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Math Help - Ideal of ring

  1. #1
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    Ideal of ring

    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.
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  2. #2
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    Quote Originally Posted by GTK X Hunter View Post
    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.
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