The set of nilpotent elements in a commutative ring is an ideal

Let R be a commutative ring. Recall that an element x \in R is called nilpotent if x^n = 0 for some nonzero natural number n. Prove that the set \mathfrak{N}(R) of all nilpotent elements in R form an ideal. \mathfrak{N}(R) is called the nilradical of R. [Hint: Use the Binomial Theorem to prove closure under addition.]

Let x,y \in \mathfrak{N}(R). Then for some nonnegative natural numbers n and m, we have x^n = y^m = 0.

Consider (x+y)^{n+m}. By the Binomial Theorem, we have (x+y)^{n+m} = \sum_{k=0}^{n+m} \binom{n+m}{k} x^k y^{n+m-k}. Note that if k \geq n, x^k = 0, and if k < n, then y^{n+m-k} = 0. Thus (x+y)^{n+m} = 0, and we have x+y \in \mathfrak{N}(R). Moreover, we have (-x)^n = (-1)^nx^n = 0, so that -x \in \mathfrak{N}(R). Since 0 = 0^1 \in \mathfrak{N}(R), \mathfrak{N}(R) is an additive subgroup of R.

Finally, since R is commutative, if r \in R then (rx)^n = r^nx^n = 0 and likewise (xr)^n = 0. Thus \mathfrak{N}(R) absorbs R on the left and the right, and hence is an ideal.

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  • Art  On November 11, 2011 at 3:24 pm

    Very nice proof.

    • nbloomf  On November 14, 2011 at 9:50 am

      I can’t really take credit for it- even without the hint given by D&F, this proof really suggests itself.🙂

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