User arsmath - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T18:43:58Z http://mathoverflow.net/feeds/user/3711 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56104/early-two-author-math-papers Early Two-Author Math Papers arsmath 2011-02-20T20:30:21Z 2013-05-08T13:52:52Z <p>The middle of the twentieth-century featured several famous papers with two authors. For example, Eilenberg and Mac Lane's papers introducing categories and Eilenberg-MacLane spaces appeared in 1945. The Feit-Thompson Odd Order Theorem appeared in 1962. Atiyah and Singer published their index theorem in 1963.</p> <p>I can't think of any important papers with two or more authors before the Eilenberg-Mac Lane collaboration, which could just be a lacuna in my historical knowledge. My question is: what are the first math papers with two or more authors? (A subsidiary question is: why were collaborations so rare before that?)</p> http://mathoverflow.net/questions/128446/general-theory-of-left-exact-localization General Theory of Left-Exact Localization? arsmath 2013-04-23T07:22:44Z 2013-04-23T07:22:44Z <p>A left-exact localization of a category is a reflective subcategory such that the reflector preserves finite limits. There are several prominent examples of such localizations, such as sheafification, and localization of module categories. Is there a general theory of such localizations? </p> <p>I don't have any particular type of result in mind, but given the prominence of the two examples I mentioned, it seems like the topic of left-exact localizations must have been studied for its own sake.</p> http://mathoverflow.net/questions/123264/market-clearing-price-vector-in-an-aggregate-demand-system/123265#123265 Answer by arsmath for Market-clearing price vector in an "aggregate demand system" arsmath 2013-02-28T20:57:51Z 2013-02-28T20:57:51Z <p>You can find this in an standard graduate-level microeconomics textbook, such as Mas Colell-Whinston-Green. A solution exists with or without the assumption of gross substitutes, and does indeed use a fixed-point theorem, such as the Brouwer or Kakutani fixed point theorem. Abraham Wald has a proof for a special case that's simpler -- I think his case includes gross substitutes, but I don't know for sure. In the gross substitutes case, the solution is unique.</p> <p>In terms of algorithms, there is an algorithm based on Sperner's lemma due to Scharf, but practically speaking any general-purpose nonlinear solver will do. Again, there might be something special about the gross substitutes case that allows a simpler algorithm, but I don't know for sure.</p> http://mathoverflow.net/questions/121406/where-in-ordinary-math-do-we-need-unbounded-separation-and-replacement/121438#121438 Answer by arsmath for Where in ordinary math do we need unbounded separation and replacement? arsmath 2013-02-11T01:23:15Z 2013-02-11T01:23:15Z <p>The Stacks Project, which is a thorough introduction to <a href="http://en.wikipedia.org/wiki/Algebraic_stack" rel="nofollow">algebraic stacks</a>, including necessary background, uses the axiom of replacement when <a href="http://stacks.math.columbia.edu/tag/000H" rel="nofollow">constructing</a> categories of schemes closed under certain operations. (I believe the purpose of this is to avoid using universes.)</p> <p>In the construction, they explicitly work with $V_\alpha$, and prove by transfinite induction that there exists a big enough $\alpha$ so that the category of schemes contained in $V_\alpha$ is closed under certain operations. </p> http://mathoverflow.net/questions/121247/differential-geometry-study-materials/121258#121258 Answer by arsmath for Differential geometry study materials arsmath 2013-02-09T00:21:14Z 2013-02-09T00:21:14Z <p>I recommend an older book, <i>Notes on Differential Geometry</i> by Noel Hicks. What I like about it is that it starts with manifolds embedded in $R^n$, and shows how all of the concepts of differential geometry naturally arise there.</p> http://mathoverflow.net/questions/120084/what-kinds-of-operations-are-well-defined-when-working-with-sets-classes-conglo/120113#120113 Answer by arsmath for What kinds of operations are well-defined when working with sets, classes, conglomerates, and yet higher order collections? arsmath 2013-01-28T14:37:05Z 2013-01-28T14:37:05Z <p>This question is considered in detail in Mike Shulman's <a href="http://arxiv.org/abs/0810.1279" rel="nofollow">Set Theory for Category Theory</a>.</p> <p>The "big picture" way to think about it is to think that each time you do a power-set-type operation, you are constructing an object of a "higher order". You just then need to postulate high enough orders to do whatever it is you want. $U_0$ is the class of all sets, $U_1$ is the class of all classes of sets, etc.