# How small can a totally positive integer be?

Consider a large, fixed $M>2$. For each $n$, let $\alpha_n$ denote the smallest algebraic integer of degree at most $n$, all of whose Galois conjugates lie in the real interval $(0,M)$.

Is there anything known on the rate at which $\alpha_n$ decays to $0$ as $n\to\infty$ ?

The exponential lower-bound $\alpha_n>\frac1 {M^{n}}$ is not hard to prove, but is it reasonable to expect that $\alpha_n$ will actually decay much more slowly, i.e. like an inverse power of $n$ ?

• I don't think $\zeta_n$ is the best (or even second-best) notation in the world for a real algebraic integer... Apr 7, 2014 at 15:24
• Ah, $\alpha_n$ is definitely an improvement. Apr 7, 2014 at 19:01

$\alpha_n$ can be exponentially small once $M$ is large enough, say $M \geq 6$.
For $m > 0$ let $\tau_m$ be the (monic, degree-$m$) polynomial such that $\tau_m(z+z^{-1}) = z^m + z^{-m}$; in other words, $\tau_m(x) = 2 T_m(x/2)$ where $T_m$ is the $m$-th Čebyšev polynomial. Then $\tau_m$ has all its roots real and contained in the interval $(-2,2)$, while there are $m+1$ points $x_0,x_1,\ldots,x_m$ with $$2 = x_0 > x_1 > x_2 > \ldots > x_m = -2$$ at which $\tau_m(x_k) = (-1)^k \cdot 2$.
Now fix an integer $M>5$, for $n$ odd let $m$ be the even number $n-1$ and consider $P(x) = x \, \tau_m (x-(M-2)) - 1$. This is a monic polynomial with $P(1)=-1$, $P(M-4)>0$, and $P$ changing sign between $x_{k-1}$ and $x_k$ for each $k=1,2,\ldots,m$. Therefore it has $m+1=n$ real roots, all in $(0,M)$, and all but one greater than $M-4$. Moreover the product of these roots is $-P(0) = 1$. Hence the remaining root is smaller than $(M-4)^{1-n}$, which decays exponentially once $M-4>1$. (In fact $M=5$ is good enough because about half the roots exceed $M-2$.)
For example, if $M=10$ and $n=7$ this recipe yields the septic polynomial $$x^7 - 48x^6 + 954x^5 - 10048x^4 + 59145x^3 - 184464x^2 + 238142x - 1,$$ with one root just below $4.2 \cdot 10^{-6}$ and the other six ranging from about $6.061$ to $9.936$.