User chris leary - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T19:03:34Z http://mathoverflow.net/feeds/user/8027 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/15028/do-you-find-your-students-are-less-competent-in-basic-algebra-and-arithmetic-and/89879#89879 Answer by Chris Leary for Do you find your students are less competent in basic algebra and arithmetic, and, if so, do you believe that this is due to overuse of calculators at an early level? Chris Leary 2012-02-29T16:33:00Z 2012-03-01T01:29:57Z <p><strong>[original answer by Chris Leary; tidied slightly by YC]</strong></p> <p>I am in sympathy with Kevin Buzzard's opinion that we mathematicians can become "grumpy old men." For several years (I was perhaps very naive), I labored under the belief that my students had a secondary school math background similar to mine. I have abandoned that belief. I have been at the same college now for over 25 years. I have noticed a decline in the preparation, but mostly in attitude, among our recent students. I wish I could say why this is the case, but I can't really.</p> <p>As far as technology is concerned, I remember an article published in some journal on technology in math education. The article appeared during the height of the calculus reform movement in the US and was based on the authors' experiences at Oklahoma State. One of their conclusions was that, in the hands of talented students, calculators et al enhance students performance, but for less talented students, and I still remember the phrase, technology "adds one more layer of obfuscation" between the student and the material.</p> <p>I believe there is something fundamentally wrong with how the US mathematics educational system functions in primary and secondary school. I don't think technology itself is the main culprit. How the technology is used is crucial.</p> <p>A bigger problem is teacher preparation. My college has a school of education and the struggles of the elementary education people with mathematics are legendary. They actively resist learning anything about the math they will be teaching and only want to learn algorithms for solving problems. Even prospective secondary school teachers are not immune. A former student of mine in abstract algebra was incensed at having to learn about factoring polynomials, claiming that she was going to be a teacher, already knew how to factor, and didn't see any value in learning about polynomial rings. Unfortunately, she displayed an amazing inability to factor quadratics on an exam. So student attitudes are sometimes working against us as well.</p> <p>What's wrong, and how to fix it, are not simple questions. I think there is a complex mixture here. Technology is a convenient target (and the crticism is not wholly unjustified). However, educational philosophy and policy, and societal factors, probably play a significant role as well. I'll stop here, because the more I think about these issues, the more discouraged I become.</p> http://mathoverflow.net/questions/10419/depressed-graduate-student/84714#84714 Answer by Chris Leary for Depressed graduate student. Chris Leary 2012-01-02T00:08:43Z 2012-01-02T00:08:43Z <p>We're all approaching this from the perspective that the correct thing for your friend is to stay in mathematics. Perhaps not. I think of three very talented friends who eventually left mathematics. One left to become a successful orthodontist, another actually got his Ph.D., but decided to become a doctor instead. The third sold all his worldly possessions and took up with the Worker's Party. I have two children in theatre, and I remember the advice they received from faculty at the schools they were considering:"You've got to want this more than anything else." I don 't think mathematics is much different. Once the desire is gone, one can try to get it back, or consider if it might be time to move on. As a mathematician, I hope your friend finds some way to get back in the groove. Time will tell.</p> http://mathoverflow.net/questions/54065/about-injective-hull/54200#54200 Answer by Chris Leary for About injective hull Chris Leary 2011-02-03T14:44:18Z 2011-02-03T14:44:18Z <p>I'll follow up on what Karl said with an example closer to my own experience. Let Z be the ring of integers and p a positive prime. Then Z/pZ is injective as a Z/pZ - module, being a vector space over a field, whence Z/pZ is its own injective envelope (hull) as a Z/pZ module. However, the injective envelope of Z/pZ as an abelian group is Z(p^{infty}), which gives witness to Karl's statement that the injective envelope over A can be much larger than the injective envelope over A/ann(M). You can play this game with A any commutative Noetherian ring with 1, ann(M) = any maximal ideal of R, and M = A/I where I is the chosen maximal ideal. Karl's example presents very limited choice for I since k[[x]] is local. I think Proposition 2.27 and Lemma 4.24 of "Injective Modules" by Sharpe and Vamos present enough to figure out what is going on in the general case.</p> http://mathoverflow.net/questions/45528/errata-for-shafarevichs-basic-algebraic-geometry/45566#45566 Answer by Chris Leary for Errata for Shafarevich's Basic Algebraic Geometry? Chris Leary 2010-11-10T16:30:21Z 2010-11-10T16:30:21Z <p>Springer printed an Errata supposedly to be included in a study edition of the text to be published in 1977. I purchased the original edition directly from Springer and they mailed me the Errata at a later date. I don't really know what transpired in the intervening years.</p> http://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor/42656#42656 Answer by Chris Leary for Rings in which every non-unit is a zero divisor Chris Leary 2010-10-18T15:27:39Z 2010-10-18T15:27:39Z <p>From a more positive perspective, you are looking at rings with the property that every regular element is a unit. I have done some work with these rings. If the ring is commutative, the condition is equivalent to the ring being a quoring, i.e., it's its own classical ring of quotients. Non-commutative rings with the property must be quorings, but the converse is not necessarily true, as seen with von Neumann regular rings. I called rings with every regular element a unit Dedekind finite because they are characterized as R-modules by the property that every monic endomorphism is an isomorphism (the Dedekind definition of finite set; I picked up the name from L. N. Stout). This is not standard terminology I believe (see Lam's book "Lectures on Modules and Rings).