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Homework answers / question archive / Prove that if R is commutative ring and N=(a1,a2,
Prove that if R is commutative ring and N=(a1,a2,..am) where each ai is a nilpotent element, then N is a nilpotent ideal, i.e N^n=0 for some positive integer n. Deduce that if the nilradical of R is finitely generated then it is a nilpotent ideal.
P.S. the set of nilpotent elements form an ideal which is called nilradical of R for a commutative ring R
Proof:
Since each ai is a nilpotent element, then we can find some positive integer ri for each ai, such that ai^ri=0. Now let n=lcm{r1,r2,...,rm} be the least multiple of r1,r2,...,rm. Then for any element x in N=(a1,a2,...,am), x has the form a1^k1 * a2^k2 *... * am^km for some integers k1,k2,...,km. Since R is commutative, then we have
x^n = a1^(k1*n) * a2^(k2*n) *... * am^(km*n) = 0
Because each ai^(ki*n) = (ai^n)^ki = 0^ki = 0.
Thus N^n=0.
Now if the nilradical N of R is finitely generated, then N=(a1,a2,...,am) for some elements a1,...,am in R. From the above proof, we know there exists some n, such that N^n=0. Thus N is a nilpotent ideal.
