User mrb - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T01:38:34Z http://mathoverflow.net/feeds/user/2189 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57337/when-should-a-supervisor-be-a-co-author When should a supervisor be a co-author? MrB 2011-03-04T10:32:15Z 2013-04-19T15:47:37Z <p>What are people's views on this? To be specific: suppose a PhD student has produced a piece of original mathematical research. Suppose that student's supervisor suggested the problem, and gave a few helpful comments, but otherwise did not contribute to the work. Should that supervisor still be named as a co-author, or would an acknowledgment suffice?</p> <p>I am interested in two aspects of this. Firstly the moral/etiquette aspect: do you consider it bad form for a student not to name their supervisor? Or does it depend on that supervisor's input? And secondly, the practical, career-advancing aspect: which is better, for a student to have a well-known name on his or her paper (and hence more chance of it being noticed/published), or to have a sole-authored piece of work under their belt to hopefully increase their chances of being offered a good post-doc position?</p> <p>[To clarify: original question asked by <a href="http://mathoverflow.net/users/2189/mrb" rel="nofollow">MrB</a> ]</p> http://mathoverflow.net/questions/109526/2-separations-in-m-connected-graphs 2-separations in M-connected graphs MrB 2012-10-13T12:35:58Z 2012-10-13T12:35:58Z <p>I am trying to prove a result about rigid graphs, which I believe holds for chordal graphs and also non-chordal but $M$-connected graphs (note: when I say $M$-connected I am referring to the rigidity matroid). Clearly, by rigidity, all graphs I mention will be at least 2-connected.</p> <p>So...I need to pinpoint exactly what special properties an $M$-connected but NOT chordal graph has, which a chordal graph does not. Is there some kind of characterisation of these?</p> <p>In particular, I believe I'm right in saying that there exist $M$-connected graphs having 2-separations (by which I mean pairs of vertices whose deletion disconnects the graph) ${u,v}\subseteq V(G)$ which are such that $uv\notin E(G)$. (As opposed to chordal graphs, in which every 2-separating pair is adjacent).</p> <p>Unfortunately I don't seem to be able to come up with any examples! I can't seem to find an $M$-connected graph having a 4-cycle (please excuse my ignorance if this is obvious---I am very new to the concept of $M$-connectedness, and don't find it intuitive).</p> <p>I would be interested to hear of any other difference between graphs which are chordal, and those which are $M$-connected and not chordal. With regards to rigidity, both types have the property that any equivalent realisation can be obtained by iteratively reflecting components around 2-separations. Chordal graphs consist of $K_2$-sums of 3-connected graphs. Is there an analagous characterisation of $M$-connected graphs?</p> <p>Finally, for generic realisations $(G,p)$ of graphs $G$, if there is anyone who is familiar with the concept of the transcendental field extension $\mathbb Q(d_G(p))$ of the rationals formed by adding all edge-distance equations defining $G$ to $\mathbb Q$...is it true that adding an edge between a non-adjacent 2-separating pair ${u,v}$ does not change $\mathbb Q(d_G(p))$? ie. is it true that $d_(u,v)\in \mathbb Q(d_G(p))$ for every 2-separation ${u,v}$??</p> <p>Thanks! (and sorry for the rambling question)</p> http://mathoverflow.net/questions/109258/does-this-isomorphism-between-galois-groups-hold-for-transcendental-extensions Does this isomorphism between Galois groups hold for transcendental extensions? MrB 2012-10-09T23:26:17Z 2012-10-10T14:00:44Z <p>Suppose $L/K$ and $M/K$ are algebraic extensions of a field $K$, such that $L\cap M=K$, and $L/K$ is a normal extension. It is well-known that, with these conditions, we have:</p> <p>$$\text{Gal}(L/K)\cong \text{Gal}(LM/M).