# Does the Galois group of a Pisot polynomial contain the alternating group?

Let $n \in \mathbb{N}$, and let $p(X) \in \mathbb{Z}[X]$ be a monic polynomial of degree $n$. Suppose that exactly one complex root of $p$ is of modulus $> 1$, and that the remaining $n-1$ roots of $p$ belong to the open unit disk in the complex plane (such a polynomial is necessarily irreducible, and is sometimes called a Pisot polynomial).

Is $\mathrm{Gal}(p(X)/\mathbb{Q}) \in \{A_n,S_n\}$ ?

Pisot polynomials are quite common, for instance Perron has shown that if $p(X) = X^n + a_{n-1}X^{n-1} + \dots + a_1X + a_0$ and $|a_{n-1}| > 1 + a_0 + a_1 + \dots + a_{n-2}$ then $p(X)$ is a Pisot polynomial.

David Speyer has beaten me to it, but for what it's worth, a simple explicit example is $$p(X)=X^4-2X^3-5X^2-4X-1$$ which is a Pisot polynomial, and has Galois group the dihedral group of order 8.

How did I find it? I started from observing that if $\vartheta$ is the positive root of $X^2-X-1$, then $\alpha=1+\sqrt{\vartheta}$ is an algebraic unit with only one conjugate inside the closed unit disk, and thus $-1/\alpha$ is a Pisot-Vijayaraghavan number, with this minimal polynomial.

Addendum: The Pisot polynomial

$$p(X)=X^4-2X^3+X-1$$

again has dihedral Galois group of order 8, corresponding to another quadratic extension of $\mathbb{Q}(\vartheta)$, and produces a root less than $2$ (which also happens to be the fourth smallest limit point of the set of PV numbers). This number is mentioned on page 1251 of David W. Boyd, Pisot and Salem numbers in intervals of the real line, Math. Comp. 32 (1978), 1244-1260.

• Great! What is the smallest such Pisot number you can find (for this example it is about $3.5$)? Can you find a pisot number $\leq 2$ with Galois group not in $\{S_n, A_n\}$. – Pablo Aug 24 '15 at 17:22
• @Pablo: I've edited one in. When you bound the size of the number and the degree, you'll be left with a finite set of polynomials, so you can (at least in principle) enumerate them all and check their Galois groups in turn. – GNiklasch Aug 24 '15 at 19:10
• This is fantastic! And one last thing: Is there an example with a polynomial satisfying the inequality on its coefficients stated in the question? – Pablo Aug 24 '15 at 20:03
• Ok, I have found one myself: $X^6+26X^5+X+1$. – Pablo Aug 24 '15 at 22:25

Any number field $K$ which has a real place is generated by a root of a Pisot polynomial. So the answer is no. For example, we can take $p$ a prime $\geq 11$, and use the real subfield of $\mathbb{Q}(\zeta_p)$, to get a Pisot polynomial whose splitting field is cyclic of degree $(p-1)/2$.

Suppose $K$ has $r$ real places, and $2s$ complex places. Let $V_i$ be $\mathbb{R}$ for $1 \leq i \leq r$, and be $\mathbb{C}$ for $r+1 \leq i \leq r+s$, and let $\sigma_i : K \to V_i$ be the $r+s$ embeddings of $K$. Then recall that $$\bigoplus \sigma_i : K \to \bigoplus V_i$$ embeds $\mathcal{O}_K$ as a discrete full rank lattice in $\bigoplus V_i$. Projecting further onto $\bigoplus_{i \neq 1} V_i$, we get a dense additive subgroup of $\bigoplus_{i \neq 1} V_i$. In particular, there is some nonzero $\theta \in \mathcal{O}_K$ such that $|\sigma_i(\theta)|<1$ for all $i \neq 1$. And, the norm of any algebraic integer is at least $1$, we have $|\sigma_1(\theta)|>1$.