$I(X)$ is a radical ideal I didn't understand why if $X\subset \mathbb A^n$ is an affine algebraic set, then $I(X)$ is a radical ideal in $k[X_1,\ldots,X_n]$.
I'm trying to understand this equivalence:

Thanks a lot

Comment made into an answer in response to Martin Brandenburg.

Because, if $f^n(x)=0$ for some $x\in \mathbb A^n$, then $f(x)=0$. This follows from the definition that $f^n(x)=(f(x))^n$. The correspondence in the other direction is the Hilbert Nullstellensatz.
Indeed, $I(X)$ is always radical, even when $X$ is not algebraic. 
