preserved upon finite intersection of open compacts, nonempty irreducibles contain a generic point).. but ... hmm wait.. maybe I should require the topological space have the property that each irreducible component must have at least a generic point? </p> http://mathoverflow.net/questions/31373/weaker-form-of-irreducible-surjections Weaker form of irreducible surjections Jose Capco 2010-07-11T08:07:28Z 2010-07-15T19:18:44Z <p>An irreducible surjection is usually defined as a continuous <strong>closed</strong> surjective map $f:X\rightarrow Y$ such that if for some closed set $C\subset X$ one has $f(C)=Y$ then $C=X$. In my dissertation I used a weaker form of this definition, removing the requirement of these functions being closed. I then defined them as quasi-irreducible surjections. Are there any applications of quasi-irreducible surjection elsewhere?</p> http://mathoverflow.net/questions/28673/need-a-reference-for-cones-and-limits-that-does-this Need a reference for cones and limits that does this... Jose Capco 2010-06-18T22:00:21Z 2010-06-18T22:16:41Z <p>Here's something that I'd like to use in my thesis.. but Im feeling too lazy to write a proof of it, I feel pretty sure this is correct though. I have a feeling that this can be found in a book on category theory. So maybe someone can point me to a reference (I have only used Adamék, Herrlich and Strecker so far).</p> <p>Conjecture: Short Statement: pullback of inverse limits is the inverse limit of pullbacks </p> <p>Long Statment: Let $I$ be a directed set and let $\mathbf D,\mathbf E: I \rightarrow \mathcal C$ be two diagrams to a complete category. Let $C$ be an object in $\mathcal C$ and suppose that we have natural sinks $\mathbf Di \rightarrow C$ and $\mathbf Ei\rightarrow C$ (Mac Lane calls these "cones"). Let $A$ and $B$ be the inverse limit of $\mathbf D$ and $\mathbf E$ respectively. We get another diagram that goes to the pullback namely $\mathbf D \times_C \mathbf E : I \rightarrow \mathcal C \times_C \mathcal C$. The claim is that the inverse limit of $\mathbf D \times_C \mathbf E$ is actually the fiber product (or pullback, however you want to call it) $A \times_C B$ (where $A\rightarrow C$ and $B\rightarrow C$ are canonical map that results from the inverse limit and the natural sinks).</p> http://mathoverflow.net/questions/10449/artin-schreier-theorem-for-rings Artin Schreier Theorem for Rings Jose Capco 2010-01-02T02:30:00Z 2010-01-26T06:44:17Z <p>This has been in my mind for quite some time. Looking at Artin Schreier Theorem for fields:</p> <p>If L is a field and K its algebraic closure and if 1&lt; [K:L] &lt; infinity then L=K[i] and L is a real closed field (Thus L has characteristic 0. Here i is just the square root of -1).</p> <p>I was wondering if a "generalized" Artin Schreier exist or if someone could refer to me to some paper that attempts this. There is a concept of real closedness and "algebraic closedness" of reduced commutative rings, but I doubt that the statement would hold.</p> <p>So one has the following conjecture:</p> <p>If L is a reduced commutative ring and K is its total integral closure (this is an equivalent notion of algebraic closure if K and L were fields) and if 1&lt;[K:L]&lt; infinity (here I mean that K is a finite L-module that is not the same as L) then L is real (thus its characteristic is 0.. and one can add that L is real closed in the sense of reduced commutative rings).</p> <p>Can one easily show this, even at least prove that L has characteristic 0?</p> http://mathoverflow.net/questions/4847/various-cartans-lemmata Various Cartan's Lemmata Jose Capco 2009-11-10T12:15:18Z 2009-12-23T21:21:17Z <p>I am a bit amazed by "Cartan's Lemma".. I have so far seen it in :</p> <p>Algebraic Geometry sources: Look at Proposition 2.9 of Freitag and Kiehl's Étale Cohomology where he used étale morphism to describe Cartan's Lemma.