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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.

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Should "annulating" (in the title) be "annihilating"? –  Michael Lugo Mar 28 '10 at 17:21
    
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? –  Dror Speiser Mar 28 '10 at 17:28
    
Thanks Michael, I corrected the title. –  Ewan Delanoy Mar 28 '10 at 19:52
    
"...allows us to view B as a subset of Z^n". Not quite, because distinct elements of B might have the same min poly. –  Kevin Buzzard Mar 28 '10 at 19:57
    
@ 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 $Z^n$. –  Ewan Delanoy Mar 29 '10 at 3:52

2 Answers 2

up vote 3 down vote accepted

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.

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@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. –  Kevin Buzzard Mar 29 '10 at 21:56
    
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. –  Dror Speiser Mar 29 '10 at 22:10
    
I follow the argument and believe you. –  Kevin Buzzard Mar 29 '10 at 22:16
    
@ 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. –  Ewan Delanoy Mar 30 '10 at 5:17

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).

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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. –  Ewan Delanoy Mar 29 '10 at 3:50
    
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. –  Kevin Buzzard Mar 29 '10 at 6:46
    
@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. –  Kevin Buzzard Mar 29 '10 at 6:49

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