What is the computational complexity of solutions over $\mathbb{Q}$ of polynomial equation with coeffiecents over $\mathbb{Z}$ What is the complexity of the following problem? (e.g. best-known running time, space, best upper bound in terms of complexity classes, etc.)


Input: A multivariate polynomial $f$ with coefficients over $\mathbb{Z}$
Decide: Is there a rational zero of $f$?


What about the complexity where $f$ is the translation of a Boolean formula under the standard translation? (viz. $t(x) = x$, $t(f \vee g) =  1-(1-t(f))(1-t(g))$ and $t(f \wedge g) = t(f) t(g)$; thus, a $\{0,1\}$ assignment satisfies $f$ iff it satisfies the polynomial equation $t(f)=1$.)
As already pointed out by Emil Jeřábek in the comments, this is Hilbert's Tenth Problem over the rationals, whose computability is a notorious open question.
In this case, note that the Boolean formula $f$ is satisfiable iff $t(f)$ has an integer zero: a satisfying assignment, when treated as 0s and 1s, gives an integer zero, and conversely, given an integer zero $\vec{\alpha}$, taking it mod 2 gives a satisfying Boolean assignment (this follows from the form of the translation $t(\bullet)$). Now, given a rational solution to $t(f)=0$, if all of the denominators involved are odd, then we may still take it mod 2 to get a Boolean solution, so we have



  Boolean solution iff integer solution iff rational solution w/ odd denominators


Therefore, deciding the existence of integer roots or of rational roots with odd denominators for polynomials of the form $t(f)$ is $\mathsf{NP}$-complete.

This leaves us with the question of how hard it is to decide existence of a rational root of $t(f)$ in general (that is, without restriction on the denominators). I do not know anything more about this question than about Hilbert's Tenth Problem over $\mathbb{Q}$ (addressed in Q1).