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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

6 Answers 6

up vote 5 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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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

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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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... –  mathic Jan 15 '11 at 6:36

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