TITLE: Can the Riemann hypothesis be undecidable?
QUESTION [53 upvotes]: The question is contained in the title; I mean the standard axioms ZFC. The wiki link: Riemann hypothesis. There are finite algorithms allowing one to decide if there are non-trivial zeroes of the $\zeta$-function in the domains whose union exhausts the whole strip $0<\Re z<1$, but this does not seem to be the obstacle for undecidability. Are there other arguments?

REPLY [14 votes]: Without worrying about reductions, if RH is false, it is provably false: suppose $\rho\in\mathbb{C}$ is a zero in the critical strip but off the critical line (say, $\Re(\rho)>\frac12$). Then for a little rectangle $R$ with (say) rational corners, containing $\rho$ in its interior but not containing the pole at $s=1$ or intersecting the critical line we'd have
$$\frac{1}{2\pi i}\oint_{\partial R} \frac{\zeta'}{\zeta}(s)\mathrm{d}s \geq 1$$
by the argument principle.
But we can approximate $\zeta(s),\zeta'(s)$ to arbitrary precisition by a finite computation, and similarly we can approximate the integral to arbitrary precision by a numerical computation.  In other words, there is a finite computation which provably approximates the integral above to within $\frac{1}{2}$.  Then the non-vanishing of the approximation proves the falsity of RH.
Similarly, RH is equivalent to estimates on the prime-counting function, for example to $$|\psi(x)-(x)|<\sqrt{x}\log^2 x$$
where $\psi(x) = \sum_{p^r<x}\log p$.
While the models can disagree about some aspects of the real numbers, I don't think they can disagree about the logarithm of an integer (in the sense that $\log(n)=-\log(1/n) = \sum_{k=1}^\infty \frac{(n-1)^k}{kn^k}$), and it's enough to consider integer $x$ in the inequality.  This means that if RH is false in a model, RH will be false in any model whose integers contains those of the original model.
.
Edited to clarify the last comment: in the prime-counting inequality, replace the term $\log p$ associated to the prime power $p^r$ with $\sum_{k=1}^{100p^{2r}} \frac{(p-1)^k}{kp^k}$.  Similarly, replace the $\log x$ with a truncation of the logarithm series.  RH is still equivalent to the inequality using the truncated logarithm, and in the new formulation both sides of the inequality are explicit rational numbers! In other words, RH is equivalent to a countable sequence of inequalities about explicit rational numbers.
Second edit: I realize that the direct formulation can also be converted to a statement about rationals.  Let $R_n$ be the axis-parallel rectangle with corners at $\frac{1}{2}+\frac{1}{n}+i$ and $1+ni$.  Let $I_n$ be a rational approximation to the zero-counting integral above around $R_n$, accurate to better than $\frac{1}{2}$ (obtained by dividing the boundary of $R$ into many points, at each point approximating $\frac{\zeta'}{\zeta}$ by a rational number, and then integrating numerically).  This is an explicit finite computation, and RH is equivalent to the infinite system of inequalities $I_n<0.5$.
Third edit: removed false statement about submodels.