What are the prime ideals of $\mathbb F_p[x]/(x^2)$? I have been told that the only one is $(x)$, but I would like a proof of this. I want to say that a prime ideal of $\mathbb F_p[x]/(x^2)$ corresponds to a prime ideal $P$ of $\mathbb F_p[x]$ containing $(x^2)$. And then $P$ contains $(x)$ since it is prime. But I don't know if prime ideals correspond to prime ideals under the correspondence theorem, and I still can't seem to prove that if they do, $P$ can't be some non-principal ideal properly larger than $(x)$.
Some context: I'm considering why the prime ideals $\mathfrak p$ of $\mathcal O_K$, (with $K=\mathbb Q(\sqrt d)$ and $\textrm{Norm}(\mathfrak p)=p$, a ramified prime) are unique. My definition of a ramified prime is that $\mathcal O_K/(p) \cong \mathbb F_p[x]/(x^2)$ and I know nothing else about these primes.