The principal ideal $(x(x^2+1))$ equals its radical. Let $\mathbb R$ be the reals and $\mathbb R[x]$ be the polynomial ring of one variable with real coefficients. Let $I$ be the principal ideal $(x(x^2+1))$. I want to prove that the ideal of the ideals variety is not the same as its radical, that is, $I(V(I))\not=\text {rad}(I)$. I've reduced this to proving that $I=\text{rad}(I)$. How can I go about that?

This $x(x^2 + 1)$ is the factorisation of that polynomial into irreducibles (over $\mathbb{R}$). Once you have such a factorisation, the radical is the factorisation with no powers. So it is radical.
EDIT — I am referencing this:
Let $f \in k[x]$ be a polynomial, and suppose that $f = f_1^{\alpha_1} \cdots f_n^{\alpha_n}$ is the factorisation of $f$ into irreducibles. Then $\sqrt{(f)} = (f_1 \ldots f_n)$. 
One inclusion should be clear, and the other can be seen by appealing to the uniqueness of factorisation in $k[x]$.
