An ideal that is radical but not prime.
Why don't you use a simple example ... Suppose the ring to be $R=\mathbf{Z}$ and take the ideal $6\mathbf{Z}$. Since the $\sqrt{m\mathbf{Z}}=r\mathbf{Z}$ where $r=\Pi_{p|m} p$ and $p$ is a prime number. Here $\sqrt{6\mathbf{Z}}=6\mathbf{Z}$ but $6\mathbf{Z}$ is not a prime ideal since 6 is not prime.
$x(x-1)$ is certainly not prime. Suppose $f^{n}\in \langle x(x-1) \rangle$. Since $f^{n}\in \langle x \rangle$, and $\langle x \rangle$ is prime, for some $m<n,f^{m}\in \langle x \rangle$. Proceed this way we can prove $f$ must be divisible by $x$, and similarly for $x-1$. This gives $f$ is divisible by $x(x-1)$. So this ideal is its own radical.
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.