Help with proof of $ \mathbb{C}[X] \simeq R $ where $R$ is a $ \mathbb{C}$-algebra without nilpotents I am trying to understand the proof of the following proposition:

Let $X \subset \mathbb{A}^n$ be closed. Let $ R $ be a finitely generated $ \mathbb{C}$-algebra without nilpotents. There exists an affine variety $ X $ such that $\mathbb{C}[X] \simeq R $ (as $ \mathbb{C}-$ algebras).

where $\mathbb{C}[X]$ is the ring of regular functions on $X$.
Proof: Let $ \alpha _1, \cdots, \alpha _n $ be generators (over $ \mathbb{C} $) of $ R $, and let $ \phi: \mathbb{C}[z_1,\cdots,z_n] \longrightarrow R$ be the surjection of algebras mapping $ z_i$ to $\alpha_ i$ . The kernel of $ φ $ is an ideal $ I \subset \mathbb{C}[z_1,\cdots,z_n]$, which is radical because $ R $ has no nilpotents. Let $ X:= V(I) \subset \mathbb{A}^n$. Then $ \mathbb{C}[X] \simeq R $.
I can't understand the sentence in bold, i.e. why if $ R $ is a finitely generated $ \mathbb{C}$-algebra without nilpotents then $ I $ is radical.
Have you any ideas?
Thank you in advance.

Well, let $f$ be a polynomial with $f^n\in I$, i.e., $\phi(f^n)=0$. Then $\phi(f)^n=\phi(f^n)=0$ and so $\phi(f)$ is nilpotent in $R$. By hypothesis $\phi(f)=0$ and so $f\in I$. Hence, $I$ is radical.
