Every radical is prime?

Question

$a$ is an ideal of $A$. $$f:A\to A/a,\ \ x∈r(a)$$

r(a) is a prime ideal?

proof 1:

$x^n\in a$ for some $n \Rightarrow (x+a)^n\in a$ for some $n \Rightarrow f(r(a))=\text{nil-radical}$ in $f(a) \Rightarrow r(a)$ is a prime ideal. (Since nil-radical is a prime ideal)

proof 2:

$x^n\in a$ for some $n \Rightarrow (xb)^n\in a$ for some $n$ ,$b\notin r(a) \Rightarrow r(a)$ is a prime ideal

I must be doing something wrong. Can someone fix the proof?

Answer 1

for proof 1: Nilradical is not a prime ideal it is intersection of all prime ideals.
for proof 2: To prove $r(a)$ is prime you should prove: $b\notin r(a)$ and $x\notin r(a)$ which is not true