I try to solve the finite system of multivariable algebraic equations with coefficients from $\mathbb Q$. It would be sufficient for me to prove that there is only finite number of solutions over $\mathbb Q$ (or even more over algebraic closure $\bar{\mathbb Q}$).

I try to do it by calculating dimension of corresponding ideal (my goal is to get zero dimension). But direct calculation is too demanding due too growths of the rational height of coefficients.

Are there any local-global principle in that case?

My idea is to calculate dimension of ideal based on reduced system over $\mathbb F_p$ for several primes $p$. Then get for example zero dimension ideal and then conclude that dimension of corresponding ideal over $\bar{\mathbb Q}$ is therefore zero. Are there theorems and approaches helping with that?

P.S. Example of system (variables: a0, a1, a2):

5/2*a0*a2^4 - 6*a1*a2^4 + 27/4*a1^2*a2^2 + 9/4*a2^4 = 0

243/32*a0*a1*a2^2 + 405/32*a0^2*a1 - 243/64*a0^2 = 0

27/8*a1^4*a2^4 - 135/8* a0*a1*a2^6 + 9*a1^2*a2^6 + 9/4*a2^8 - 9/4*a0^2*a1^5 = 0
  • $\begingroup$ Do you mean that you have explicit equations? $\endgroup$ – jmc Jun 27 '18 at 17:25
  • $\begingroup$ @jmc, yes I added an example (simplified). $\endgroup$ – Maxim Jun 27 '18 at 21:29
  • $\begingroup$ By 'dimension $0$ ideal', do you mean that $R/I$ has Krull dimension zero? $\endgroup$ – R. van Dobben de Bruyn Jun 27 '18 at 21:40

There are easy examples of finite type $\mathbb Z$-algebras $R$ such that $R \otimes \mathbb F_p$ has Krull dimension $0$ for any given finite set of primes $p$, but $R \otimes \mathbb Q$ has positive dimension.

Example. Let $n \in \mathbb Z_{\geq 1}$, and set $$R = \mathbb Z[x,y]/(nx-1) = \mathbb Z\left[\tfrac{1}{n}\right][y].$$ Then $R \otimes \mathbb F_p \cong \mathbb F_p[x,y]/(nx-1)$ is the zero ring if $p \mid n$ (hence has Krull dimension $-\infty$, I guess). On the other hand, $R \otimes \mathbb Q \cong \mathbb Q[y]$ has Krull dimension $1$.

Example. It is easy to modify the above example so that $R \otimes \mathbb F_p$ has Krull dimension $0$ instead of $-\infty$: just take the product $$R = \mathbb Z\big[\tfrac{1}{n}\big][y] \times \mathbb Z \cong \mathbb Z[x,y,z]/\big((nx-1,z)\cdot(x,y,z-1)\big).$$ Indeed, $(nx-1,z) + (x,y,z-1) = (1)$, so the Chinese remainder theorem shows that the quotient is the product ring. (Geometrically, we embed a disjoint union onto the $z = 0$ and $z = 1$ hyperplanes.)

Thus, $R \otimes \mathbb F_p \cong \mathbb F_p$ for $p \mid n$, whereas $R \otimes \mathbb Q \cong \mathbb Q[y] \times \mathbb Q$.

Remark. On the other hand, if $R \otimes \mathbb F_p$ has Krull dimension $0$ for infinitely many primes $p$, then it is true for all but finitely many primes, and $R \otimes \mathbb Q$ also has Krull dimension $0$. This follows by constructibility of the dimension function; cf. EGA IV$_3$, Prop.

However, this is probably not very useful for computational purposes...


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