</p> <p>Once you are forming classes of classes, etc., you are strictly outside what is provably consistent with ZFC, but set theorists routinely consider much, much stronger theories that do not seem to harbor contradiction. So it's probably okay.</p> http://mathoverflow.net/questions/120090/are-integers-real/120093#120093 Answer by arsmath for Are integers real? arsmath 2013-01-28T09:34:09Z 2013-01-28T09:34:09Z <p>We do in fact have many different sets isomophic to $\mathbb{Z}$. The correct definition of $\mathbb{Z}$ is a set with some operations on it that satisfy some axioms. This is easier to see with $\mathbb{N}$, which is a set with a 0, a "plus one" operations, that satisfies mathematical induction. There is only one such set up to oeration-preserving isomorphism.</p> <p>$\mathbb{R}$ can be axiomatized similarly, as a complete Archimedean ordered field (see the synthetic approach section <a href="http://en.wikipedia.org/wiki/Axiomatic_theory_of_real_numbers" rel="nofollow">at Wikipedia</a>). Again there is only one such set up to operation-preserving isomorphism. You can identify in that set a copy of $\mathbb{Z}$, so you can treat $\mathbb{Z}$ as a subset. But this is a matter of notation. If you had a compelling reason to make it disjoint, you could require that, instead.</p> <p>We say "the integers" or "the reals" because there is only one, up to isomorphism. Any theorem you prove about $\mathbb{Z}$ or $\mathbb{R}$ that doesn't use the internal representation will transfer to any isomorphic copy, so we don't need to know which one.</p> <p>There is a whole mathematical notion of types that actually underpins languages with more-complicated type systems, such as ML or Haskell. You can think of $\mathbb{Z}$ and $\mathbb{R}$ as a type in a type system. Some typing systems have a notion of "subtype", but let's suppose that you don't. Then $\mathbb{Z}$ and $\mathbb{R}$ are types, and there's a designated monomorphism $i$ from $\mathbb{Z}$ to $\mathbb{R}$. Then 2 + 3.5 is an overloaded operation that is syntactic sugar for $i(2) + 3.5$. I don't know if anyone has ever worked out a clear account of what mathematicians do from this point of view, but what they do is not very complicated.</p> <p>If you do want to allow subtypes, there's a notion of order-sorted algebra that allows the kind of overloading you probably have in mind. I don't know of a canonical link, but introductions are easy to find.</p> http://mathoverflow.net/questions/119561/mathematics-with-the-negation-of-ac/119567#119567 Answer by arsmath for Mathematics with the negation of AC arsmath 2013-01-22T13:58:54Z 2013-01-22T13:58:54Z <p>It's tempting to think that if you add the negation of axiom of choice, then you can prove things like "all sets of reals are Lebesgue measurable," but it's not quite that easy. To really get a grip on what AC adds to ZF, I suggest familiarizing yourself with Gödel's <a href="http://en.wikipedia.org/wiki/Constructible_universe" rel="nofollow">constructible universe</a>, at least in broad outlines. Gödel iteratively constructs (using only ZF) a hierarchy of sets, where (roughly) at each stage he adds the sets that set theory requires. This is a model of ZF. He then proves a surprising result: in this model, the axiom of choice holds. </p> <p>That means that if the negation of the axiom of choice is true, then floating out there, somewhere, outside this model, is a collection of sets that lacks a choice function. Nothing tells us where this mysterious collection is -- if you stick to the types of constructions Gödel uses, you'll miss it.</p> <p>What's interesting, then, is not so much the negation of AC, but the addition of axioms that let you <i>do</i> things that <i>imply</i> that the axiom of choice is false. One example would be the axiom "all sets of reals are Lebesgue measurable". Another, more dramatic one is the <a href="http://en.wikipedia.org/wiki/Axiom_of_determinacy" rel="nofollow">axiom of determinacy</a>, which implies that all sets of reals are measurable, and much else besides.</p> http://mathoverflow.net/questions/118874/structure-of-f-g-modules-over-a-non-commutative-ring/119083#119083 Answer by arsmath for Structure of f.g. modules over a non-commutative ring arsmath 2013-01-16T16:53:38Z 2013-01-16T16:53:38Z <p>The question is thoroughly explored in Chapter 3 of Nathan Jacobson's <i>Theory of Rings</i>. I took a quick look, and it looks like the analogous results go through in the noncommutative case. For example, Theorem 19 in Chapter 3 states that a finitely-generated module over a noncommutative principal ideal domain is a direct sum of cyclic modules.