TITLE: An ideal that is radical but not prime.
QUESTION [13 upvotes]: I'm preparing for an exam and, as part of this preparation, 

I'm looking for an ideal $I$ in an integral domain $R$ that is radical but not prime. 

Here is an example I'm fooling around with:
Let $R=\mathbb{R}[x]$ and let $I=(x(x-1))$. I'm having trouble showing that this ideal is in fact radical. My intuition is to consider the quotient ring $\mathbb{R}[x]/(x(x-1))$ and determine if it is reduced, that is, whether or not it has trivial nilradical. However, this has only led me in circles so far. $(x(x-1))$ is clearly not a prime ideal, so it suffices to show it's radical.
Any help would be appreciated.

REPLY [3 votes]: Not sure how much algebraic geometry you have at your disposal, but since your attempt used polynomials maybe this will work well for you. It's certainly not as simple as the $6\mathbb{Z}$ answer, but how about an intentional algebro-geometric construction? Let's take $\mathbb{C}[x,y]$ which suffices to be an integral domain, and will help us out for being algebraically closed. We will use the following results:


$\textbf{I}(V)$ is a radical ideal.
$\textbf{I}(V)$ is prime if and only if $V$ is irreducible.


So the game plan now is to construct a reducible variety $V$, then take $\textbf{I}(V)$, which cannot be prime but must be radical.
The variety $V_1 = \textbf{V}(y-x^2)$ is the variety who's real part corresponds to a classic upward opening parabola, and $V_2 = \textbf{V}(y+x^2)$ is the variety who's real part is a downward opening parabola. It is easy enough to form the variety $V_3 = V_1 \cup V_2$ by taking $V_3 = \textbf{V}((y-x^2)(y+x^2))$. Thus $\textbf{I}(V_3)$ is radical, but cannot be prime.