Radical and Ideal

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Originally Posted by tttcomrader

Proof.

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

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

Yes, this is good.

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

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

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

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