Hello all, let $n$ be an integer $\geq 2$ and let $\alpha$ be an algebraic number
of degree $n$. Let $R$ be the ring of algebraic integers in ${\mathbb Q}(\alpha)$, and
let $B$ be the subset of $R$ containing the elements whose degree is exactly
$n$. Any $\beta \in B$ has a minimal polynomial 
$X^n+b\_{n-1}X^{n-1}+ \ldots + b\_1X+b\_0$. Identifying this latter polynomial
with the uple $(b\_0,b\_1, \ldots ,b\_{n-1})$ allows us to view $B$ as a subset
of ${\mathbb Z}^n$. I define a combinatorial subvariety $V$ of dimension
at most $r$ of ${\mathbb Z}^n$ to be a subset of $Z^n$ such that
there is a set of indices $I \subseteq \lbrace 1,2, \ldots , n \rbrace$ with
$|I|=n-r$ and the projection $p:V \to {\mathbb Z}^{n-r}, 
(v\_1,v\_2, \ldots ,v\_n) \mapsto (v\_i)\_{i\in I}$ is constant. 


My question is : what is the smallest $r$ such that
there is an infinite subset $B' \subset B$ corresponding to
a subvariety of dimension at most $r$ ?


In other words, we are asking for infinitely many elements in $B$,
whose minimal polynomials are ``as similar as possible".


An easy case is when $\alpha=a^{\frac{1}{n}}$ for some
$a \in {\mathbb Q}$, because the rational multiples of $\mathbb \alpha$
correspond to a subvariety of dimension 1, so that $r=1$ in this case.



Infinite collection of elements of a number field with very similar annihilating polynomials
For $n>4$, almost all fields of degree $n$ will have $r>1$:


Fix a field $K$ with discriminant $D\_0$. Fix the $n-1$ coefficients $b\_{n-1},...,b\_{i+1}, b\_{i-1},..., b\_0$. The discriminant of the polynomial $x^n+b\_{n-1}x^{n-1}+...$ is a polynomial $D(b\_i)$ in the single variable $b\_i$, and is of degree at least $4$.


If this polynomial is squarefree, as it will be for almost all $n-1$ fixed coefficients, then the hypersurface $D\_0y^2 = D(b\_i)$ has genus at least $1$, and hence finitely many integer points.


But, every polynomial defining the same field must have the same discriminant up to a square factor, and hence $r > 1$.


Going back on my comment above: since the degree of the discriminant (multivariate) polynomial is large (linear in the number of variables) the equation $D(b\_0,...,b\_{n-1}) = D\_0y^2$ will probably have only a finite number of solutions for most $D\_0$, if $r$ is much smaller than $n$.


Therefore, my new pessimistic conjecture is that for almost all fields you will have $r \gg n$.


Note: $r \le n-1$ - in any number field there are always an infinite number of algebraic integers with trace 0.


