Counting the ideals of $\frac{\mathbb{R}[X]}{(X^2)}$ I want to ask you guys if I'm on the right track:
Here's the question:
Suppose $a \in \mathbb{R}$. Count the ideals of $\frac{\mathbb{R}[X]}{(X^2-a)}$.
Give an example of a ring with exactly 3 prime ideals. Explain.
So I'm going to begin by first assuming $a=0$. I'll treat $a<0$ and $a>0$ later.
Now $\frac{\mathbb{R}[X]}{(X^2)} = aX+ b +(X^2)$. Where of coure $(X^2)=\{a_nX^n+ \dots a_2X^2 \}$. Now let's take a look at some ideals of $\frac{\mathbb{R}[X]}{(X^2)}$. $\mathbb{R}[X]$ so I suppose that $\frac{\mathbb{R}[X]}{(X^2)}$ will be one too. Therefore all ideals can be writen as $(a+bX + (X^2))$ with $a,b \in \mathbb{R}$. It's trivial that $(c+(X^2))=\frac{\mathbb{R}[X]}{(X^2)}$. Now look at $(aX + b + (X^2))$ with $a, b \neq 0$. So multiply by the right  $cX +d + (X^2)$ and I guess you can quickly derive that $(aX+ b + (X^2))=R$ Wich leaves $(X)$ as the only other ideal. 
What do you think of this line of reasoning? Is my assumption that $\frac{\mathbb{R}[X]}{(X^2)}$ is an PID true? It seems so logical... 
Thanks!

$\Bbb R[x]/(x^2)$ isn't a principal ideal domain because $x^2=0$. (For commutative rings, $R/I$ is a domain exactly when $I$ is a prime ideal.)
It is a principal ideal ring though, because it's a quotient of a principal ideal ring. Your reasoning that $(x)$ is the only ideal between $x^2$ and $R[x]$ is correct, but could be greatly simplified. The ideals of $\Bbb R[x]$ correspond to the monic polynomials that generate them, and containment between these ideals corresponds to divisiblility. Ideals of $\Bbb  R[x]/(p(x))$ correspond (by an isomorphism theorem) to divisors of $p(x)$. The only ideal between the two is $(x)$ since $x$ is (up to equivalence) the only proper irreducible divisor of $x^2$.
So, the ring has exactly three ideals, but only one of them is prime.
One thing you can use to your advantage is that the prime ideals coincide with the maximal ideals in Artinian rings, and quotients of $\Bbb R[x]$ by nonzero ideals are going to be Artinian. Luckily, it's very easy to control the maximal ideals of quotients of $\Bbb R[x]$. You found one where $(x)$ was the unique maximal ideal, since $x$ was the only divisor of $x^2$. Can you imagine a way to get three different irreducible divisors?