</p> <p>Complex Analysis: Look <a href="http://www.joensuu.fi/matematiikka/kurssit/complex/luku10.pdf" rel="nofollow">here</a></p> <p>Algebra and Ring Theory: Look <a href="http://books.google.com/books?id=hJ4AcMMPW48C&amp;lpg=PA223&amp;ots=7RVgxhF8UR&amp;dq=cartan%27s%20lemma&amp;pg=PA223#v=onepage&amp;q=cartan%27s%20lemma&amp;f=false" rel="nofollow">here</a></p> <p>I do need a little help in explaining how all these Cartan's Lemmas relate with each other</p> http://mathoverflow.net/questions/9356/real-spectrum-of-ring-of-continuous-semialgebraic-functions/9360#9360 Answer by Jose Capco for Real spectrum of ring of continuous semialgebraic functions Jose Capco 2009-12-19T08:45:30Z 2009-12-19T08:58:51Z <p>The real spectrum of ring of continuous functions (not necessarily semialgebraic) looks exactly like its prime spectrum (i.e. they are homeomorphic), this is also the same for ring of continuous semialgebraic functions (see "<a href="http://books.google.com/books?id=rJk0yn2TROYC&amp;dq=semialgebraic+function+rings&amp;printsec=frontcover&amp;source=bl&amp;ots=ci-7yimvD5&amp;sig=CnSi2RCoOuAMnhcSm6c1JCOOOZ4&amp;hl=en&amp;ei=IZQsS67sMJiXkQWq8LWECQ&amp;sa=X&amp;oi=book%5Fresult&amp;ct=result&amp;resnum=3&amp;ved=0CA0Q6AEwAg" rel="nofollow">Semi-algebraic Function Rings via Reflectors of Partially Ordered Rings</a>" by Schwartz and Madden, Corollary 7.8 ). And $U$ should be homeomorphic to $\text{Spec}_r (S^0(U))$ when we consider their constructible topology (see the book of Schwartz and Madden, Proposition 7.9).</p> http://mathoverflow.net/questions/9224/constructing-a-convex-valuation-ring-ordered-group-of-rank-n Constructing a convex valuation ring/ordered group of rank n Jose Capco 2009-12-18T02:02:39Z 2009-12-18T02:57:29Z <p>I know at least one method of constructing a convex valuation ring of rank n (but it is rather complicated).. What are the easiest methods of doing this.. given a natural number n I want to have a valuation ring (preferably convex) whose rank is n. I have heard you can do this with polynomials and power series, but I am not really sure how this is done.</p> http://mathoverflow.net/questions/3024/characterisation-for-separable-extension-of-a-field Characterisation for separable extension of a field Jose Capco 2009-10-28T09:19:16Z 2009-12-09T04:48:39Z <p>Can someone verify this for me.. or tell me what reference shows me this... is this true:</p> <p>Let k be a field then a field extension K of k is separable over k iff</p> <p>for any field extension L >= k the Jacobson radical of the tensor product K (x)_k L is trivial.</p> <p>I got this idea by looking at some definitions of separable algebras (which is not my field of research.. but somehow this definition got me intruiged). Anyone knows if this is true and why so? or maybe a reference or two about it?</p> http://mathoverflow.net/questions/5357/theorems-for-nothing-and-the-proofs-for-free/6410#6410 Answer by Jose Capco for Theorems for nothing (and the proofs for free) Jose Capco 2009-11-21T21:40:58Z 2009-11-21T21:40:58Z <p>Artin-Schreier Theorem: If k is a field of characteristic p and strictly contained in its algebraic closure K and such that [K:k] is finite THEN (was surprising for me..) p is actually 0 and K = k(sqrt(-1)) and k is a real closed field!</p> <p>A not so well known but deserving result from the "failed" thesis of Abhyankar: If K and L are algebraically closed fields contained in another algebraically closed field, then the compositum KL is not necessarily algebraically closed.</p> http://mathoverflow.net/questions/6227/which-topological-spaces-have-the-property-that-their-sheaves-of-continuous-funct/6339#6339 Answer by Jose Capco for Which topological spaces have the property that their sheaves of continuous functions are determined by their global sections? Jose Capco 2009-11-21T01:21:56Z 2009-11-21T01:43:06Z <p>I tried a bit of thinking, but I haven't worked all the details. I have a hint though that may lead to the answer of your question. You may want to regard the continuous functions over an open set as a ring. This ring is reduced and commutative (thus there is a so-called <em>rational completion</em>) and we could then look at rational completion of them and this may lead to an answer.</p> <p>A good and downloadable reference of this is found <a href="http://at.yorku.ca/i/a/a/l/94.htm" rel="nofollow">here</a>. A classical reference (and also the best one) is the book of Lambek "<a href="http://books.google.com/books?id=LQPid5RSkJsC&amp;dq=Lambek+rings+of+modules&amp;printsec=frontcover&amp;source=bl&amp;ots=KwwoMK3b7w&amp;sig=jQXc%5FedvqN9LQSilXu7kQxfqg20&amp;hl=en&amp;ei=7T8HS6DxApO1sgbT1vCzCg&amp;sa=X&amp;oi=book%5Fresult&amp;ct=result&amp;resnum=3&amp;ved=0CBAQ6AEwAg#v=onepage&amp;q=&amp;f=false" rel="nofollow">Lectures on Rings and Modules</a>" by Lambek (please don't confuse it with the book of Lam, who happens to have the same first 3 letters in his last name, entitled "Lectures on Modules and Rings"), see for instance sections 2.3 and 4.4 of the book.