</p> http://mathoverflow.net/questions/118829/difference-between-generalized-gradient-and-subgradient/118850#118850 Answer by arsmath for Difference between 'generalized gradient' and 'subgradient' ? arsmath 2013-01-13T22:19:42Z 2013-01-13T22:19:42Z <p>Generalized gradients generalize subdifferentials. Subdifferentials are defined globally, and rely on the simple geometry of convex functions. If in a neighborhood a linear subspace only touches the graph of a convex function without crossing it, then you know for a fact it won't cross the graph later. So the intuition between the subdifferential is that you take each linear subspace of that is strictly below the convex function, and slide it up until it touches it. The subdifferential tells you the relation between the slopes of linear subspaces, and the points on the graph where the linear subspace will touch without crossing.</p> <p>For a general non-convex function, a linear subspace that touches the graph at one point can cross it again at another point. So you need a definition that works locally, only in a small neighborhood of the point. The definition of generalized gradient gives you that. It's reasonably easy to find information online about it. For example, the page on <a href="http://www.encyclopediaofmath.org/index.php/Clarke_generalized_derivative" rel="nofollow">Clarke generalized derivative</a> at Encyclopedia of Mathematics gives a quick introduction. Clarke's book, <i>Optimization and nonsmooth analysis</i>, is nice. A more recent book reference is Rockafeller and Wets, <i>Variational analysis</i>, which thoroughly covers this and related topics.</p> <p>I quickly looked over the paper you link to, and I don't see why they need generalized gradients instead of just subdifferentials, though I didn't look too closely.</p> http://mathoverflow.net/questions/118815/small-implications-of-the-axiom-of-replacement Small Implications of the Axiom of Replacement arsmath 2013-01-13T15:10:37Z 2013-01-13T17:25:39Z <p>The axiom of replacement implies the existence of sets larger than usual in mathematical practice, but can be used to prove theorems about sets of real numbers, such as Borel determinacy. This is interesting because it suggests there's some sort of recursive procedure that makes sense for sets of reals, but is not provable in ZC alone. This procedure seems like it would be of independent interest from the question of the existence of sets beyond $V_{\omega + \omega}$ in the cumulative heirarchy.</p> <p>Is there a weaker axiom or recursive set of axioms that can be added to ZC that imply exactly the implications of replacement that hold for sets in $V_{\omega + \omega}$, one that explains the kind of additional constructions that replacement permits you to make? </p> http://mathoverflow.net/questions/2525/different-definitions-of-the-dimension-of-an-algebra/118767#118767 Answer by arsmath for Different definitions of the dimension of an algebra arsmath 2013-01-12T22:21:44Z 2013-01-12T22:21:44Z <p>There's actually a Gelfand-Kirillov transcendence degree for a noncommutative algebra over a field that generalizes the classical transcendance degree. Gelfand and Kirillov introduced it to prove that the field of fractions for the $n$th Weyl algebra is different for different $n$. <a href="http://www.ams.org/journals/tran/1996-348-07/S0002-9947-96-01702-3/S0002-9947-96-01702-3.pdf" rel="nofollow">Here's a paper</a> by James Zhang that gives the definition and calculates it for several examples.</p> <p>There's actually a pretty nice theory for when you can form a field of fractions. A Noetherian domain always has a field of fractions, in a construction that closely resembles the commutative case. (This is a special case of <a href="http://en.wikipedia.org/wiki/Goldie%27s_theorem" rel="nofollow">Goldie's Theorem</a>.) For a non-Noetherian domains, things are more complicated: the domain can fail to have a field of fractions, and when it does, it won't usually have a unique smallest field it embeds into (unlike the commutative or noncommutative Noetherian case). </p> http://mathoverflow.net/questions/117791/what-axioms-are-stronger-than-the-axiom-of-choice/117793#117793 Answer by arsmath for What axioms are stronger than the Axiom of choice? arsmath 2013-01-01T18:06:31Z 2013-01-01T18:06:31Z <p>The <a href="http://en.wikipedia.org/wiki/Axiom_of_constructibility" rel="nofollow">Axiom of Constructibility</a> implies the Axiom of Choice, as well as many other results independent of ZFC, such as the continuum hypothesis.