User greg marks - MathOverflow

Answer by Greg Marks for When is a topological group Hausdorff (separated)?

2012-06-13T17:31:03Z

You can probably find this result in a million places, one of which is N. Bourbaki, General Topology, Part 1, Chapter 3, Section 1.2, p. 223, Proposition 2.

Answer by Greg Marks for Why is it a good idea to study a ring by studying its modules?

2011-03-28T19:10:03Z

In homage to Serge Lang, I might suggest that your friend pick up any book on Morita theory and solve all the exercises. Less facetiously, I might point out that many interesting ring-theoretic properties can be characterized, sometimes unexpectedly, by properties of the (right) module category over the ring. For instance, consider what it means for all right $R$-modules to be injective, or what it means for all right $R$-modules to be projective, or what it means for all right $R$-modules to be flat, or what it means for all direct sums of injective right $R$-modules to be injective, or what it means for all direct products of projective right $R$-modules to be projective, or what it means for all flat right $R$-modules to be projective, or ...

Answer by Greg Marks for Group ring and left zero divisor.

2011-03-12T22:39:46Z

The condition $ab=0 \Rightarrow ba=0$ defines what are often called reversible rings, which, for example, have the property that the set of nilpotent elements is an ideal that coincides with the prime radical. A full matrix ring can't have this property, so you can construct counterexamples by taking any finite field $K$ and nonabelian group $G$ to which Maschke's theorem applies. An alternative example of a non-reversible group algebra is $K[G]$ where $K$ is the field of two elements and $G$ is the dihedral group of order $8$. Here the set of nilpotent elements does coincide with the prime radical (the ring is local artinian), but one can find elements $a,b \in K[G]$ with $ab=0$ but $ba \neq 0$.

Answer by Greg Marks for What are the reasons for considering rings without identity?

2011-02-11T17:15:15Z

For work related to radicals of rings, the Köthe Conjecture, etc., it's very useful to consider "rngs" (Louis Rowen's term for rings without identity).

Answer by Greg Marks for Direct sum of injective modules over non-Noetherian rings

2011-02-01T00:44:43Z

A module is called $\Sigma$-injective if a direct sum of arbitrarily (equivalently, countably infinitely) many copies of that module is injective. So it suffices to find an example of a $\Sigma$-injective module over a non-noetherian ring. Apart from silly examples such as a direct product of two rings one of which is one-sided noetherian and the other of which is not, the main theorem of C. Megibben, “Countable injective modules are $\Sigma$-injective,” Proc. Amer. Math. Soc. 84 (1982), no. 1, 8–10, says what the title indicates. This gives all sorts of examples of $\Sigma$-injective modules over non-noetherian rings.

Answer by Greg Marks for Mathematical "urban legends"

2011-01-28T20:38:41Z

I've heard stories about von Neumann chatting up recent Ph.D.'s and solving their thesis problems in his head. An incident along those lines is recounted in Sylvia Nasar's biography of John Nash. Perhaps someone here can shed some light on which von Neumann stories are purely mythological?

Answer by Greg Marks for Cool problems to impress students with group theory

2011-01-28T19:45:20Z

Quintic equations?

Answer by Greg Marks for Teaching proofs in the era of Google

2010-12-02T01:33:25Z

Representing someone else’s work as one’s own is plagiarism. Students can be expelled for it, and tenured professors can be fired for it.

This issue is addressed eloquently and emphatically in Section I of the Ethical Guidelines of the American Mathematical Society.

It is true that it’s easier to commit plagiarism now than before computers existed, just as it’s easier to rob a bank now than before automobiles existed.

Assuming that you’re permitting your students to look up solutions, and that they are providing proper citations of sources as necessary, there might be considerable pedagogical benefit to this exercise. For instance, they might discover that certain useful ideas and techniques occur over and over in solutions to some sort of problem.