</p> <p>A few years ago, I had written a small entry in Planetmath that characterized <a href="http://planetmath.org/?method=l2h&amp;from=objects&amp;name=RationalExtension&amp;op=getobj" rel="nofollow">rational extensions of commutative reduced rings</a>. And you can use that as an easy definition.</p> http://mathoverflow.net/questions/6010/dense-section-of-sheaves-of-modules Dense section of sheaves of modules Jose Capco 2009-11-18T20:25:50Z 2009-11-18T23:39:46Z <p>Here is something that isn't yet very clear to me. Say, I've got a commutative ring A. I consider the affine scheme from A, so it's a sheaf of rings over Spec A. </p> <p><strong>EDIT</strong>: And additionally let's say Spec A is Hausdorff. </p> <p>Now additionally let's say I know an A-module M and from that I can make a sheave of modules over O_Spec(A), call it M~. All standard stuffs till now. But now I want to have one more information, namely I have an open subset of Spec(A), say U, that is dense in Spec(A). And I know additionally that the stalks M~_x are isomorphic to O_x for all x in U.. Can one conclude that M and A are A-module isomorphic? (if so can one follow the same argument for general schemes with modules over them?) What are the conditions by which one can conclude this?</p> http://mathoverflow.net/questions/5841/spectra-of-rings-that-are-projective-module-over-a-subring Spectra of rings that are projective module over a subring Jose Capco 2009-11-17T17:49:57Z 2009-11-17T19:23:11Z <p>This question is motivated from my <a href="http://mathoverflow.net/questions/4317/integrally-closed-factor-rings-and-projective-modules" rel="nofollow">last question</a> here. I wonder if one has a ring A and an over-ring of this ring say B, and if we know that B is a projective A-module can we have a particular idea of how Spec B would look like if we know how Spec A would look like?</p> <p>This question does make sense to me. Because for instance given a local ring A, then B's are some form of copies of A. If A were zero dimensional reduced and commutative then Spec B would look like copies of clopen subsets of Spec A (because projective modules over von Neumann regular rings are isomorphic to direct sum of principal ideals). So what else do we know?</p> http://mathoverflow.net/questions/5485/how-many-mathematicians-are-there/5487#5487 Answer by Jose Capco for How many mathematicians are there? Jose Capco 2009-11-14T08:45:14Z 2009-11-14T08:45:14Z <p>Current count of <a href="http://genealogy.math.ndsu.nodak.edu/" rel="nofollow">Mathematics Genealogy Project</a> is 137672 (I am assuming that the PhD students that graduated are ranked as "research mathematicians"). But the problem is.. Mathematics Genealogy is mostly for universities of developed countries. There could be some really good university in Russia, China or Korea out there that doesn't give us the correct statistics. Another problem is.. Mathematics Genealogy Project counts even the dead mathematicians (like Hilbert, Hasse, Kepler and so on).. and I am assuming you want a report of living mathematicians.. but hey, I'm quite surprised by the number even 200k is pretty low for the living! </p> http://mathoverflow.net/questions/5386/typo-grammar-checker-for-latex/5394#5394 Answer by Jose Capco for Typo/grammar checker for LaTeX Jose Capco 2009-11-13T17:03:04Z 2009-11-13T17:03:04Z <p>Yes, I was looking for that a week ago. I bumped into <a href="http://excalibur.sourceforge.net/" rel="nofollow">Excalibur</a>. I'm not sure how good it is though. If I get the time I want a progam that does both, the problem is how to exclude the maths when doing the check. Problem with grammar is worst, you need a program that will treat some maths as objects of a sentence.. that should difficult.