Let I be an ideal of a commutative ring R with identity. Prove that the radical of I is an ideal of R and each prime ideal that contains I, also contains $\displaystyle \sqrt {I} $. Also prove that $\displaystyle \frac { \sqrt {I} }{I} $ is the nilradical of $\displaystyle \frac {R}{I} $

Proof.

Now the radical of I is $\displaystyle \sqrt {I} = \{ r \in R : r^n \in I \ , \ n \geq 1 \} $

So pick $\displaystyle i \in \sqrt {I} $ and $\displaystyle j \in R $, so $\displaystyle i^n = k \in I $, $\displaystyle (ij)^n=i^nj^n=kj^n \in I$ since I is an ideal. Therefore $\displaystyle \sqrt {I} $ is also an ideal.

Now, suppose that Q is a prime ideal of R such that $\displaystyle I \subseteq Q $. Pick $\displaystyle i \in \sqrt {I} $, so then $\displaystyle \exists n $ such that $\displaystyle i^n \in I $ So I need to show that $\displaystyle i \in Q $... How should I start?