</p> http://mathoverflow.net/questions/117705/the-practical-impact-of-set-theoretic-axioms-on-measure-theory The Practical Impact of Set-Theoretic Axioms on Measure Theory arsmath 2012-12-31T09:37:42Z 2012-12-31T19:50:00Z <p>The set-theoretic evidence is that we could probably safely add axioms to make many more sets measurable. For example, we could add axioms that would make <a href="http://www.encyclopediaofmath.org/index.php/Projective_set" rel="nofollow">projective sets</a> measurable.</p> <p>I'm curious what would be the implications for working analysts of such a move. I can see two potential ways in which it could potentially have an impact:</p> <ul> <li>Currently, proving measurability of sets is a somewhat fussy activity. With the additional freedom provided by extra constructions, the existing theory would become much simpler.</li> <li>There are existing theories that are already straining at the limits of what can proved measurable in ZFC. These theories could be usefully extended.</li> </ul> <p>I could also see that it potentially having no real impact. I'd be curious to hear which if any of these possibilities actually holds.</p> http://mathoverflow.net/questions/114395/result-that-follows-from-zfc-and-not-zf-but-are-strictly-weaker-than-choice/114418#114418 Answer by arsmath for Result that follows from ZFC and not ZF but are strictly weaker than choice arsmath 2012-11-25T13:15:07Z 2012-11-25T13:15:07Z <p>Eric Schecter's <i>Handbook of Analysis and Its Foundations</i> is an excellent source for questions like this in analysis. He has some <a href="http://www.math.vanderbilt.edu/~schectex/ccc/index.html" rel="nofollow">information on the subject</a> on his home page.</p> <p>One example he gives that's not equivalent to countable choice or the Boolean prime ideal theorem is the Hahn-Banach Theorem.</p> http://mathoverflow.net/questions/109957/what-is-the-definition-of-the-picard-group-of-a-non-necessarilly-commutative-ri/109966#109966 Answer by arsmath for what is the definition of the Picard group of a (non necessarilly commutative) Ring? arsmath 2012-10-18T01:54:27Z 2012-10-18T01:54:27Z <p>Amnon Yekutieli has studied the derived Picard group in a noncommutative setting. See <a href="http://arxiv.org/abs/math/9810134" rel="nofollow">Dualizing Complexes, Morita Equivalence and the Derived Picard Group of a Ring</a>.</p> http://mathoverflow.net/questions/44593/still-difficult-after-all-these-years Still Difficult After All These Years arsmath 2010-11-02T17:42:32Z 2011-12-31T22:42:29Z <p>I think we all secretly hope that in the long run mathematics becomes easier, in that with advances of perspective, today's difficult results will seem easier to future mathematicians. If I were cryogenically frozen today, and thawed out in one hundred years, I would like to believe that by 2110 the Langlands program would be reduced to a 10-page pamphlet (with complete proofs) that I could read over breakfast.</p> <p>Is this belief plausible? Are there results from a hundred years ago that have not appreciably simplified over the years? From the point of view of a modern mathematician, what is the hardest theorem proven a hundred years ago (or so)?</p> <p>The hardest theorem I can think of is the <a href="http://en.wikipedia.org/wiki/Riemann_mapping_theorem" rel="nofollow">Riemann Mapping Theorem</a>, which was first proposed by Riemann in 1852 and (according to Wikipedia) first rigorously proven by Caratheodory in 1912. Are there harder ones?</p> http://mathoverflow.net/questions/39762/vopenkas-principle-at-small-cardinals Vopenka's Principle at Small Cardinals arsmath 2010-09-23T15:13:52Z 2011-12-31T22:38:12Z <p>I'm trying to understand Vopěnka's Principle, which is a large cardinal axiom. One version of the principle is that there does not exist a proper class of directed graphs such that there are no homomorphisms between any two graphs in the class. This is a large cardinal axiom because it implies the existence of a proper class of measurable cardinals. If Vopěnka's principle is true, then it is not provable in ZFC, since it implies the consistency of ZFC. (It's falsity may be provable in ZFC.)</p> <p>I presume that since it functions as a large cardinal axiom, then it must fail at small cardinals (i.e. cardinals whose existence is provable within ZFC, such as $\aleph_\omega$). Is there an explicit construction for any such cardinal $\kappa$ there exists a set of graphs of size $\kappa$ such that Vopěnka's Principle fails for that set? In other words, there are no two homomorphisms between the two graphs in that set? </p> <p>I can come up with a construction for $\aleph_0$, but that's it. (For $\aleph_0$, the set of directed cycle graphs with a prime number of vertices does the trick, I think.)