Answer by Greg Marks for term for a "faithful" module

2010-11-30T01:45:47Z

For flat modules $M$, the condition you cite is equivalent to $M$ being faithfully flat. But some modules that are not flat also satisfy your condition; for example, let $k$ be a field, let $A = k[[ x_1, \ldots, x_n ]]$ where $n>1$, and let $M$ be the maximal ideal of $A$.

function that sums to zero over cube vertices

2010-11-27T01:00:33Z

Does anyone have an answer to the three-dimensional analogue of the 2009 Putnam Competition A1 problem, viz., if $f\colon \mathbb{R}^3 \rightarrow \mathbb{R}$ satisfies $\sum_{i=1}^8 f(a_i) = 0$ whenever $a_1, \ldots, a_8$ are the vertices of a cube, must $f$ be identically zero?

A few thoughts: Since the cube is not self-dual above dimension $2$, the solution (well, at least, my solution) to the $2$-dimensional problem doesn’t generalize. To try to show that the answer to the $3$-dimensional problem is “no,” one might try letting $\Omega$ be the set of ordered pairs $(X, f_X)$ where $X \subseteq \mathbb{R}^3$ is a subset and $f_X\colon \mathbb{R}^3 \rightarrow \mathbb{R}$ is a function that sums to zero over any eight points of $X$ that form the vertices of a cube. Ordering $\Omega$ in the usual way, we can invoke Zorn’s Lemma to obtain a maximal $(X, f_X) \in \Omega$. If $X \neq \mathbb{R}^3$, then every $x \in \mathbb{R}^3 \setminus X$ must, in two “incompatible” ways, be the eighth vertex of a cube whose remaining vertices lie in $X$. But I don't see how a contradiction arises from this (and of course one could make the same argument in two dimensions, where a contradiction does not arise).

Answer by Greg Marks for How to introduce notions of flat, projective and free modules?

2010-11-19T19:56:07Z

You might perhaps mention Lazard's theorem.

Answer by Greg Marks for Fundamental groups of topological groups.

2010-10-22T23:40:28Z

This question occurred as Advanced Problem 5889 in the Amer. Math. Monthly 80 (1973), no. 1, 82. It was listed as still unsolved five years later, in vol. 85, no. 10, p. 834, of the Monthly; however, my recollection is that it mysteriously vanished from the Monthly's "unsolved" list the next time this got updated, but without a solution having appeared in the interim.

Answer by Greg Marks for Rings in which every non-unit is a zero divisor

2010-10-22T23:09:09Z

Pace Chris Leary, the standard terminology is that a module all monomorphisms of which are automorphisms is said to be cohopfian (or co-Hopfian, if you're checking MathSciNet). A Dedekind-finite (a.k.a. directly finite) module usually means a module whose left invertible endomorphisms are also right invertible, equivalently, a module that is not isomorphic to any proper direct summand of itself.

T. Y. Lam, in his book Lectures on Modules and Rings, pp. 320–322, calls a noncommutative ring in which every regular element (i.e. neither right nor left zero-divisor) is a unit a classical ring, and he provides various examples. One that has not already been mentioned here is that any right (or left) self-injective ring is classical. Right self-injective rings need not have the property that every element that is merely not a left zero-divisor is a unit; interestingly, for right self-injective rings the latter condition is equivalent to the ring being Dedekind-finite (in the sense of the preceding paragraph), and also equivalent to the ring having stable range 1 (see Y. Suzuki, “On automorphisms of an injective module,” Proc. Japan Acad. 44 (1968), 120–124, and G. F. Birkenmeier, “On the cancellation of quasi-injective modules,” Comm. Algebra 4 (1976), no. 2, 101–109).

Answer by Greg Marks for Demonstrating that rigour is important

2010-09-08T16:47:39Z

In response to the request for an example of a statement that was widely but erroneously believed to be true: does Gauss's conjecture that $\pi(n) < \operatorname{li}(n)$ for every integer $n \geq 2$, disproved by Littlewood in 1914, qualify?

