Uniqueness of prime ideals of $\mathbb F_p[x]/(x^2)$ 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.

Prime ideals do correspond under the correspondence theorem, so your argument suffices.
To see this,
Let $I\subset P\subset R$ be  any prime in $R$ containing an ideal $I$, then
$R/P \cong \frac{R/I}{P/I}$ by the 3rd isomorphism theorem. Since $P$ is prime, $R/P$ is an integral domain, hence so is $\frac{R/I}{P/I}$. 
Thus, $P/I$ is a prime ideal in $R/I$.
Now start with a prime ideal $Q\subset R/I$, lift it to an ideal containing $I\subset Q'\subset R$, and apply the same argument to see that $Q'$ is a prime ideal in $R$.
For uniqueness, you are correct in saying that if $(x^2)\subset P$, then $(x)\subset P$, as $P$ is prime. But $(x)$ is maximal, so this forces $(x) = P$ by definition of a maximal ideal. To see that it is maximal, note $F[x]/(x)\cong F$ is a field.
