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Let $P_n(t)$ be polynomials with integer coefficients with $d_n = \deg(P_n(t))$ going to infinity when $n$ goes to infinity and with nonzero discriminants $disc(P_n(t)) \neq 0$.

Question: Is $$ \lbrace disc({P_n(t)})\rbrace ^{\frac{1}{d_n}} $$ bounded when $n$ goes to infinity ?

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What do the braces mean? – Mariano Suárez-Alvarez Jan 14 '11 at 5:20
Can you provide a reference where Serre asks this? – Mariano Suárez-Alvarez Jan 14 '11 at 5:22
braces: Nothing special, same as "(" ")" : my latex is not fantastic... reference: No. – Luis H Gallardo Jan 14 '11 at 5:35
Possibly the reference is Minorations de discriminants, note of October 1975, in Oeuvres, vol. 3, Springer, 1986, pp. 240-243. – Gerry Myerson Jan 14 '11 at 6:10
Are these monic integer polynomials? – Andreas Thom Jan 14 '11 at 10:13
up vote 10 down vote accepted

Some people seem a little confused by the wording of this question. The better way of phrasing the question is: Is there a lower bound on the root discriminants of polynomials of degree $d$ as $d \rightarrow \infty$. The survey paper by Odlyzko (noted by Gerry in the comments above):

mentions this as a question of Serre and others (it is a pretty obvious question to ask), and also points out that essentially nothing is known about this question. I don't think the situation has changed considerably in the intervening $20$ years.

As Franz notes, the existence of class field towers is no way implies the existence of polynomials with bounded discriminant - "most" fields are not monogenic.

Some have wondered - perhaps optimistically - whether asymptotically there is even a bound $$|\mathrm{disc}(P)| > (cd)^d$$ For irreducible $P$, where one can take $c$ to be some fixed constant $ > 1$ unless $P$ is essentially a cylotomic polynomial. If this was true, then it would imply that the Mahler measure of an algebraic integer that is not a root of unity is bounded away from $1$, answering a question of Lehmer.

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I think the polynomial $x^d-x-1$ disproves the statement in the last paragraph (the possible refinement of Lehmer's conjecture). If I am not mistaken, its discriminant is $d^d + (-1)^{d}(d-1)^{d-1}$. The Chebyshev polynomials (normalized: take the minimum polynomial of $\zeta_n+\zeta_n^{-1}$) are another counterexample, although this is closer to the cyclotomic examples and may still be considered "essentially cyclotomic." – Vesselin Dimitrov Aug 8 '14 at 15:12

I don't think that this is too easy. By Golod and Shafarevich, there exist towers of number fields with constant "root discriminant". Whether these fields can be generated by roots of polynomials for which the index is bounded is, I fear, open. It might very well be the case that there aren't any considerable improvements over the family $P_n(t) = t^n-2$.

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To reiterate Serre's question (Open Problem 2.5 in Odlyzko's survey): Must there be only finitely many polynomials having root discriminant below a given bound?

With this answer I just want to note that the much stronger statement formulated in the last paragraph of user631's answer, which concerned the existence of a $c > 1$ such that all irreducible polynomials of discriminant $\leq (cd)^d$ are essentially cyclotomic, is false without additional assumptions. Since we may take $c = M(f)^2$ as Mahler proved, this is a natural statement to consider: its truth would have strengthened Lehmer's conjecture.

A counterexample is given by $x^d-x-1$, whose discriminant has absolute value $d^d + (-1)^{d}(d-1)^{d-1}$. We may take more generally any sequence of irreducible trinomials with $\pm 1$ coefficients and degree going to infinity. This follows by the explicit calculation of the discriminant of the general trinomial; the statement and simple derivation of the formula is given as Theorem 2 in Swan's 1962 paper Factorization of polynomials over finite fields in the Pacific Journal of Mathematics. (Another reference is Prasolov's book Polynomials, which reproduces the same calculation).

A counterexample of a rather different kind (in particular, having unbounded Mahler measure under all integer translations) is provided by the minimum polynomial of the generator $\zeta_n+\zeta_n^{-1} = 2\cos(2\pi/n)$ of the integer ring of the maximal totally real subfield of $\mathbb{Q}(\zeta_n)$. Further counterexamples would include the minimum polynomials of the generators over $\mathbb{Z}$ of the integer rings of other bounded index monogenic subfields, if such exist, of either $\mathbb{Q}(\zeta_n)$ or the splitting fields of the $\pm 1$ trinomials considered in the previous paragraph.

However, one may wish to restrict to reciprocal polynomials; for Lehmer's problem this would be sufficient. For these the statement seems very interesting, and could well be true; by the formula in Swan's article, it certainly holds for all trinomials. The above examples are of no use here: the only reciprocal $\pm 1$ trinomials are $x^{2n} \pm x^n +1$, and those have only cyclotomic factors.

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Start with Poitou or Martinet, look at their Mathscinet reviews and the papers which cite them.

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You can certainly choose the polynomials $P_n(t)$ so that your sequence is unbounded: there are polynomials of every degree of arbitrarily large discriminant.

You can also choose them so that your sequence is bounded. This is highly nontrivial, but the existence is established by the existence of infinite Hilbert class field towers. Take a number field $K_0$, let $K_1$ be the maximal abelian unramified extension of $K_0$, let $K_2$ be the maximal abelian unramified extension of $K_1$, etc. The root discriminants of the minimal polynomials generating these fields are all the same.

Typically this doesn't go on forever, but it can. Examples were constructed by Golod and Shafarevich. I am not an expert in the subject, but you can google "infinite class field tower" or look at this for a related question.

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For example, to get an unbounded sequence, you could take $P_n(t) =t^n +1$. then disc $P_n$ =$n^n$. – John Bentin Jan 14 '11 at 12:14

There is a lot of work on discriminants of number fields, which is not far from discriminants of polynomials with integer coefficients. Have you tried typing "discriminant bounds" into Google or Math Reviews?

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I'm by no means an expert, but I would think you can let $\operatorname{disc}(P_n(t))^{1/d_n}$ approach infinity by letting the leading coefficient grow fast enough.

For example, just to be cute, let $n > 2$, let $F_n$ the $n$-th Fibonacci number and let $G_n = \prod_{i < j \leq n}(F_i-F_j)$. Now, pick another sequences $s_n = (nG_n)^{n^2}$ and define our polynomials $P_n(t) = s_n(t - F_1)(t - F_2)\cdots(t - F_n)$ for all $n > 2$ (so $d_n = n$ for us). Then

$\operatorname{disc}(P_n(t))^{1/d_n} = \left[s_n^{2n-2}\prod_{i < j \leq n} (F_k - F_j)^2\right]^{1/n} = \left[(nG_n)^{n^2(2n-2)}\prod_{i < j \leq n} (F_k - F_j)^2\right]^{1/n}$

But this is just

$n^{n(2n-2)}G_n^{2n-2}\prod_{i < j \leq n} (F_k - F_j)^{2}$

which tends to infinity quite quickly.

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I would guess that degree should be understood as degree of the splitting field, which in you case is always $1$. – Mariano Suárez-Alvarez Jan 14 '11 at 7:23
Fair enough. Something seemed a little off about me solving Serre could not do in only a few minutes... – Alex Jan 15 '11 at 6:36

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