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 $n$ ? 



How small can a totally positive integer be?
$\alpha\_n$ 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 $9.936$.


