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?

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.
Yes, this is good.

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?
Let $\displaystyle R$ be a commutative (unitary) ring. An ideal $\displaystyle P$ is called a prime ideal iff $\displaystyle ab\in P \implies a\in P \text{ or }b\in P$.

Lemma: If $\displaystyle P$ is a prime ideal and $\displaystyle a_1\cdot ... \cdot a_k \in P$ then $\displaystyle a_1\in P$ or $\displaystyle a_2\in P$ or ... or $\displaystyle a_k \in P$.
Proof: The proof is by induction. If $\displaystyle n=2$ that is true because that is the definition of what it means being a prime ideal. So say it is true for $\displaystyle n=k$. Now say $\displaystyle a_1\cdot ... \cdot a_k \cdot a_{k+1} \in P$ then it means $\displaystyle \left( a_1\cdot ... \cdot a_k \right) \cdot a_{k+1} \in P$ then by property of being a prime ideal means $\displaystyle a_1\cdot ... \cdot a_k \in P$ or $\displaystyle a_{k+1} \in P$. By the inductive step we get $\displaystyle a_1\in P$ or ... or $\displaystyle a_k \in P$ or $\displaystyle a_{k+1}\in P$. And that completes the proof.

Therefore given $\displaystyle Q$ and ideal containing $\displaystyle I$ let $\displaystyle a\in \sqrt{I}$ then it means $\displaystyle a^n \in I \subseteq Q$ for some $\displaystyle n\geq 1$. However, $\displaystyle a^n = a\cdot ... \cdot a \in Q \implies a\in Q$ by above.