</p> http://mathoverflow.net/questions/45448/aleph-0-as-a-large-cardinal Aleph 0 as a large cardinal arsmath 2010-11-09T15:44:45Z 2011-12-31T22:35:48Z <p>The first infinite cardinal, $\aleph_0$, has many large cardinal properties (or would have many large cardinal properties if not deliberately excluded). For example, if you do not impose uncountability as part of the definition, then $\aleph_0$ would be the first inaccessible cardinal, the first weakly compact cardinal, the first measureable cardinal, and the first strongly compact cardinal. This is not universally true ($\aleph_0$ is not a Mahlo cardinal), so I am wondering how widespread of a phenomenon is this. Which large cardinal properties are satisfied by $\aleph_0$, and which are not?</p> <p>There is a philosophical position I have seen argued, that the set-theoretic universe should be uniform, in that if something happens at $\aleph_0$, then it should happen again. I have seen it specifically used to argue for the existence of an inaccessible cardinal, for example. The same argument can be made to work for weakly compact, measurable, and strongly compact cardinals. Are these the only large cardinal notions where it can be made to work? (Trivially, the same argument shows that there's a second inaccessible, a second measurable, etc., but when does the argument lead to more substantial jump?)</p> <p>EDIT: Amit Kumar Gupta has given a terrific summary of what holds for individual large cardinals. Taking the philosophical argument seriously, this means that there's a kind of break in the large cardinal hierarchy. If you believe this argument for large cardinals, then it will lead you to believe in stuff like Ramsey cardinals, ineffable cardinals, etc. (since measurable cardinals have all those properties), but this argument seems to peter out after a countable number of strongly compact cardinals. This doesn't seem to be of interest in current set-theoretical research, but I still find it pretty interesting.</p> http://mathoverflow.net/questions/59157/reference-request-independence-of-the-ultrafilter-lemma-from-zf/59202#59202 Answer by arsmath for Reference Request: Independence of the ultrafilter lemma from ZF arsmath 2011-03-22T16:40:10Z 2011-03-22T16:40:10Z <p>If you're just interested in context, rather than proofs, Eric Schechter's book <i>Handbook of Analysis and Its Foundations</i> talks about the relation between existence of ultrafilters, various weak notions of choice, and standard results in analysis.</p> http://mathoverflow.net/questions/55415/syntactically-capturing-complexity-classes/55418#55418 Answer by arsmath for Syntactically capturing complexity classes arsmath 2011-02-14T15:25:18Z 2011-02-14T15:25:18Z <p>This isn't exactly the same flavor as the results you mentioned, but there is an area known as <a href="http://en.wikipedia.org/wiki/Descriptive_complexity_theory" rel="nofollow">descriptive complexity</a> which tries to find syntactic characterizations of complexity classes in terms of properties definable in different logical languages.</p> <p><a href="http://en.wikipedia.org/wiki/Fagin%27s_theorem" rel="nofollow">Fagin's theorem</a> says that the class NP corresponds to existential second-order logic. It's sufficient to restrict yourself to graphs. In that case, computing a graph property is in NP if and only if it's given by an existential second-order formula.</p> <p>For P, there is a partial result that says in the presence of a linear order, P is equivalent to first-order logic with an additional <a href="http://en.wikipedia.org/wiki/Least_fixed_point" rel="nofollow">least fixed point</a> operator.</p> <p><a href="http://arxiv.org/abs/1001.2572" rel="nofollow">This paper</a> by Martin Grohe begins with a survey of the area.</p> http://mathoverflow.net/questions/54007/is-there-any-straightforward-way-to-substitute-for-gaussian-brownian-assumptions/54017#54017 Answer by arsmath for Is there any straightforward way to substitute for Gaussian/Brownian assumptions in financial mathematics? arsmath 2011-02-01T20:30:58Z 2011-02-02T13:59:40Z <p>Without additional assumptions, the answer is basically no, not in any great generality. The derivation of Black-Scholes requires that you can perfectly hedge movements in the option using a stock and a bond. If the underlying stock price process has jumps, then you have jumps in the value of the option, and you can't hedge those jumps using only two assets. (There is one exception &mdash; if the process is Poisson, then you can hedge the jumps, but as soon as you have jumps of more than one size then you're stuck.)