Answer by Greg Marks for Refereeing a Paper

2010-08-26T18:48:04Z

Re question 5 ("How long should one spend trying to understand an argument?"): an interesting case was Hales's proof of the Kepler Conjecture, in which the 12-member team of Annals referees spent five years before resigning themselves to being unable to completely certify the validity of the proof.

Answer by Greg Marks for Interesting applications (in pure mathematics) of first-year calculus

2010-08-11T20:25:00Z

A few years ago I gave a departmental colloquium talk, aimed at beginning M.A. students, on "An application of calculus to ring theory." A slightly facetious little abstract can be found here. The example establishing the main results—very well known to workers in commutative ring theory—was the ring of germs at $0$ of class $C^{\infty}$ functions on $\mathbb{R}$. A bit of calculus is needed in verifying the requisite properties.

Answer by Greg Marks for endomorphism ring of a finite-length module

2010-08-05T20:56:00Z

What is true is that the endomorphism ring of any finite length module is semiprimary.

Answer by Greg Marks for Examples of badly behaved derivatives

2010-08-05T20:24:18Z

In Y. Katznelson and K. Stromberg’s paper “Everywhere differentiable, nowhere monotone, functions,” Amer. Math. Monthly 81 (1974), no. 4, 349–354, there is a construction, based on somewhat similar ideas to those described in Majer’s above mentioned article, whereby for any two disjoint countable sets $A, B \subset \mathbb{R}$ there exists a differentiable function whose derivative equals $1$ on $A$ and is less than $1$ on $B$. So if $A$ and $B$ are both dense in $\mathbb{R}$ then this derivative must be discontinuous at every point of $B$.

Answer by Greg Marks for When and how is it appropriate for an undergraduate to email a professor out of the blue?

2010-08-05T19:00:26Z

The reasons I became a math professor are that I love mathematics and I love helping students. I constantly try to elicit questions from my own students. I would be delighted, not annoyed, to receive e-mail from a student who was sufficiently interested in something I had said or written to send me comments or questions.

Answer by Greg Marks for Terminology: Algebras where long strings of products are 0?

2010-06-30T05:23:32Z

It seems to me that Jordan and Dotsenko are giving different answers from one another, and I agree with Dotsenko’s. The condition Thurston has stated is the definition of $A_{+}$ being nilpotent. “Locally nilpotent” is a weaker condition. There are many examples of nonunital rings $A_{+}$ that are locally nilpotent (meaning for any finite set $a_1, \ldots, a_k \in A_{+}$ there exists $n \in \mathbb{N}$ such that $a_{i_1} \cdots a_{i_n} = 0$ provided every $i_j \in \lbrace 1, \ldots, k\rbrace$) but not nilpotent (meaning there exists $n \in \mathbb{N}$ such that $a_1 \cdots a_n = 0$ provided every $a_i \in A_{+}$, which is the condition Thurston stated). Even better: two nice (and quite different) examples of a locally nilpotent prime nonunital ring can be found in E. I. Zelmanov, “An example of a finitely generated primitive ring,” Sibirsk. Mat. Zh. 20 (1979), no. 2, 423, 461, and J. Ram, “On the semisimplicity of skew polynomial rings,” Proc. Amer. Math. Soc. 90 (1984), no. 3, 347–351. (Of course, if one merely wants an example where $A_{+}$ is locally nilpotent but not nilpotent—and so does not satisfy Thurston’s condition—one could take something like $A_{+} = \bigoplus_{i=2}^{\infty} 2\mathbb{Z}/2^i\mathbb{Z}$.)

N.B. Mathematical Reviews incorrectly lists the title of Zelmanov’s paper as “An example of a finitely generated primary ring.” It’s listed correctly in Zentralblatt. Possibly the problem lies in the translation from the original Russian; the condition primitive in the English translation of the paper (Siberian Math. J. 20 (1979), no. 2, 303–304) is what we would today call prime.