$$</p> <p>However, suppose we now complicate matters by specifying that $M/K$ (and hence also $LM/L$) is not algebraic but transcendental ($L/K$ remaining normal). In this case, am I right in thinking that this identity still holds? </p> <p>If not, can anyone provide a counterexample, or is there an obvious reason why this doesn't work? What if we specify that $K$ has characteristic zero? </p> http://mathoverflow.net/questions/109151/does-there-exist-a-3-connected-chordal-graph-which-is-not-globally-rigid Does there exist a 3-connected, chordal graph which is not globally rigid? MrB 2012-10-08T14:23:07Z 2012-10-08T17:32:09Z <p>The question is in the title! I know that a globally rigid graph is 3-connected and redundantly rigid, so my question could be rephrased as: "does there exist a graph which is 3-connected and chordal but not redundantly rigid?"</p> <p>It seems fairly intuitive to me that there does not, but my intuition about graphs has a fairly bad record...</p> <p>Thanks in advance.</p> http://mathoverflow.net/questions/78802/is-a-non-disjoint-union-of-connected-matroids-always-connected Is a non-disjoint union of connected matroids always connected? MrB 2011-10-21T22:19:00Z 2011-10-22T01:11:45Z <p>This is perhaps an easy question, but...</p> <p>Let $M$ be a matroid on a ground set $E$, and let $A$ and $B$ be non-disjoint subsets of $E$ such that $M|A$ and $M|B$ are both connected. Is $M|(A\cup B)$ then necessarily connected? Clearly this is true for graphic matroids, but I can't find any results in the literature regarding the general case.</p> http://mathoverflow.net/questions/74989/a-family-of-polynomials-with-symmetric-galois-group A family of polynomials with symmetric galois group MrB 2011-09-09T11:35:38Z 2011-10-01T04:39:40Z <p>Consider the following family of polynomials in $K[x,y]$, where $K$ has characteristic zero:</p> <p>$f_n(x,y)=(x+y)^n+(x-1)y^n,$</p> <p>for $n\geq 3$. I can prove that $f_n(x,y)$ has an irreducible factor of degree $n-1$ in $x$. I also know that the galois group of $f_n$ over $K(y)$ is the symmetric group of degree $n-1$, but am having trouble proving this.</p> <p>Here is an alternative form for $f_n$: make the substitution $x\rightarrow xy$ and divide by $y^n$. this gives:</p> <p>$g_n(x,y)=(x+1)^n+yx-1.$</p> <p>Substituting $-n$ for $y$ in $g_n(x,y)$ we get:</p> <p>$g_n(x,-n)=x^2h(x),$</p> <p>where $h(x)$ is separable (EDIT: $h(x)$ is the subject of <a href="http://mathoverflow.net/questions/75023/is-this-pleasing-polynomial-irreducible" rel="nofollow">this</a> question).</p> <p>Alternatively, substituting $x\rightarrow x-1$ into $g_n(x,y)$ gives us a polynomial which factors as:</p> <p>$(x-1)(x^{n-1}+x^{n-2}+\ldots +x^2+x+y+1)$</p> <p>It seems as though it shouldn't be hard to show that some specialisation of $y$ into this gives a polynomial with galois group $S_{n-1}$ over $K$...but I'm well and truly stuck.</p> <p>Any advice much appreciated! </p> http://mathoverflow.net/questions/75023/is-this-pleasing-polynomial-irreducible Is this pleasing polynomial irreducible? MrB 2011-09-09T18:19:48Z 2011-09-09T18:46:00Z <p>Let:</p> <p>$f(x)=x^n+2x^{n-1}+3x^{n-2}+4x^{n-3}+\ldots + (n-1)x^2+nx+(n+1)$.</p> <p>Is $f(x)$ irreducible?</p> <p>In light of the answers to <a href="http://mathoverflow.net/questions/18094/polynomial-with-the-primes-as-coefficients-irreducible" rel="nofollow">this</a> question, I now know that this is true when $n+1$ is prime. What about when $n+1$ is composite? I have checked a lot of cases and it seems to be true.