</p> <p>The additional assumption is some rule to determine how the option value jumps when the stock price jumps. One rule is that the jumps are "idiosyncratic risk", and therefore are not hedged. This is called the Merton jump-diffusion model. There's plenty of material online about this model. From a quick Google search, <a href="http://www.math.nyu.edu/~benartzi/Slides5.2.pdf" rel="nofollow">these slides</a> look pretty good. </p> http://mathoverflow.net/questions/53138/commutative-ring-of-finite-global-dimension Commutative Ring of Finite Global Dimension arsmath 2011-01-24T22:19:13Z 2011-01-24T23:15:22Z <p>The only examples of commutative rings of finite <a href="http://en.wikipedia.org/wiki/Global_dimension" rel="nofollow">global dimension</a> I know are either:</p> <ul> <li>Dedekind domains (and fields as a degenerate special case)</li> <li>Regular local rings</li> <li>Rings constructed from the previous examples by taking direct sums, or forming the rings of polynomials over a ring of finite global dimension.</li> </ul> <p>Are there other examples? In particular, are there other examples that are finite-dimensional over a field $k$?</p> <p>(Examples of rings of finite global dimension are easier to come by in the noncommutative case, but I'm specifically curious about the commutative case.)</p> http://mathoverflow.net/questions/52186/self-dual-complete-category Self-dual Complete Category arsmath 2011-01-15T21:35:15Z 2011-01-22T01:20:39Z <p>Is there an example of a self-dual complete category that is not a partially-ordered set?</p> http://mathoverflow.net/questions/14253/monoid-ring-and-some-structure-within-it-how-is-it-called/52627#52627 Answer by arsmath for monoid ring and some structure within it - how is it called? arsmath 2011-01-20T15:12:23Z 2011-01-20T15:12:23Z <p>In your decomposition in terms of $g_3$, I'm assuming $r_3 \in R$, otherwise what I'm about to say is completely useless.</p> <p>If you have a monoid ring that has dimension $n$, then the monoid itself must have exactly $n$ elements. What must be happening is that there is an "extra" relation implied by the other relations, but wasn't taken into account when you came up with your list of possible words. I can't think of a good example, so here's a bad one. Let's suppose you had the monoid with two generators $x$ and $y$, and the relation $x^3 = y$. You could take as your list of words any word that doesn't contain $x^3$, like $x, y, xy, x^2y, xyx, \ldots$, but then you're missing relations like $x^2 y = y x^2$.</p> <p>There's a standard procedure for determining whether you are "missing" relations in that way. It's goes by the names "diamond lemma" or sometimes "noncommmutative Grobner basis". The procedure doesn't always terminate, but when it does it gives you a way of telling whether you have a basis of words for the ring.</p> <p>It's easy to apply to the example you give, but in that case there are no "missing" relations, and thus the ring is infinite-dimensional over $R$, and there is no such $g_3$.</p> http://mathoverflow.net/questions/50971/how-to-make-ext-and-tor-constructive/51000#51000 Answer by arsmath for How to make Ext and Tor constructive? arsmath 2011-01-03T09:36:41Z 2011-01-03T09:36:41Z <p>This is not exactly what you're asking, but <i>computationally</i> there is no problem with Ext and Tor over finitely-generated commutative algebras over a field. The algorithms are described in Eisenbud's commutative algebra book, as well as other places. For schemes, you can use the Cech definitions. I don't know to what extent these depend on choice to prove that they compute the right thing, since the theory relies on injective resolutions at various points, even though they are avoidable in calculations.</p> <p>It seems to me that you could constructively handle injective modules the same way that you constructively handle the algebraic closure of a field. (Does the existence of the algebraic closure of $\mathbb{Q}$ require choice?) I don't think you literally need the whole injective module, just as in the way you usually don't literally need the whole algebraic closure of a ring. You adjoin roots of polynomials/solutions of linear equations as necessary. I don't know if you can give a constructive proof that this stops, though.</p> http://mathoverflow.net/questions/50703/defining-multiplication-in-polynomials-over-rings-of-matrices/50721#50721 Answer by arsmath for Defining Multiplication in Polynomials over Rings of Matrices arsmath 2010-12-30T12:44:24Z 2010-12-30T12:44:24Z <p>I have two answers for you, depending on what you have in mind.