</p> http://mathoverflow.net/questions/3044/tools-for-collaborative-paper-writing/5327#5327 Answer by Jose Capco for Tools for collaborative paper-writing Jose Capco 2009-11-13T08:52:06Z 2009-11-13T16:01:17Z <p>Sadly, I haven't collaborated yet.. but I would use svn for collaboration. If it's supposed to be a top-secret collaboration, then I'd go for a commercial web svn repository (lots out there that are very cheap). If you think no one could really steal the work from you or that it's ok for people to read it, I would use a free svn repository. The pain is when your partner isn't a techy person, you can't even ask him to merge new versions of works with some sort of merger.. you are supposed to do all the merging from his end. </p> <p>As for myself, I would not use svn. I usually have papers 15 pages long or so, and I just keep backups everytime I make a milestone and then zip the whole thing (like every month or so).. in case something goes wrong I go back to the backup, it's like a small scale inefficient svn.. but you don't want to kill a bird with an atomic bomb. Always worked for me, I'm not sure for writing books.. probably an svn would become necessary, but I won't be suprised if my backup technique work as well.. it worked for my PhD dissertation anyway :p</p> <p><strong>EDIT</strong>: Most of us mathematician wouldn't have our entire .tex work exceeding 200Mb. For that I would recommend the svn web repositories like (these support closed source, i.e. private projects): <a href="http://www.xp-dev.com/" rel="nofollow">XP-Dev</a> and <a href="http://unfuddle.com/" rel="nofollow">Unfuddle</a> These two are free if you don't exceed 200Mb and you can choose to have the project (you are allowed max 2 or 1 projects for these free repositories.. but I would just keep them in one big folder and name them as 1 project). I like the look and feel of Unfuddle, but XP-Dev isn't that bad either. The advantage of SVN is that it caters for both windows and linux users (for windows I'd recommend TortoiseSVN Client). In case you exceed the 200Mb (which for our purpose is hardly believable.. unless you have lots of images in your LaTeX documents), then I would recommend <a href="http://www.assembla.com" rel="nofollow">Assembla</a> because it's very cheap (like \$3/Month/user + \$0.3/100Mb/User) and reliable and you can buy as much space as you want.</p> http://mathoverflow.net/questions/5139/how-have-mathematicians-been-raised/5178#5178 Answer by Jose Capco for How have mathematicians been raised? Jose Capco 2009-11-12T09:45:08Z 2009-11-12T09:45:08Z <p>This is not something i would advocate, but since you asked for only descriptive answers.. apparently a reputable statistics for very talented mathematician has been their inability (or that they did not have the chance) to socialize in their early years. I'm sure there are many out there who are exceptions. But the biographies of most of the very gifted ones that I have read show this pattern of life.</p> http://mathoverflow.net/questions/5143/pushouts-in-the-category-of-schemes/5165#5165 Answer by Jose Capco for Pushouts in the Category of Schemes Jose Capco 2009-11-12T07:24:26Z 2009-11-12T09:32:05Z <p>Consider a commutative local ring R, say a valuation domain, with maximal ideal M. Consider the fiber product $R \times_M R$ (I wrote M instead of R/M), coming from the pullback in commutative rings $R\rightarrow R/M$. Then the corresponding prime spectra of this fibered product (in rings) is actually a form of gluing of the same same (affine) scheme Spec R along the closed point M. So this is the case where this happens. </p> <p>So I think you can do such things for affine Schemes. For affine schemes, you can at least reverse the topology (they are sometimes called inverse spectrum) and you can form a sheaf over this topology similar to the canonical structure sheaf, but the closed points becomes the generic points in this topology. I cannot recall correctly, but I think the stalks of this sheaves become integral domains (so it is some form of dual to the affine schemes, local becomes integral and so on)</p> http://mathoverflow.net/questions/4317/integrally-closed-factor-rings-and-projective-modules Integrally closed factor rings and projective modules Jose Capco 2009-11-05T23:52:43Z 2009-11-09T18:37:33Z <p>I have a weird vision that comes from reading a <a href="http://www.informaworld.com/smpp/content~db=all~content=a780098125" rel="nofollow">paper</a> by Raphael and Desrochers..</p> <p>Let R be commutative unitary semiprime ring such that for any integral and essential element a of R, R[a] is a projective R-module. I conjecture that for any minimal prime ideal P of R, one has R/P is an integrally closed domain.