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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.

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  • $\begingroup$ You may look at $\mathbb F_p[x]/(x^2)$ as an $\mathbb F_p$- vector space (or algebra), then you would then realise it is a two dimensional $\mathbb F_p$- vector space. $\endgroup$
    – Math137
    Apr 22, 2015 at 12:22
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    $\begingroup$ This question should be useful: math.stackexchange.com/questions/54943/… $\endgroup$
    – lokodiz
    Apr 22, 2015 at 12:23

2 Answers 2

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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.

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and I still can't seem to prove that if they do [correspond], $P$ can't be some non-principal ideal properly larger than $(x)$.

Assuming you convince yourself of the correspondence of prime ideals (which is just fine) here's an elementary way to see that there can only be one prime ideal in this ring.

Now $(x)$ is a maximal ideal of $\Bbb F[x]$ for any field $F$. This is easily seen since $F[x]/(x)\cong F$. The ideal $(x^2)=(x)^2$ therefore is a power of a maximal ideal.

We can say a little by abstracting:

Proposition: If $R$ is any ring and $M$ is a maximal ideal, $R/M^n$ has exactly one prime ideal, namely $M/M^n$.

Proof: A prime ideal of $R/M^n$ looks like $P/M^n$ where $P$ is a prime ideal of $R$ containing $M^n$. Plainly $M^n\subseteq P$, and by primeness $M\subseteq P$. But $M$ is maximal, so $M=P$. Thus there is only one prime ideal containing $M^n$, and only one prime ideal of $R/M^n$.

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