Prove $\text{rad}(I)/I$ is isomorphic to $\mathfrak{N}(R/I)$

Question

I want to know if this is the correct way to do it. Define $\varphi:\text{rad}(I) \longrightarrow \mathfrak{N}(R/I)$ by $\varphi(r)= r^n+I$,then ker$\varphi = I$, so therefore by the 1st isomorphism theorem, we have:

$$\text{rad}(I)/I \cong \mathfrak{N}(R/I) $$

it satisfies the condition for homomorphism, since

$$\varphi(r_1r_2) = (r_1r_2)^n+I = \varphi(r_1)\varphi(r_2),$$ $$\varphi(r_1+r_2)= (r_1^n+r_2^n) + I = \varphi(r_1)+\varphi(r_2). $$

Do I need to prove something else, or do I miss something? Any help would be appreciate it , thanks

Answer 1

$x \in rad(I) \Leftrightarrow x^n \in I,$ for some $n \geq 1.$ So, $x + I \in rad(I)/I \Leftrightarrow x^n + I = 0$ in $R/I$ for some $n \geq 1 \Leftrightarrow (x + I )^n = 0$ in $R/I$ for some $n \geq 1 \Leftrightarrow x + I \in \mathfrak{N}(R/I).$