Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The only precise statement (coming from a reliable source) of the "arithmetic Nullstellensatz" I can find is in Gowers's book, stating that two polynomials with integral coefficients have the same roots mod every $m$ iff they differ by a sign. I would like to know the general form of this result, and see some reference where I can read about it and some applications (perhaps). All help is appreciated.

share|improve this question
    
The polynomials $X^p - X$ and $p(X^p-X)$ provide a counterexample; is primitivity intended? And "multiplicity" of root doesn't have any meaning mod $m$ for $m$ not prime (esp. not square free), but is it intended that the polynomials are irreducible over $\mathbf{Q}$? Anyway, the Jacobson property of finite type $\mathbf{Z}$-algebras is a useful "arithmetic" version of the Nullstellensatz (but it has no arithmetic content, since it is a general fact with $\mathbf{Z}$ replaced by any Dedekind domain with infinitely many primes, or any Jacobson ring at all). –  user54268 Aug 26 at 5:01

1 Answer 1

One "arithmetic version" of the Nullstellensatz states that if $f_1, ..., f_s$ belong to $\mathbb{Z}[X_1,...,X_n]$ without a common zero in $\mathbb{C}^n$, then there exist $a \in \mathbb{Z} \setminus {0}$ and $g_1,...,g_s$ in $\mathbb{Z}[X_1,...,X_n]$ such that $a = f_1g_1 + ... + f_sg_s$. Finding degree and height bounds for $a$ and $g_1, ..., g_s$ has received some attention, see for example here.

share|improve this answer
    
Dietrich: I was the one who asked the question there. I am somehow curious whether there is some big picture result that generalizes the result from Gowers book. Don't worry, I have already googled for some time –  Polynomial ring Aug 25 at 21:10
    
Half of your Z's are $Z$ and the other half are $\mathbb Z$. I'm guessing that's a typo. –  André Henriques Aug 26 at 8:54
    
@AndréHenriques thank you - I fixed it. –  Dietrich Burde Aug 26 at 9:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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