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 nontrivial 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?

I do not know anything about zerofinding algorithms for $\zeta$, so I will make only one small remark which doesn't require such knowledge: If the Riemann Hypothesis is false, then it is provably false (in ZFC, or any similar system). This is because Robin's theorem tells us that the Riemann hypothesis is equivalent to the assertion that, for every natural $n \geq 5041$, the sum of the divisors of $n$ is less than $e^{\gamma} n \log{\log{n}}$; since there are programs which calculate this latter quantity to arbitrary precision, and thus can verify whether this inequality holds for any given $n$, we find that the Riemann hypothesis is a $\Pi_1$ statement: it is equivalent to the assertion that some computer program never outputs "NO" on any input. (Although not familiar with the proofs of Robin's theorem, etc., I assume they can be carried out in ZFC, and thus establish the relevant equivalence within ZFC.). There may be more direct ways to establish that the Riemann hypothesis is a $\Pi_1$ statement, such as by knowledge of algorithms which enumerate to arbitrary precision the zeros of $\zeta$, but at any rate, there is this one. Accordingly, if the Riemann hypothesis is false, then the relevant computer program does output "NO" on some input, from which it would follow that ZFC proves that that computer program outputs "NO" on that input, and thus ZFC would prove the Riemann hypothesis to be false. The possibility still remains, however, as far as I know, that the Riemann hypothesis may be true but unprovable in ZFC. 


Yes. That being said, it's pure speculation. We can as well talk about the decidability of any other famous open problem, but in my experience that usually doesn't lead to anything new. To me, it's highly unlikely that is undecidable, but of course, we can't exclude it. Compare it for example to the rationality of $2*\pi^{4/3}  e^{3/2}$. 


If negation of the Riemann hypothesis is not provable in Peano arithmetic, then the Riemann hypothesis is true. 


In an interview with Martin Davis that appeared in the notices of the AMS, he speculates by the end (pp. 570) with the possibility of RH being undecidable. He was explaining that every $\Pi_1^0$ is equivalent to a statement asserting about a particular polynomial equation with integer coefficients that that equation has no natural number solutions, and that RH was a statement of that kind, as worked out in "Hilbert’s tenth problem: Diophantine equations: positive aspects of a negative solution", by Martin Davis, Yuri Matijasevic, and Julia Robinson, in Mathematical developments arising from Hilbert problems, Proc. Sympos. Pure Math., AMS, 1976. He then goes on: "I am certainly no analyst, but the reason I think the Riemann Hypothesis is a good candidate for undecidability by elementary methods is that it is sitting right in the middle of classical analysis, and it has been attacked by brilliant mathematicians—Paul Cohen spent a lot of time on it—and the existing methods just don’t seem to resolve it. It’s hard to believe it isn’t true. And why shouldn’t it be one of those propositions that require set theoretic methods? That would be great!" As he previously explained, there are some mathematical propositions that "have a very simple form involving solvability of specific Diophantine equations and that require settheoretic methods for their resolution". The fact that RH might be of that character, as he remarks, had already been conjectured by Gödel at the Gibbs lecture. "And that wouldn’t surprise me in the least", he claims. 


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 nonvanishing of the approximation proves the falsity of RH. Similarly, RH is equivalent to estimates on the primecounting 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 $\log(n)=\log(1/n) = \sum_{k=1}^\infty \frac{(n1)^k}{kn^k}$), 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 primecounting inequality, replace the term $\log p$ associated to the prime power $p^r$ with $\sum_{k=1}^{100p^{2r}} \frac{(p1)^k}{kp^k}$. 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 axisparallel rectangle with corners at $\frac{1}{2}+\frac{1}{n}+i$ and $1+ni$. Let $I_n$ be a rational approximation to the zerocounting 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. 


protected by François G. Dorais♦ Dec 23 '15 at 15:24
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