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?

Let $f\in{\rm rad}( I)$, that is, $f^n\in I$. You can also think over the complex field: $f=a_n(x-z_1)(x-z_2)\dots (x-z_n)$ for some $z_j\in\Bbb C$. Then $f^n\in I$ means that among the complex roots of $f^n$ we also find the roots of $x(x^2+1)$, i.e., $-i,0,i$. But that means we already had $f\in I$.