Answer by Greg Marks for Isomorphism between direct sum of modules

2010-06-30T00:02:50Z

There are even counterexamples in the case $A = {\mathbb Z}$: at the end of B. Jónsson’s paper “On direct decompositions of torsion-free abelian groups,” Math. Scand. 5 (1957), 230–235, an example is given of torsion-free, finite-rank abelian groups $B \not\cong C$ such that $B \oplus B \cong C \oplus C$.

A further counterexample, which I believe has been pointed out independently by L. S. Levy, R. Wiegand, and R. G. Swan: let $A$ be the coordinate ring of the real 2-sphere and ${}_AM$ the module for the tangent bundle; then $M \oplus M$ is free of rank $4,$ but $M$ is not free of rank $2$.

In the positive direction, K. R. Goodearl has proved (“Direct sum properties of quasi-injective modules,” Bull. Amer. Math. Soc. 82 (1976), no. 1, 108–110, Theorem 3) that if $M$ and $N$ are quasi-injective modules over a ring (commutative or not), then $M^n \cong N^n$ implies $M \cong N$ for any positive integer $n$.

Your question is related to an important open problem in noncommutative ring theory, the “separativity” problem for von Neumann regular rings: if $R$ is a von Neumann regular ring (or more generally an exchange ring), and $A$ and $B$ are finitely generated projective left $R$-modules with the property that $A \oplus A \cong A \oplus B \cong B \oplus B$, must we have $A \cong B$? An affirmative answer would resolve several major open problems, as explained in P. Ara, K. R. Goodearl, K. C. O’Meara, and E. Pardo’s paper “Separative cancellation for projective modules over exchange rings,” Israel J. Math. 105 (1998), 105–137.

Answer by Greg Marks for Finite number of minimal ideals

2010-06-29T00:01:10Z

I'm not sure how to answer your exact question ("the necessary condition that guarantees..."?), but here are a few minor observations. A direct product of commutative rings has only finitely many minimal (by definition nonzero) ideals if and only if each component ring has only finitely many and all but finitely many of the component rings have none. So it suffices to consider indecomposable commutative rings with only finitely many ideals. There are all sorts of indecomposable commutative rings with no minimal ideals. Now suppose the indecomposable commutative ring has a positive finite number of minimal ideals. The socle of such a ring has square zero; thus, the socle is a nonunital subring with the structure of an additive abelian group with zero multiplication on it (of course, this additive abelian group need not have only finitely many minimal subgroups). One can construct all sorts of examples of this sort. For instance, let $A$ be an indecomposable commutative ring with no minimal ideals, and let $M$ be an $A$-module with only finitely many simple submodules. (For example, one might take $M$ to be a uniserial $A$-module.) Let $R = A \oplus M$ as an additive group, with multiplication given by $(a_1, m_1) (a_2, m_2) = (a_1 a_2, a_1 m_2 + a_2 m_1)$. Then $R$ is an indecomposable commutative ring with only finitely many minimal ideals.

Answer by Greg Marks for elementary classification of artinian rings

2010-06-25T20:34:12Z

Sorry if this is merely a reformulation of what has already been said (and doubtless it is a "standard proof"), but perhaps a suggestive hint for students would be to show that if an ideal $P$ of a commutative noetherian ring $R$ is maximal for the property that $R/P$ is non-artinian, then $P \subset R$ must be prime. A sort of philosophical underpinning for this hint is offered in a pretty paper by T. Y. Lam and M. L. Reyes, "A prime ideal principle in commutative algebra," J. Algebra 319 (2008), no. 7, 3006-3027.

Answer by Greg Marks for Finite number of minimal ideals

2010-06-02T20:49:20Z

Please clarify: are you asking about commutative rings or noncommutative rings? (The tag "ac.commutative-algebra" suggests the former; your reference to semisimplicity suggests the latter.) If the question is about noncommutative rings, then presumably by "ideals" you mean two-sided ideals.