</p> <p>You want to add an $x$ to the ring of 2x2 matrices, such that while $x$ commutes with multiples of the identity, it doesn't commute with anything else. You can adjoin a noncommuting indeterminate by using what's called the free product. You take the two rings $M_2(\mathbb{R})$ and $\mathbb{R}[x]$, and then you form the free product over $\mathbb{R}$. Wikipedia has an entry on the <a href="http://en.wikipedia.org/wiki/Free_product" rel="nofollow">free product</a> of groups. The construction for rings is fairly similar.</p> <p>That construction has one weakness: $x$ will not satisfy any relations. There relations that all 2x2 matrices will satisfy, but $x$ in the free product won't. Rings all of whose elements satisfy identities are known as <a href="http://en.wikipedia.org/wiki/Polynomial_identity_ring" rel="nofollow">polynomial identity rings</a>. For example, any four 2x2 matrices satisfy an identity of degree four (described <a href="http://gilkalai.wordpress.com/2009/05/12/the-amitsur-levitski-theorem-for-a-non-mathematician/" rel="nofollow">in this blog post</a>). So any three elements of $M_2(\mathbb{R})$ and $x$ should satisfy that relation. (I don't know if all possible relations are generated from specializing this one relation. There could be other relations that rely on specific properties of specific matrix elements.)</p> http://mathoverflow.net/questions/50458/topological-groups-and-families-of-pseudometrics Topological Groups and Families of Pseudometrics arsmath 2010-12-26T23:51:25Z 2010-12-28T20:10:07Z <p>The topology on a topological group is generated by a family of pseudometrics. The only proof I know passes through uniform spaces (by which I mean the entourage definition): A topological group has a uniformity and by a theorem of Weil, every uniformity comes from a family of pseudometrics. Is there a direct construction of the pseudometrics that bypasses Weil's theorem?</p> http://mathoverflow.net/questions/42446/metric-spaces-as-algebraic-systems/50457#50457 Answer by arsmath for metric spaces as algebraic systems arsmath 2010-12-26T23:30:18Z 2010-12-26T23:30:18Z <p>I have to believe that this construction was previously known, though I can't point to a precise reference. The maps that preserve the relations are exactly the non-expansive maps. The category of metric spaces with non-expansive maps is a reasonably standard category (for example, Adamek, Herrlich, and Strecker include it as one of the standard examples, <b>Met</b>). It occurred to me years ago that you could write this category in terms of relations as above, so it must have occurred to many people over the years, and someone must have written it down somewhere. I did a couple of Google searches, but I didn't find anything directly relevant. The category was first introduced by John Isbell, if that helps.</p> http://mathoverflow.net/questions/50432/a-noetherian-proof-of-zariskis-main-theorem/50435#50435 Answer by arsmath for A noetherian proof of Zariski's Main Theorem? arsmath 2010-12-26T16:18:36Z 2010-12-26T16:18:36Z <p>There's a purely algebraic proof in some <a href="http://www.math.lsa.umich.edu/~hochster/615W10/615.pdf" rel="nofollow">lecture notes</a> by Mel Hochster. He explains the translation into the language of varieties, as well.</p> http://mathoverflow.net/questions/131021/calculate-the-tendency-of-a-set-of-samples Comment by arsmath arsmath 2013-05-18T07:36:25Z 2013-05-18T07:36:25Z These are pretty close to the same thing -- the beta coefficient from the regression can be written in terms of the correlation -- so there's not much to choose from. The question of what makes a good model for your application depends on your application. I would have to think there's a large medical literature on modelling normal and abnormal heart behavior, so I suggest looking there. http://mathoverflow.net/questions/112715/why-when-classification-of-simple-objects-is-simple-e-g-unknown-classifica Comment by arsmath arsmath 2013-05-10T16:51:37Z 2013-05-10T16:51:37Z There seems to be some dividing line where tightly constrained cases have a series/sporadic structure, and relaxing the constraint too much destroys this. For example, the classification of finite simple Moufang loops adds one additional series to the list of finite simple groups, while relaxing this slightly to Bol loops destroys any chances of classification. http://mathoverflow.net/questions/128446/general-theory-of-left-exact-localization Comment by arsmath arsmath 2013-04-23T22:51:59Z 2013-04-23T22:51:59Z Thanks, Ricardo, that does help. (The terminology is slightly different than I used above, so if anyone is curious it's section 5.6.) http://mathoverflow.net/questions/128446/general-theory-of-left-exact-localization Comment by arsmath arsmath 2013-04-23T10:31:04Z 2013-04-23T10:31:04Z I don't see where Krause addresses left-exactness. http://mathoverflow.net/questions/126609/how-to-refer-to-a-theorem-that-you-have-shown-to-be-wrong