Definition and basic properties of the radical of an ideal

Let R be a commutative ring and let I \subseteq R be an ideal. Define the radical of I, \mathsf{rad}(I), by \mathsf{rad}(I) = \{ x \in R \ |\ x^n \in I\ \mathrm{for\ some}\ n \geq 1 \}. Prove that \mathsf{rad}(I) is an ideal of R containing I and that \mathsf{rad}(I)/I = \mathfrak{N}(R/I).


First, we certainly have I \subseteq \mathsf{rad}(I) since for all a \in I, a^1 \in I. In particular, \mathsf{rad}(I) is nonempty. Now let a,b \in \mathsf{rad}(I) with a^n,b^m \in I. Now (a+b)^{m+n} = \sum_{k=0}^{m+n} \binom{m+n}{k} a^k b^{m+n-k} by this previous exercise. Note that if k \geq n, then a^k \in I, and if k < n, then b^{m+n-k} \in I, since I is an ideal. Thus every term in the expansion of (a+b)^{n+m} is in I, and thus (a+b)^{m+n} \in I. We also have, for all r \in R, (ra)^n = r^na^n \in I, and in particular (-a)^n \in R. Thus \mathsf{rad}(I) is an ideal of R.

Now note the following. x + I \in \mathsf{rad}(I)/I if and only if x \in \mathsf{rad}(I), if and only if x^n \in I for some n \geq 1, if and only if x^n + I = 0 in R/I for some n \geq 1, if and only if (x+I)^n = 0 in R/I for some n \geq 1, if and only if x+I \in \mathfrak{N}(R/I). Thus \mathsf{rad}(I)/I = \mathfrak{N}(R/I).

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