Comment by Greg Marks

2012-07-06T18:19:22Z

This story implicitly answers the problem of the student who shrugs upon hearing the theorem on bounded entire functions. By the time it occurs in a class it's very natural, but presented with it cold, the reaction of any student of normal inquisitiveness would be astonishment and an attempt at a counterexample. A student who isn't excited about celebrated classical results is perhaps best advised to find a field of study more to his liking. Incidentally, a result that has allured many a student into mathematics is, so to speak, a non-existence non-theorem: the continuum hypothesis.

Comment by Greg Marks

2012-07-06T18:03:23Z

My first reaction on seeing the title of this question was to recall Rota's "Ten Lessons I Wish I Had Been Taught": completely erase the blackboard, begin writing in the upper left corner, ....

Comment by Greg Marks

2011-08-27T18:29:52Z

For the general problem, can you see why the augmentation ideal is nilpotent?

Comment by Greg Marks

2011-07-30T21:09:51Z

The poor devils could follow Kiran Kedlaya's example: mit.edu/~kedlaya/about-my-name.html

Comment by Greg Marks

2011-04-27T19:21:59Z

Easy exercise (including for more general $R$ and $A$).

Comment by Greg Marks

2011-04-04T01:41:03Z

Simple but amusing application: that multiplicative order is the minimal number of perfect shuffles required to restore a deck of $p \pm 1$ cards (the $\pm$ depending on which of the two ways of perfectly shuffling we're talking about).

Comment by Greg Marks

2011-03-14T02:08:25Z

You are very kind to credit me, but I doubt I deserve it. I am embarrassed to say that the very interesting paper of Gutan and Kisielewicz had slipped my mind. Gutan and Kisielewicz completely resolve the group algebra reversibility question for the case of torsion groups. They remark that the torsion-free case is connected with the zero divisor problem, in the sense that it will be very difficult to produce a non-reversible group algebra with a torsion-free group, since this will of course resolve the zero divisor problem in the negative.

Comment by Greg Marks

2011-03-03T20:36:08Z

Well said. Another example of one of those useful ring-theoretic properties is that every prime ideal is completely prime (for $q$ not a root of unity). This would seem to go without saying, but since the question was asked--the study of any mathematical object is generally advanced by being able to view the object as a special case of something with a well-developed theory.

Comment by Greg Marks

2011-02-28T21:21:37Z

Actually, every countable field of characteristic zero does embed in the field of complex numbers. (Embed it in its algebraic closure and think about transcendence degrees.)

Comment by Greg Marks

2011-02-23T23:25:38Z

It's interesting that the answer is "yes" for Hausdorff topological spaces, and "yes" for groups, but "no" for Hausdorff topological groups.

Comment by Greg Marks

2011-02-23T22:48:55Z

Also, pleasantly (in contrast with the identical character tables of nonisomorphic extraspecial $p$-groups of the same order), nonisomorphic groups have different group determinants in characteristic $0$. See E. Formanek, D. Silbey, "The group determinant determines the group," Proc. Amer. Math. Soc. 112 (1991), no. 3, 649-656.

Comment by Greg Marks

2011-02-02T00:57:17Z

(Farcical exercise for students: why doesn't that also show that a commutative ring with a faithful artinian module must be artinian?)

Comment by Greg Marks

2011-02-01T01:10:29Z

I hope Lenstra at least installed an extra case fan.

Comment by Greg Marks

2011-02-01T00:01:23Z

Modulo standard results (such as can be found, for example, in Eisenbud's book "Commutative Algebra"), Manny Reyes's comment completely resolves this question.

Comment by Greg Marks

2011-01-29T00:21:35Z

I've heard a version of this story with von Neumann as protagonist. A student desperately trying to evaluate a definite integral at wit's end shows up at von Neumann's office for any suggestion of an approach. Von Neumann thinks for a few moments and tells him the value of the integral. The student, amazed and dismayed at the absence of any details, respectfully requests that von Neumann "explain it a different way." Von Neumann thinks some more and cheerfully announces that both ways yield the announced value. (I thought this story apocryphal even before hearing the Wiener version.)