# roots of reciprocal polynomials

Consider the set $S = \{x \in \mathbb{R} \left| f(x) = 0\right. \},$ where $f$ is a reciprocal monic irreducible polynomial with integer coefficients (reciprocal means that the sequence of coefficients reads the same left to right or right to left; for example, the characteristic polynomials of symplectic matrices are reciprocal (conversely, every polynomial in italics (and of even degree) is the characteristic polynomial of some matrix in $Sp(2k, \mathbb{Z}).$ The question is:

Is $S$ dense in $\mathbb{R}?$

• It doesn't have a direct bearing, but is related: have you heard of Lehmer's conjecture on the Mahler measure of polynomials? – KConrad May 8 '15 at 22:19
• Instead of "reciprocal", I think the usual terminology for this condition is to say either that the polynomial is "self-reciprocal" or that it is "palindromic". – Dave Witte Morris May 8 '15 at 22:25
• @DaveWitteMorris Actually, I think "reciprocal" is common usage, at least for people who work in Diophantine approximation. Thus it is standard to say that "Chris Smyth proved that Lehmer's conjecture is true for roots of non-reciprocal polynomials." (There's nothing wrong with "self-reciprocal", but if one is speaking of a single polynomial, the "self" seems redundant.) – Joe Silverman May 8 '15 at 23:17
• @JoeSilverman Indeed, I haven't seen self-reciprocal used much... – Igor Rivin May 8 '15 at 23:19
• @KConrad Indeed I have, and this was, in fact, the reason for the question. I REALLY want to know whether there is some interval where the Mahler measure of the algebraic integers therein is bounded away from $1.$ If the answer to my question is "no", which I strongly doubt, then this would give a positive answer by Chris Smyth's result alluded to by Joe Silverman. – Igor Rivin May 8 '15 at 23:21

Yes, also if $f$ has to be irreducible. Elaborating on Dave Witte Morris' comment, I take $\alpha=3-2\sqrt{2}$, $\beta=2-\sqrt{3}$, both clearly algebraic units whose inverses are their Galois conjugates $\bar{\alpha}=3+2\sqrt{2}$, $\bar{\beta}=2+\sqrt{3}$. But now the inverse of any power $\alpha^n$ is $\bar{\alpha}^n=\overline{\alpha^n}$, again the Galois conjugate (if $n \neq 0$), and similarly for $\beta$. And a product $\alpha^m\beta^n$ ($m \neq 0$, $n \neq 0$) has three Galois conjugates $\bar{\alpha}^m\beta^n$, $\alpha^m\bar{\beta}^n$, $\bar{\alpha}^m\bar{\beta}^n$, one of which is its inverse and the other two are the inverses of each other.

So your set $S$ certainly contains $X=\{\alpha^m\beta^n| m \neq 0$ or $n \neq 0\}$. Now $\log \alpha$ and $\log \beta$ are $\mathbb{Q}$-linearly independent and so $\{m\log\alpha+n\log\beta|m,n \in \mathbb{Z}\}$ is dense in $\mathbb{R}$ (and removing $0$ doesn't change that). It follows that $X$ is dense in $\mathbb{R}_{>0}$. Since the inverse and the Galois conjugates of $-\gamma$ are the negatives of the inverse and the Galois conjugates of $\gamma$ respectively, the set $S$ also contains $-X$, which is dense in $\mathbb{R}_{<0}$. Your statement now follows.

• Why is your statement about the Galois conjugates of $\alpha^m \beta^m$ true? – Igor Rivin May 9 '15 at 15:50
• @Igor: We have $\alpha^m\beta^n\bar{\alpha}^m\bar{\beta}^n=(\alpha\bar{\alpha})^m(\beta \bar{\beta})^n=1$. Similarly $\bar{\alpha}^m\beta^n\alpha^m\bar{\beta}^n=1$. – Gabriel Dill May 9 '15 at 16:30
• Ah, so. I was thinking on completely different lines, but this works :) – Igor Rivin May 10 '15 at 0:34

Algebraic integers are dense (even of the form $a+b\sqrt{2}$). If $c\ne 0$ is real, find algebraic integer $u$ close to $c+1/c$, that is, $u=v+1/v$ for $v$ close to $c$ and $h(u)=h(v+1/v)=0$ for some monic $h\in \mathbb{Z}[x]$ of degree $d$, then $v$ is a root of $x^dh(x+1/x)$.

• Another way of looking at this is that $S$ is the set of algebraic units (that is, the set of algebraic integers whose reciprocal is also an algebraic integer). It is easy to see that this is dense in $\mathbb{R}$. (For example, take $\{\alpha^m \beta^n\}$ where $\alpha$ and $\beta$ are multiplicatively independent.) – Dave Witte Morris May 8 '15 at 22:40
• Sorry, I forgot the "irreducible" assumption in the hypotheses, otherwise it's trivial... – Igor Rivin May 8 '15 at 22:59
• Well, are not polynomials of the form $x^2((x+1/x-a)^2-2b^2)=(x^2-ax+1)^2-2b^2x^2$ irreducible often enough? – Fedor Petrov May 8 '15 at 23:18
• @DaveWitteMorris I don't think that $S$ is the set of algebraic units. Not every unit is the root of a reciprocal polynomial. For example, the roots of $x^3+2x^2+1$ are units. Being reciprocal means that the reciprocals of each root is a Galois conjugate of one of the other roots. – Joe Silverman May 8 '15 at 23:21
• @JoeSilverman $x^3+2x^2+1$ divides $(x^3+2x^2+1)(x^3+2x+1)$ which is reciprocal. But of course, it is reducible. – Fedor Petrov May 8 '15 at 23:25