</p> <p>(Unnecessary background information: this is linked to <a href="http://mathoverflow.net/questions/74989/a-family-of-polynomials-with-symmetric-galois-group" rel="nofollow">this</a> recent question of mine. At the point where I say: </p> <p>"$g_n(x,−n)=x^2h(x)$," </p> <p>the $h(x)$ in question has the property that substituting $x-1$ for $x$ puts it in the form of $f(x)$ above. If I can show that $h(x)$ is irreducible, then I will have shown that the galois group of the original polynomial is doubly transitive. Not that I am trying to draw attention back to my original question!)</p> http://mathoverflow.net/questions/109258/does-this-isomorphism-between-galois-groups-hold-for-transcendental-extensions Comment by MrB MrB 2012-10-10T12:46:45Z 2012-10-10T12:46:45Z Oh dear, I do apologise...I appear to have mixed up my $M$'s and my $L$'s. As it was the question made no sense, so sorry for wasting your time. Serves me right for posting questions so late at night... I have edited the question - hopefully it is clear now. http://mathoverflow.net/questions/57337/when-should-a-supervisor-be-a-co-author/109259#109259 Comment by MrB MrB 2012-10-10T01:38:54Z 2012-10-10T01:38:54Z fair enough...deleted! http://mathoverflow.net/questions/109151/does-there-exist-a-3-connected-chordal-graph-which-is-not-globally-rigid/109157#109157 Comment by MrB MrB 2012-10-08T15:36:36Z 2012-10-08T15:36:36Z thanks, much appreciated. http://mathoverflow.net/questions/78802/is-a-non-disjoint-union-of-connected-matroids-always-connected/78809#78809 Comment by MrB MrB 2011-10-22T03:31:01Z 2011-10-22T03:31:01Z How does one show transitivity of $\sim$? http://mathoverflow.net/questions/78802/is-a-non-disjoint-union-of-connected-matroids-always-connected/78809#78809 Comment by MrB MrB 2011-10-22T01:06:03Z 2011-10-22T01:06:03Z Thanks for the swift response. Surprising that such a fundamental result doesn't seem to appear in any textbooks. http://mathoverflow.net/questions/74989/a-family-of-polynomials-with-symmetric-galois-group/74993#74993 Comment by MrB MrB 2011-09-14T15:03:35Z 2011-09-14T15:03:35Z yes good point...I should have said &quot;direct product of cyclic groups&quot; http://mathoverflow.net/questions/74989/a-family-of-polynomials-with-symmetric-galois-group Comment by MrB MrB 2011-09-14T08:55:19Z 2011-09-14T08:55:19Z Yes you're right...it was supposed to be $(x+y)^n+(x-1)y^n$. Thanks for pointing this out - I have edited the question. http://mathoverflow.net/questions/74989/a-family-of-polynomials-with-symmetric-galois-group/74993#74993 Comment by MrB MrB 2011-09-09T13:52:53Z 2011-09-09T13:52:53Z No..I don't think so. $x^n-1$ has cyclic galois group, so lots of cycle-types would be missing. http://mathoverflow.net/questions/74989/a-family-of-polynomials-with-symmetric-galois-group/74993#74993 Comment by MrB MrB 2011-09-09T13:01:10Z 2011-09-09T13:01:10Z Thanks. Originally I had thought that I could prove it that way. Unfortunately it is not an irreducible trinomial though! So I don't think any of the various results apply. http://mathoverflow.net/questions/57337/when-should-a-supervisor-be-a-co-author Comment by MrB MrB 2011-03-04T14:10:37Z 2011-03-04T14:10:37Z I also vote to reopen! I have so far found the answers very helpful. And as far as I can see the conversation was proceeding in a very good-natured fashion, without argument. As for &quot;subjective&quot;, surely any question asking for views is going to be by nature subjective? eg. common questions on MO include the very subjective &quot;which books do you think are good for...&quot;, &quot;what is your favourite theorem on...&quot; etc. I don't see how this is different