most recent 30 from http://mathoverflow.net 2013-05-25T01:53:09Z http://mathoverflow.net/feeds/user/19965 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100755/a-graded-ring-r-is-graded-local-iff-r-0-is-a-local-ring/101162#101162 rschwieb 2012-07-02T17:33:35Z 2012-07-02T17:33:35Z

<p>An elementary argument seems to work as well, even for monoids as mentioned previously.$\dagger$</p> <p>Suppose $M$ and $N$ are distinct maximal homogeneous right ideals. Then $M+N=R$, and there exists $m+n=1$ with $m\in M$ and $n\in N$. Because of the grading, the grade zero parts must be such that $m_0+n_0=1$, and because $M$ and $N$ are both proper and homogeneous, neither $m_0$ nor $n_0$ can be units of $R_0$. This implies $R_0$ is not local.</p> <p>By contrapositive then, we have shown if $R_0$ is local, then $R$ is graded local.</p> <p>$\dagger$ I convinced myself that there are maximal proper homogenous ideals, and that the sum of homogeneous ideals is again homogeneous. I hope those lemmas were not heat induced delirium.</p>

http://mathoverflow.net/questions/95779/bimodule-version-of-ibn rschwieb 2012-05-02T15:25:31Z 2012-05-02T15:37:20Z

<p>Hello all,</p> <p>Does anyone have an example in mind of a ring $R$ for which $R^n\cong R^m$ <em>as $R,R$ bimodules</em> for some positive integers $n\neq m$?</p> <p>I would be a little surprised if someone showed no such thing could exist, but that would also be a welcome answer. </p> <p>Thanks!</p> <p>P.S.: Naturally such a ring could not have <a href="http://en.wikipedia.org/wiki/Invariant_basis_number" rel="nofollow">IBN</a>. I don't recall deciding whether or not the "easiest" ring without IBN (the endomorphism ring of an $\aleph_0$-dimensional vector space $V$) precluded this, so that is a starting point.</p>

http://mathoverflow.net/questions/84032/short-course-suggestions-for-high-school-students/84044#84044 rschwieb 2011-12-21T21:41:22Z 2011-12-21T21:41:22Z

<p>Some elementary graph theory with the intent of solving traversal or traveling salesman type problems is pretty easy at that level. Introducing incidence matrices can also be a foothold for learning matrix multiplication.</p>

http://mathoverflow.net/questions/78366/koethe-conjecture/83759#83759 rschwieb 2011-12-18T01:33:17Z 2011-12-18T01:33:17Z

<p>Not solved. Many special cases though. The paper cited above looks nice, and another short survey is given in Lam's First Course in Noncommutative rings, around page 164. There are a bunch of equivalent statements there, but maybe the above paper covers them all.</p>

http://mathoverflow.net/questions/10788/name-for-semiring-with-weakened-annihilation-law/83547#83547 rschwieb 2011-12-15T18:49:20Z 2011-12-15T20:10:16Z

<p>If you were looking for a field with convenient terminology for its structures, I should warn you that the field of semirings is pretty bad :)</p> <p>In </p> <p>"Graphs, dioids and semirings: new models and algorithms" by Michel Gondran, Michel Minoux</p> <p>they use "presemiring" to mean a set with two associative binary opearations, + and X where + is commutative and X distributes over + on both sides.</p> <p>To make a presemiring a semiring, they require identity elements for both operations, and require that the additive identity 0 is absorbing, as you describe.</p> <p>I wouldn't put too much stock in the one book though. Really there are so many authors throwing around so many terms about these things it's probably useless to find out which is the most common.</p>

http://mathoverflow.net/questions/74315/simple-rings-are-v-rings/83540#83540 rschwieb 2011-12-15T18:13:13Z 2011-12-15T18:13:13Z

<p>Not all simple rings are V-rings. See</p> <p>Osofsky, B. L. On twisted polynomial rings. J. Algebra 18 1971 597–607. </p> <p>In the middle of page 606, an example (example b) is given of a simple domain that is not a V-ring.</p> <p>Interestingly, at the end of the paper, Dr. Osofsky comments that it "seems highly unlikely" that simple rings can have <em>both</em> injective and noninjective simple modules.</p>

http://mathoverflow.net/questions/26607/what-does-the-semiring-of-ideals-of-a-ring-r-tell-us-about-r/83466#83466 rschwieb 2011-12-14T20:20:34Z 2011-12-14T20:55:21Z

