ac.commutative algebra - 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.