</p> http://mathoverflow.net/questions/33816/closed-range-for-a-continuous-linear-transformation Closed range for a continuous linear transformation Chris Leary 2010-07-29T17:38:01Z 2010-07-29T18:38:58Z <p>I have a Banach space $B$ and a continuous linear transformation $F:B \rightarrow B\times B$. One of the induced transformations $F(1):B \rightarrow B$ and $F(2):B \rightarrow B$ into the factors of the product has closed range. Must F have closed range? I have the max norm on the product, i.e., $\|F(x) \| = max{\|F(1)(x)\|, \|F(2)(x)\|}$ for $x$ in $B$. I was hoping to use the minimum moduli of the $F(i)$ to provide an affirmative answer. </p> http://mathoverflow.net/questions/130139/principal-i-deal-domain-and-gcd Comment by Chris Leary Chris Leary 2013-05-09T03:38:18Z 2013-05-09T03:38:18Z You might want to try this on math.stackexchange.com. The idea is, as David said, to work as you would in $\mathbb{Z}$. The set of all linear combinations of your elements is an ideal of $R$, hence principal. Call the generator $g'$. You want to show that $g$ and $g'$ generate the same principal ideal. Or, if you are familiar with sums of ideals, you want $(g)=(a)+\ldots+(z)$. http://mathoverflow.net/questions/105138/is-primary-decomposition-still-important/105187#105187 Comment by Chris Leary Chris Leary 2012-08-22T20:52:21Z 2012-08-22T20:52:21Z Very nicely done. I tried to give more than plus one, but the site is too smart. I would've given several more if I could. http://mathoverflow.net/questions/28612/do-names-given-to-math-concepts-have-a-role-in-common-mistakes-by-students/28621#28621 Comment by Chris Leary Chris Leary 2012-02-29T03:47:26Z 2012-02-29T03:47:26Z Or my favorite from Lang's &quot;Algebra&quot;: &quot;a ring $R$ is said to be simple if it is semi-simple and ... .&quot; http://mathoverflow.net/questions/89406/example-of-reduced-modules Comment by Chris Leary Chris Leary 2012-02-24T16:26:50Z 2012-02-24T16:26:50Z $A = \mathbb{Z}_{n}$ is torsion and $C = \mathbb{Z}$ is tosion free. What can you say about $Hom(A,C)$ in this case? With the same $C$ and $A = \mathbb{Q}$, a divisible abelian group, what can you say about $Hom(A,C)$? http://mathoverflow.net/questions/87992/minimal-divisible-group/88023#88023 Comment by Chris Leary Chris Leary 2012-02-10T13:32:56Z 2012-02-10T13:32:56Z Your $E$ is what would be called the divisible hull (or envelope) of $A$. Essentially by definition, $E$ is essential over $A$. This is the result of several equivalent conditions that are summarized in Section 4.2 of Lambek's &quot;Lectures on Rings and Modules.&quot; For abelian groups, or modules over a PID, injective is equivalent to divisible. http://mathoverflow.net/questions/17778/books-you-would-like-to-see-translated-into-english/17781#17781 Comment by Chris Leary Chris Leary 2012-01-03T04:28:05Z 2012-01-03T04:28:05Z I don't know how easy the French is in EGA or SGA, but the French in Grothendieck's &quot;Sur quelques points d'algebre homologique&quot;, the Tohoku paper, is terribly opaque. I read French well, but that paper was an exceptional challenge. I would welcome an English translation. http://mathoverflow.net/questions/37610/demonstrating-that-rigour-is-important/37670#37670 Comment by Chris Leary Chris Leary 2012-01-01T23:54:44Z 2012-01-01T23:54:44Z If one wants to carry this to the extreme, any divergent series with the property that the n-th term goes to zero will converge on a calculator as the terms will eventually fall below the underflow value for the calculator, and hence be considered to be zero. http://mathoverflow.net/questions/80056/using-slides-in-math-classroom/80498#80498 Comment by Chris Leary Chris Leary 2011-11-09T18:05:37Z 2011-11-09T18:05:37Z In lower level math courses at a US university, most students do not want to be in the course to begin with. Providing slides, or even lecture notes, on-line would make it very tempting for them not to come to class. This is a consequence that must be weighed in any decision to use slides. If a student is physically present, you still have an opportunity to reach them. http://mathoverflow.net/questions/79717/does-a-notion-of-convex-graph-make-sense Comment by Chris Leary Chris Leary 2011-11-02T20:44:22Z 2011-11-02T20:44:22Z This is an intriguing idea. I was wondering if non-planarity of the graph could be a complication. I could be wrong, but it seems that a lot depends on the number of edges, or, the length, of paths between vertices. If so, would it be of benefit to phrase the axioms or definitions in terms of this? http://mathoverflow.net/questions/75417/exact-value-of-a-series Comment by Chris Leary Chris Leary 2011-09-14T17:24:34Z 2011-09-14T17:24:34Z The first thing that came to mind is that you might need Cesaro summability or something (cf. exercise 12-37 in Apostol, Mathematical Analysis), but I'm not sure that gets you where you where you want to go. It seems reasonable that this might be possible. Have you tried Hardy's Divergent Series? There may be a simple answer (one way or the other), but I'm not seeing it. http://mathoverflow.net/questions/33816/closed-range-for-a-continuous-linear-transformation/33824#33824 Comment by Chris Leary Chris Leary 2010-07-29T19:41:13Z 2010-07-29T19:41:13Z Does the answer change if the linear maps $F(i)$ must both be nonzero? http://mathoverflow.net/questions/33816/closed-range-for-a-continuous-linear-transformation/33824#33824 Comment by Chris Leary Chris Leary 2010-07-29T19:03:05Z 2010-07-29T19:03:05Z Thanks much, Bill. It is embarassing not to have considered mapping into one of the factors (I should have known better).