Ideal of ring

Quote:

Originally Posted by GTK X Hunter

Let $R$ be a commutative ring and $N$ be the ideal of $R$ . Show that $\sqrt{N} = \{a\in R | a^n \in N \text{ for some } n\in \mathbb{Z^+}\}$ is also the ideal of $R$ .

Let $a, b \in \sqrt{N}$ . Then, $a^i \in N$ and $b^j \in N$ for some $i, j \in \mathbb{Z^+}$ .
Basically you need to argue the followings in order to show that $\sqrt{N}$ is indeed an ideal of R.

1. Since $0 \in N$ , it follows immediately that $0 \in \sqrt{N}$ .
Since $a^i \in N$ , we see that $(-a)^i = (-1)^i(a)^i \in N$ . Thus $-a \in \sqrt{N}$ .

2. $a+b \in \sqrt{N}$ .
Since $a^i \in N$ and $b^j \in N$ for some $i, j \in \mathbb{Z^+}$ , we see that $(a+b)^{i+j} \in N$ (Use a binomial exapansion ).

3. $ab \in \sqrt{N}$ .
Argue that $(ab)^{ij} = (a^i)^j(b^j)^i \in N$ .

4. $ra \in \sqrt{N}$ or $ar \in \sqrt{N}$ for any $r \in R$ .
Argue that $(ra)^{i} = (r^i)(a^i) \in N$ or $(ar)^{i} = (a^i)(r^i) \in N$ . Since R is a commutative ring (its ideals are two-sided), you need to show either of them.