Infinite collection of elements of a number field with very similar annihilating polynomials

Question

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.

Answer 1

The answer "is" that the smallest $r$ is what it is, and what it is could well depend on $\alpha$. Let me also raise the possibility that there might be no simple "formula" relating $r$ to $\alpha$. This in some sense is the "problem" with questions like this ("given some data, compute some number $r$: what 'is' $r$?")---they are not really questions (in my mind, at least). Who knows though, perhaps someone can find some extra structure. For example can one always take $r=1$? That's a proper question ;-) I'd be surprised though!

But on a more positive note let me say that in my (rather long) answer to

Integers not represented by $ 2 x^2 + x y + 3 y^2 + z^3 - z $

I show in passing that if $\alpha$ is a root of $z^3-z+1$ then there are infinitely many integers $C$ such that $z^3-z+C$ is irreducible and has a root in $\mathbf{Q}(\alpha)$, giving a perhaps slightly less trivial example. The integers $C$ are the odd solutions to $27C^2-4=23D^2$ and there are infinitely many of these (the smallest two being 1 and 599).

Answer 2

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.