Consequences and state-of-the-art of NP ≠ ZPP? Consider the complexity classes $\mathsf{NP}$ and $\mathsf{ZPP}$. Whether the two classes are equal is an open question, but as far as I know, $\mathsf{NP} = \mathsf{ZPP}$ is not known to imply $\mathsf{P} = \mathsf{NP}$. For example, there are many inapproximability results depending on $\mathsf{NP} \neq \mathsf{ZPP}$, but what evidence do we have for believing they would indeed be not equal?

In other words, what would be the consequences of $\mathsf{NP} = \mathsf{ZPP}$? Is it safe to claim that likewise for $\mathsf{P}$ and $\mathsf{NP}$, many theorists also believe $\mathsf{NP}$ is not equal to $\mathsf{ZPP}$?
Yes, it is safe to claim that most theorists believe that $\mathsf{NP}$ is not equal to $\mathsf{ZPP}$ -- and that most theorists believe that $\mathsf{NP}$ is not equal to $\mathsf{BPP}$.

One consequence of $\mathsf{NP} = \mathsf{ZPP}$ is that computational cryptography is impossible.  This is described as the world Algorithmica in Impagliazzo's famous five worlds:


A Personal View of Average-Case Complexity.  Russell Impagliazzo.  1995 Complexity Conference. (A PDF version is here).


Note that as far as practical consequences go, Impagliazzo says it doesn't matter much whether we have $\mathsf{P} = \mathsf{NP}$ or $\mathsf{NP} \subseteq \mathsf{BPP}$ (he even calls the latter the "moral equivalent" of $\mathsf{P} = \mathsf{NP}$).

See also What if P = NP? for additional practical consequences (assuming the constants hidden by asymptotic notation are not too large).

I don't know whether there are any surprising implications for complexity classes.