Comment by arsmath arsmath 2013-04-05T12:58:11Z 2013-04-05T12:58:11Z I would go with a neutral term. I suggest &quot;claim&quot;. http://mathoverflow.net/questions/124998/was-the-early-calculus-inconsistent/125667#125667 Comment by arsmath arsmath 2013-04-02T21:04:35Z 2013-04-02T21:04:35Z I don't agree with the opening paragraph of this answer. &quot;Consistent&quot; is a perfectly clear English word that means, in this context, &quot;not self-contradictory&quot;. If a person is inconsistent, they are not saying or doing the same things over time. http://mathoverflow.net/questions/120729/cosheafification/124929#124929 Comment by arsmath arsmath 2013-03-20T22:35:21Z 2013-03-20T22:35:21Z It's way up there in the hierarchy. Vopenka himself thought it was false -- that's why he proposed it as an argument against the largest of large cardinal axioms -- but I think most set theorists think it's independent. Even if it's fine, it means that you're saying you have no control over the size of cosheafification: it could turn a countable set into a set of cardinality so large that it dwarfs all sets that appear in day-to-day mathematics. http://mathoverflow.net/questions/120729/cosheafification/124929#124929 Comment by arsmath arsmath 2013-03-19T07:00:46Z 2013-03-19T07:00:46Z Your easier-to-check version of the adjoint functor theorem relies on a large cardinal axiom, which is therefore not provable in ZFC unless ZFC is inconsistent. http://mathoverflow.net/questions/124057/one-big-hilbert-scheme Comment by arsmath arsmath 2013-03-14T21:43:53Z 2013-03-14T21:43:53Z I think he means so that he can accept it as the answer. http://mathoverflow.net/questions/121406/where-in-ordinary-math-do-we-need-unbounded-separation-and-replacement/123583#123583 Comment by arsmath arsmath 2013-03-05T10:47:40Z 2013-03-05T10:47:40Z Game theory is not ordinary mathematics? Gale-Stewart games arose as an attempt to extend the determinacy of finite-length zero-sum games to infinite games. David Gale was a mathematical economist who's most famous for the Gale-Shapley stable matching algorithm that helped Shapley win the Nobel Prize in economics. Once determinacy of Gale-Stewart games for open and closed sets was shown, it's a natural question to wonder if it holds for sets higher in the Borel hierarchy. The theory of infinite games just turned out to be really, really hard. http://mathoverflow.net/questions/123302/a-problem-in-functional-analysis-that-erdos-solved-in-2-lines Comment by arsmath arsmath 2013-03-01T20:18:08Z 2013-03-01T20:18:08Z This should be closed in favor of the stack exchange question, in light of the quality of the answers there. It's hard to do much better than have the actual source of the anecdote answer. http://mathoverflow.net/questions/123264/market-clearing-price-vector-in-an-aggregate-demand-system/123265#123265 Comment by arsmath arsmath 2013-03-01T10:35:43Z 2013-03-01T10:35:43Z It's not a convex optimization problem, so you can't plausibly solve a problem with a million variables, but for problems with 20 variables, non-convex solvers will work. Google &quot;Computational General Equilibrium&quot; for more information. http://mathoverflow.net/questions/121565/mathematicians-whose-works-were-criticized-by-contemporaries-but-became-widely-ac/121581#121581 Comment by arsmath arsmath 2013-02-13T09:28:15Z 2013-02-13T09:28:15Z And then two years later, computer scientists figure out how to promote the assistant to the boss, and now that mathematics has been automated, all human mathematicians become instantly obsolete. Alternate scenario: thanks to the proof assistants, the world's supply of constructive theorems can be supplied by one guy in his basement and constructivism dies as an active research project. Since constructive proof assistants are a solved problem, all of the grant money in computer science goes to the unsolved problem of proof assistants for ZFC. http://mathoverflow.net/questions/121565/mathematicians-whose-works-were-criticized-by-contemporaries-but-became-widely-ac/121581#121581 Comment by arsmath arsmath 2013-02-12T20:17:44Z 2013-02-12T20:17:44Z Intuitionistic logic is of <i>mathematical</i> interest. Some things are intuitionistic, the way some things are nonabelian groups. http://mathoverflow.net/questions/121565/mathematicians-whose-works-were-criticized-by-contemporaries-but-became-widely-ac/121597#121597 Comment by arsmath arsmath 2013-02-12T15:18:49Z 2013-02-12T15:18:49Z Is this true? Of course not everyone is going to switch mathematical styles instantly, but was there any actual resistance? (The evidence that Paul Gordan said anything like &quot;That is not mathematics, that is theology&quot; is pretty thin.)