# Explicit bounds for transfer results in algebraic geometry

Assume you have an ideal $I\subseteq\mathbb{Z}[X_1,\ldots,X_n]$ of the polynomial ring in $n$ variables over the integers. For any field $\Bbbk$, I can consider the ideal $I(\Bbbk):=I\otimes_{\mathbb{Z}}\Bbbk\subseteq\Bbbk[X_1,\ldots,X_n]$. It is well-known that $1\notin I(\mathbb C)$ implies that for prime numbers $p\gg 0$, we also have $1\notin I(\overline{\mathbb F}_p)$ where $\overline{\mathbb F}_p$ denotes an algebraic closure of the field with $p$ Elements.

Geometrically, it means that the scheme $Z(I)$ which is defined over any field has $\overline{\mathbb F}_p$-rational points for large enough $p$ if (and actually, only if) it has $\mathbb{C}$-rational points. Even more geometric, if the system of equations in $I$ has a solution over the complex numbers, then it admits solutions in characteristic $p$ for large $p$.

I am interested in explicit bounds on $p_0\in\mathbb N$ such that the above holds for all prime numbers $p>p_0$. For this purpose, let us assume that $I$ is generated by polynomially many polynomials of degree less or equal to $d$, for some fixed $d$. Then $p_0=p_0(n)$ depends on the number of variables involved.

In the 1978 paper Bounds on transfer principles for algebraically closed and complete discretely valued fields by Scott Brown, we get a roughly driple exponential bound:

Theorem If $S$ is a first-order statement in the language with plus, minus, times, and constant symbols $0, 1, 10, 11, 100,\ldots$ whose representation is $m$ characters long, and $p>2^{2^{2^{3m}}}$, then $S$ holds in the theory of algebraically closed fields of characteristic $p$ if and only if it holds in the theory of algebraically closed fields of characteristic zero.

Observe that since we fixed the degree $d$, the number $m$ from that theorem will be a polynomial in $n$, because for a fixed degree $d$, the number of monomials in $n$ variables of degree $d$ grows polynomially and I assumed that $I$ is generated by polynomially many elements.

Edit. I now realize that I also have to assume that the coefficients of my generators are not too large, so let us assume that as well.

Question. Is there a reference for a substantially better bound? I have reasons to believe that a doubly exponential bound (with possibly some polynomial in $n$ in the exponent) should be possible.

• You are not interested in general first-order statements, but only existential statements. This should indeed remove one exponential. Moreover, this is most likely stated somewhere in Brown’s paper, as the general argument probably proceeds by induction on quantifier nesting depth. However, I’ve exhausted my Google books page view limit. Feb 19, 2015 at 13:20
• @EmilJeřábek: Yes, that are my thoughts and problems precisely ;) - unfortunately, Google books is also my only way to look at the document in question. Feb 19, 2015 at 17:05
• probably nothing better is known in general. AMS's MathSciNet lists few papers citing the paper by Brown, and all of them are number-theoretic papers dealing with much less general problems, it seems. Feb 19, 2015 at 21:22

Namely, say $I = (f_1, \ldots, f_s) \subset \mathbf{Z}[x_1, \ldots, x_n]$ with $f_i$ of degree $d$. Then if modulo $p$ these polynomials generate the unit ideal (this you can test over the field with $p$ elements or its algebraic closure, does not matter which), then we can write $$1 = \sum a_i f_i \bmod p$$ with $\deg(a_i) \leq B = \max(3, d)^n$. This is due to Kollar. Ok, so we look at the map $$\mathbf{Z}[x_1, \ldots, x_n]_{\leq B}^{\oplus s} \longrightarrow \mathbf{Z}[x_1, \ldots, x_n]_{\leq B + d}/\mathbf{Z},\quad (a_i) \longmapsto \sum a_if_i \mod \mathbf{Z}$$ This is given by a huge matrix whose entries are bounded. Moreover, the map is injective. Then an easy argument with determinants shows that the map is injective for all $p$ large enough.
How large? Well, just eyeballing what the argument would give you, I think it would not be more than $$C^{s{n + B \choose n}}$$ where $C$ is the bound for the coefficients of the polynomials. I am very bad at doing bounds, so I may be off by quite a large factor. You can also bound $s$ by ${n + d \choose n}$ but I think you should be able to do much better for $s$.