</p> <p>Does anyone have a counter example to this? </p> <p>PS: In case someone is unfamiliar.. a is an essential element of R iff aR[a] ∩ R ≠ 0</p> http://mathoverflow.net/questions/4580/pacing-for-learning-new-material/4641#4641 Answer by Jose Capco for Pacing for learning new material Jose Capco 2009-11-08T15:45:25Z 2009-11-08T15:45:25Z <p>Here are my advise that are mostly based on experience:</p> <p>If I start a totally new math, especially in the graduate level. I'd give my self at least 1.5-2 years (especially if it's an area in which a lot of lot of reading is involved.. say algebraic geometry). One of the things I find important, is not getting frustrated that you haven't learned to the level you need to learn. It is indeed frustrating, but when I look back in my PhD years.. I indeed took 2 years of just reading before even being able to start any new ideas of my own. You just don't have enough knowledge to make a ground-breaking mathematics and until you do be patient and learn it and try new ideas and create new examples (out of the book). I personally hardly answer excercises in the book, but created my own questions and tried to answer them first. If you were able to make new ideas and even publish a paper or two during these 1.5-2years then thats a bonus but you shouldn't feel incapable during that time.</p> <p>The other advise I'd give is the references. Never stick to one or even just two single reference. Especially if they are the references that are difficult to digest. You should collect as many of the references in that area of mathematics as possible. If this is math that people have done already, chances are there are many many references about it that you don't probably know yet. And never read linearly through the references (esp. textbooks). I don't know of any professional mathematicians that has actually finished reading an entire book that he has not written himself. Switch from the different references as much as possible and try to get as much goodies from each as possible. There is no ONE book in homological algebra and different mathematician find different book suitable, you should find one that writes in a style your prefer and every now and then look at the other books as well. There is NO one book in commutative algebra, you can read certain characterization of testing for flatness of modules/algebras in commutative algebra books but you can hardly find ALL of them in ONE single book and some of them don't even have all of the proof.</p> <p>Third advise, is to collaborate or speak as often as possible with people very knowledgeable in the topic. Attend seminar and conferences in that area of mathematics, even if you don't understand a pea. Chances are you learn something new or you learn about a question you think you find interesting in that area that is unanswered. There are some people who are knowledgeable in some area and make me feel like sh*t when I speak to them, I tend to avoid them.. but sometimes I mingle nevertheless. For me, true authentic mathematicians must good educators as well, so that if they find someone not knowledgeable in one thing they actually help him become knowledgeable instead of making him feel bad about it.</p> http://mathoverflow.net/questions/82361/how-would-you-define-a-hyper-cylinder Comment by Jose Capco Jose Capco 2011-12-01T11:32:08Z 2011-12-01T11:32:08Z I'd use it for pathplanning problem in an n-dimensional manifold. I would like to define a situation why even probabilistic complete pathplanning algorithms or analytic path planning algorithms would have difficulty (like the &quot;bottleneck&quot; phenomenon). This is a &quot;current research question&quot; in real algebraic geometry, but I'd like to start with something palatable in this site because I do not expect everyone to be specialized in this and still I want a broad discussion and perspectives from people from other fields. http://mathoverflow.net/questions/82083/when-is-the-tensor-product-of-two-fields-a-field Comment by Jose Capco Jose Capco 2011-11-29T10:13:11Z 2011-11-29T10:13:11Z Didn't Abhyankhar once worked on this (he once worked on compositum of algebraically closed field when he was a student)? http://mathoverflow.net/questions/82151/do-noetherian-rings-have-the-amalgamation-property Comment by Jose Capco Jose Capco 2011-11-29T08:41:16Z 2011-11-29T08:41:16Z Maybe I understand this incorrectly. But can't you just take the trivial ring as the terminal object of this category? http://mathoverflow.net/questions/36638/effective-algorithm-to-test-positivity/81166#81166 Comment by Jose Capco Jose