1
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Should "annulating" (in the title) be "annihilating"? $\endgroup$ Mar 28, 2010 at 17:21
  • $\begingroup$ I suspect there is a uniform bound on such r: given an r, there is the hypersurface $D(b_0,...,b_{n-1})/D_0 = y^2$, where $D$ is the discriminant of the monic polynomial, and $D_0$ will be the discriminant of a fixed monic polynomial. Not all solutions to this equation will be in the same field, but probably infinitely many. I believe there are conjectures on the number of integer solutions to hypersurfaces in $r$ variables, can anyone who knows shed more light? $\endgroup$ Mar 28, 2010 at 17:28
  • $\begingroup$ Thanks Michael, I corrected the title. $\endgroup$ Mar 28, 2010 at 19:52
  • $\begingroup$ "...allows us to view B as a subset of Z^n". Not quite, because distinct elements of B might have the same min poly. $\endgroup$ Mar 28, 2010 at 19:57
  • $\begingroup$ @ Kevin : this does not matter because at most $n$ many elements share the same min poly. So an infinite subset of $B$ will always yield an infinite subset of $Z^n$. $\endgroup$ Mar 29, 2010 at 3:52

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ @Dror: the heart of this answer is surely right but I'm not convinced that D(b_i) is always a polynomial of degree greater than 4 [try quintics with i=0 for example], and even if it is then the hypersurface might be singular and have lots of rational points, like y^2=x^101 or something. But in some sense this is a moraly correct approach, and it will surely suffice to find one explicit number field with r>1. $\endgroup$ Mar 29, 2010 at 21:56
  • $\begingroup$ Sorry sorry, I meant genus at least 1, Siegel's theorem covering 1. The "almost all" takes out the singular hypersurfaces. I believe some very easy sieve theory can prove this since, mod p, most assignments of the other coefficients should give a squarefree polynomial. $\endgroup$ Mar 29, 2010 at 22:10
  • $\begingroup$ I follow the argument and believe you. $\endgroup$ Mar 29, 2010 at 22:16
  • $\begingroup$ @ Dror : very nice. I think that we can always take r=1 when n=3, because Kevin's computation on $x^3-x+1$ generalizes. $\endgroup$ Mar 30, 2010 at 5:17
5
$\begingroup$

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).

$\endgroup$
3
  • 1
    $\begingroup$ I completely disagree with your comment that the question is not meaningful. For any given $\alpha$, the (smallest) $r$ is a defnite value. It may be that $r$ is very hard to compute in terms of $\alpha$ in general, but you cannot say that the question is meaningless. Thanks for your example with $z^3-z-1$. I guess there is a parametrization of the solutions of $27C^2-4=23D^2$ by some linear-recurrence sequence. $\endgroup$ Mar 29, 2010 at 3:50
  • $\begingroup$ Well OK :-) Yes, I agree that the smallest $r$ is a definite value. My point is that if the question is a question, then an answer to it could be "the value of $r$ is whatever it comes out to be". The parametrisation of the solutions to 27C^2-4=23D^2 can be obtained by the theory of Pell's equation: the solutions grow exponentially but there are infinitely many of them. $\endgroup$ Mar 29, 2010 at 6:46
  • $\begingroup$ @Ewan: aah yes, I understand what you're saying: yes, the solutions to the equation are generated by a degree 2 linear recurrence relation, and 2/3 of them are odd. $\endgroup$ Mar 29, 2010 at 6:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.