<p>I've been interested in this lately. Hopefully you have seen this?</p> <p>Golan, Jonathan S.(IL-HAIF)</p> <p>Semirings for the ring theorist. </p> <p>Rev. Roumaine Math. Pures Appl. 35 (1990), no. 6, 531–540. </p> <p>Golan cautions that treating semirings as 'poor man's rings' is not always good. They can really be different animals altogether. I think someone noted above that the semiring of ideals is additively idempotent. In a sense, this is as far as you can get from an additive group.</p> <p>The compensation for the loss of the additive group is the complete lattice structure compatible with the multiplication. That is, if A\leq B, then AC\leq BC and CA\leq CB.</p> <p>Ring theorists have been saying things about rings via the lattice of one sided ideals for years! The lattices of onesided ideals are almost as nice, except you lose compatibility of multiplication with the order, and there is no longer a twosided identity for the semiring. These are called <em>quantales</em> in some places.</p>

http://mathoverflow.net/questions/42647/rings-in-which-every-non-unit-is-a-zero-divisor/83454#83454 rschwieb 2011-12-14T18:53:46Z 2011-12-14T18:53:46Z

<p>I have studied a noncommutative version of this. There is such a thing called a <em>right cohopfian ring</em> in the sense that if the right annihilator of r is zero, then r is a unit. If you add commutativity and look at the contrapositive, you get that nonunits are zero divisors.</p> <p>I don't think this terminology has caught on, but here is the rationale. A "cohopfian object" is one for which injections are surjections. Looking on elements of the ring as maps sending x-->rx, we are saying that if such a map is injective, it is surjective.</p> <p>Right Artinian, right perfect and strongly-pi regular rings (commutative VNR rings are strongly pi regular) are all right and left cohopfian. Finding a one-sided cohopfian ring seems tough, but Varadarajan did it here:</p> <p>"Varadarajan, K. Hopfian and co-Hopfian objects. Publ. Mat. 36 (1992), no. 1, 293–317."</p> <p>I think someone has noted above that right cohopfian rings have ot be Dedekind finite, and it is interesting that Dedekind finite=right Hopfian=left Hopfian.</p> <p>Too bad I didn't see this a year ago :)</p>

http://mathoverflow.net/questions/1083/do-good-math-jokes-exist/4050#4050 rschwieb 2012-06-26T14:57:52Z 2012-06-26T14:57:52Z

Bwahaha! I like this one... :)

http://mathoverflow.net/questions/95779/bimodule-version-of-ibn/95780#95780 rschwieb 2012-05-02T17:16:18Z 2012-05-02T17:16:18Z

@BenjaminSteinberg Thank you for the connection. To date I have not had the opportunity to learn anything about Hochschild cohomology.

http://mathoverflow.net/questions/95779/bimodule-version-of-ibn/95780#95780 rschwieb 2012-05-02T15:59:37Z 2012-05-02T15:59:37Z

Nice! I should have thought about it longer! The &quot;center&quot; of a bimodule - I think I had one-sided module blinders on.

http://mathoverflow.net/questions/84058/are-all-commutative-rngs-ideals-of-commutative-rings/84060#84060 rschwieb 2011-12-22T02:59:27Z 2011-12-22T02:59:27Z

The &quot;Dorroh extension&quot; above completely answers the question, but I wanted to tack on this interesting paper I ran across recently: Burgess, W. D.(3-OTTW); Stewart, P. N.(3-DLHS) The characteristic ring and the &quot;best'' way to adjoin a one. J. Austral. Math. Soc. Ser. A 47 (1989), no. 3, 483–496. In addition to the Dorroh extension, I think they argue for another economical adjunction of one. I'm not really up to speed on the categorical properties of either extension, but it all looks pretty interesting.

http://mathoverflow.net/questions/62464/rings-with-all-modules-projective/62465#62465 rschwieb 2011-12-17T18:37:36Z 2011-12-17T18:37:36Z

I think in condition 5) &quot;*simple* right ideals&quot; was intended.

http://mathoverflow.net/questions/29745/semisimple-ish-rings/29748#29748 rschwieb 2011-12-17T12:58:55Z 2011-12-17T12:58:55Z

The type of right ideals which <i>do not</i> have such a complement are exactly the superfluous (or small) right ideals. As the excellent answer above shows, rings with $J(R)=0$ are the rings without small right or left ideals. Going one step further, semisimple rings (right Artinian +$J(R)=0$) are the rings without essential (or large) right ideals. It's interesting that &quot;no essential right ideals&quot; implies it's dual relative &quot;no superfluous right ideals&quot;. It's someone akin to right Artinian implying right Noetherian in rings.

http://mathoverflow.net/questions/57877/artin-jacobson-semisimple-rings-are-semisimple-constructively-too rschwieb 2011-12-16T20:39:42Z 2011-12-16T20:39:42Z

Apologies in advance, because I am not familiar with what is not allowed in constructive proofs. Are we to use only the definitions above, without the usual alternative characterizations of VNR and Artinian semisimple rings?

http://mathoverflow.net/questions/58395/a-problem-for-finite-dimensional-commutative-algebra/58399#58399 rschwieb 2011-12-16T14:44:56Z 2011-12-16T14:44:56Z

&quot;Commutative zero dimensional Gorenstein rings&quot; also have another popular name: Frobenius algebras(/rings). If memory is serving me correctly, then among commutative Artinian local rings, the rings with a unique minimal ideal are exactly the Frobenius rings.