Capco 2011-11-17T19:57:06Z 2011-11-17T19:57:06Z I believe Schm&#252;dgen, Schweighofer and Prestel were addressing this at some point. I think in a paper Schweighofer even introduced some algorithms and discussed the complexity involved. But as this is not my area of expertise, I can't really say much more than this. http://mathoverflow.net/questions/81011/prime-ideals-in-the-ring-of-germs-of-continuous-functions Comment by Jose Capco Jose Capco 2011-11-17T19:34:01Z 2011-11-17T19:34:01Z I believe you will find the free reference by Lambek, Gillman and Fine useful: <a href="http://at.yorku.ca/i/a/a/l/94.htm" rel="nofollow">at.yorku.ca/i/a/a/l/94.htm</a> http://mathoverflow.net/questions/29100/real-algebraic-geometry-vs-algebraic-geometry/29103#29103 Comment by Jose Capco Jose Capco 2011-11-17T11:24:09Z 2011-11-17T11:24:09Z I find the book by Brumfiel very helpful especially in real algebra. Though it has much less real geometry than BCR (Bochnak-Coste-Roy). BCR covers Nash Manifolds and other things in real geometry (like curve-selection lemma and separation of semialgebraic closed sets) http://mathoverflow.net/questions/81054/julia-sets-using-other-fields Comment by Jose Capco Jose Capco 2011-11-16T11:57:43Z 2011-11-16T11:57:43Z I mean, I think, its easy to restrict the julia set to the real algebraic numbers and arrive to some non-compact set. But why would it be interesting to look at Julia sets this way? http://mathoverflow.net/questions/70209/fractal-questions-weierstrass-mandelbrot/70229#70229 Comment by Jose Capco Jose Capco 2011-07-14T08:18:06Z 2011-07-14T08:18:06Z Thanks for the detailed info. Do you happen to have a pictorial link to the Kie&#223;wetter fractal? I wasn't able to find it by doing a google image search. http://mathoverflow.net/questions/70209/fractal-questions-weierstrass-mandelbrot Comment by Jose Capco Jose Capco 2011-07-13T11:00:58Z 2011-07-13T11:00:58Z thanks fixed- A common typo for me. http://mathoverflow.net/questions/63423/checkmate-in-omega-moves/63507#63507 Comment by Jose Capco Jose Capco 2011-05-03T18:37:45Z 2011-05-03T18:37:45Z Well initially I thought that the question was: given a number n, find position in an infinite board with finite pieces where I have mate in n (but not less). But as kevin pointed out I understood the question wrongly. I didnt deleted the example from the answer because I thought it has some point of interest here. http://mathoverflow.net/questions/63423/checkmate-in-omega-moves/63517#63517 Comment by Jose Capco Jose Capco 2011-04-30T17:12:21Z 2011-04-30T17:12:21Z I love the left board its just beautiful :) http://mathoverflow.net/questions/63423/checkmate-in-omega-moves/63507#63507 Comment by Jose Capco Jose Capco 2011-04-30T17:03:52Z 2011-04-30T17:03:52Z Yes correct. Sorry, now I see it :) http://mathoverflow.net/questions/63423/checkmate-in-omega-moves/63517#63517 Comment by Jose Capco Jose Capco 2011-04-30T15:30:19Z 2011-04-30T15:30:19Z this could work Im convinced that the left board will help with a solution. But what if white tries to cover the g7 square in the right board? Maybe we should secure the above and below part so white cannot cover the g6 square or do d2 and and move around the board with the southeast B. If white is able to go Be9 without being taken it covers the g7 square and that would be a problem. But I think you can secure e9 and other squares here so this wont happen http://mathoverflow.net/questions/63423/checkmate-in-omega-moves/63507#63507 Comment by Jose Capco Jose Capco 2011-04-30T15:02:02Z 2011-04-30T15:02:02Z if white king remains on the last file all the time the pushing black pawn stops unless there is a king support to cover the pawns (because the rook should either cover the pawn or a rank where the king cannot climb up to). The rook can force the king until its first rank but then it should either remain on the second rank or in another rank allowing the king to climb up or just. Or force a stalemate. With rook and pawns alone in that situation you cant promote http://mathoverflow.net/questions/63423/checkmate-in-omega-moves/63507#63507 Comment by Jose Capco Jose Capco 2011-04-30T07:46:15Z 2011-04-30T07:46:15Z In this particular example for the finite board situation there is no queen promotion and still mating in less than N-1 moves. For an infinite board position (with the pawn boundaries) I wouldn't understand queen promotion here