User mrb - MathOverflow

When should a supervisor be a co-author?

2011-03-04T10:32:15Z

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?

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?

[To clarify: original question asked by MrB ]

2-separations in M-connected graphs

2012-10-13T12:35:58Z

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.

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?

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).

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).

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?

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 ${u,v}$??

Thanks! (and sorry for the rambling question)

Does this isomorphism between Galois groups hold for transcendental extensions?

2012-10-09T23:26:17Z

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:

$$\text{Gal}(L/K)\cong \text{Gal}(LM/M).$$

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?

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?

Does there exist a 3-connected, chordal graph which is not globally rigid?

2012-10-08T14:23:07Z

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?"

It seems fairly intuitive to me that there does not, but my intuition about graphs has a fairly bad record...

Thanks in advance.

Is a non-disjoint union of connected matroids always connected?

2011-10-21T22:19:00Z

This is perhaps an easy question, but...

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.

A family of polynomials with symmetric galois group

2011-09-09T11:35:38Z

Consider the following family of polynomials in $K[x,y]$, where $K$ has characteristic zero:

$f_n(x,y)=(x+y)^n+(x-1)y^n,$

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.

Here is an alternative form for $f_n$: make the substitution $x\rightarrow xy$ and divide by $y^n$. this gives:

$g_n(x,y)=(x+1)^n+yx-1.$

Substituting $-n$ for $y$ in $g_n(x,y)$ we get:

$g_n(x,-n)=x^2h(x),$

where $h(x)$ is separable (EDIT: $h(x)$ is the subject of this question).

Alternatively, substituting $x\rightarrow x-1$ into $g_n(x,y)$ gives us a polynomial which factors as:

$(x-1)(x^{n-1}+x^{n-2}+\ldots +x^2+x+y+1)$

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.

Any advice much appreciated!

Is this pleasing polynomial irreducible?

2011-09-09T18:19:48Z

Let:

$f(x)=x^n+2x^{n-1}+3x^{n-2}+4x^{n-3}+\ldots + (n-1)x^2+nx+(n+1)$.

Is $f(x)$ irreducible?

In light of the answers to this 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.

(Unnecessary background information: this is linked to this recent question of mine. At the point where I say:

"$g_n(x,−n)=x^2h(x)$,"

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!)

Comment by MrB

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.

Comment by MrB

2012-10-10T01:38:54Z

fair enough...deleted!

Comment by MrB

2012-10-08T15:36:36Z

thanks, much appreciated.

Comment by MrB

2011-10-22T03:31:01Z

How does one show transitivity of $\sim$?

Comment by MrB

2011-10-22T01:06:03Z

Thanks for the swift response. Surprising that such a fundamental result doesn't seem to appear in any textbooks.

Comment by MrB

2011-09-14T15:03:35Z

yes good point...I should have said "direct product of cyclic groups"

Comment by MrB

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.

Comment by MrB

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.

Comment by MrB

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.

Comment by MrB

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 "subjective", surely any question asking for views is going to be by nature subjective? eg. common questions on MO include the very subjective "which books do you think are good for...", "what is your favourite theorem on..." etc. I don't see how this is different