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Let $p(x)$ be the chromatic polynomial of a special graph. Performing a certain type of operation on the graph changes $p$ by shifting it and adding a constant, say to: $q(x)=p(x+a) + b$.

I have noticed that this operation always results in $q$ having the same discriminant and splitting field as $p$. This would of course be obvious if we were just shifting $p$ by $a$, but it seems unusual given that we are also adding $b$.

I realise it is difficult to comment without knowing more details, but all I really want to know is whether or not this might be significant. The polynomials in question are irreducible cubics. How common is it that two cubics - which are not just shifted versions of each other - have the same discriminant? Does anybody have any idea what this might signify?

(edit: since posting this I have noticed that the chromatic polynomial in question is of the form $p(u,v,x)$, where $u$ and $v$ are given constants, and the discriminant of $p$ is a symmetric polynomial in $u$ and $v$. The graph operation switches $u$ and $v$...this is why the discriminant remains the same. So my question could be formulated as: what is the significance - if any - of a discriminant which is a symmetric polynomial in 2 variables?)

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Frank Thorne, in your above reply you state that "it is known that the number of cubic fields of discriminant n is $O(n^{1/3 + \epsilon})$, due to Ellenberg and Venkatesh" - could you give the reference for that? Sadly, I can not find it! Thanks and best regards, Gero – user15047 May 12 '11 at 13:05
up vote 4 down vote accepted

[Substantial edit: As I mentioned previously, cubic fields can have the same discriminant but not be isomorphic; but I've revised my answer to better address the author's question.]

There are a lot of polynomials that will generate the same cubic field. Here is the basic idea, due to Delone and Faddeev. Write down the cubic form $f(x, y) = ax^3 + bx^2 y + cxy^2 + dy^3$, where setting $y = 1$ gives a generating polynomial for the cubic field. Then there is an action of $GL_2$ on such cubic fields; if $\gamma$ is a 2x2 matrix with entries $\alpha, \beta, \gamma, \delta$ then $\gamma f(x, y) = f(\alpha x + \gamma y, \beta x + \delta y)$, optionally with a factor of $(det \ \gamma)^{-1}$.

Anyway, if two cubic forms are $GL_2(\mathbb{Q})$-equivalent then the corresponding polynomials (if irreducible and nondegenerate) generate the same field. Moreover, if two cubic forms are $GL_2(\mathbb{Z})$-equivalent then the corresponding polynomials generate the same order over $\mathbb{Z}$.

See here, page 10 of Section 4, for a good reference.

[Previous answer:]

I think the underlying question is how often nonisomorphic cubic fields have the same discriminant. This happens, but it doesn't seem to be very common.

You might find it interesting to browse through this database which lists cubic (and other) fields, their discriminants, and their minimal polynomials.

Theoretically, I believe it is known that this happens infinitely often, although I don't recall the reference. For upper bounds, it is known that the number of cubic fields of discriminant n is $O(n^{1/3 + \epsilon})$, due to Ellenberg and Venkatesh; the "correct" bounds are believed to be much sharper.

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Thanks, I will have a look at that database. Actually the polynomials in question have the same splitting field. This is what is puzzling me: given that the link between a graph's structure and its chromatic polynomial is very poorly understood, it seems significant that the chromatic roots of two related graphs would have the same imaginary/irrational part. – Adam Nov 19 '10 at 12:23
Hm. I seem to have gotten two upvotes despite not properly understanding your question! Please see substantial revision above. – Frank Thorne Nov 24 '10 at 3:11

If you write your cubic as $p(x) = x^3 + rx^2+sx+t$, then the discriminant is $\Delta=r^2s^2-4s^3-4r^3t-27t^2+18rst$, which is quadratic in $t$. So the fact that $p(x+a) + b$ has the same discriminant means that $b$ satisfies a quadratic polynomial with coefficients determined by $r,s,t$ and $a$. In other words, for a given translation $a$, at most two values of $b$ will give you the same discriminant.

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