Answer to Question #218734 in Abstract Algebra for Komal

Question

If N is nil ideal of a ring R and A is ideal of R such that A contained in N then prove that A and N/A are also nil ideal

Answer 1

Let $p\in N/A$. Then $p=q+A$ where $q\in N$. Since $N$ is a nil ideal, $\exist k\in \mathbb{Z}$ such that $q^k=0$. But $0\in A$. So $q^k\in A$, which implies that

$$p^k=(q+A)^k=q^k+A=0+A$$

the zero element of $N/A$. Hence $N/A$ is a nil ideal.