Let $A$ is a commutative ring with $1$ and $a$, $b$ are the elements of $A$ such that $a+b=1$ and there is some $n>0$ , $(ab)^n=0$. Prove: $(1)=(a^n)+ (b^n)$.
I would say it's kind of easy one, but somehow I got stuck on it too long. Please help me !
There is a more general phenomenon here. Suppose that $A$ is a commutative ring with unity, and $I, J \subset A$ are ideals. Then if $I + J = A$, then $I^{n} + J^{m} = A$ for any $n, m > 0$.
The proof is as follows. If $I + J = A$, then there exist $i \in I, j \in J$ such that $i + j = 1$. Then $(i+j)^{m+n} = 1^{m+n} = 1$. By the binomial theorem,
$$(i+j)^{m+n} = \sum_{k = 0}^{m+n} {m+n \choose k} i^{k}j^{m+n-k}$$
For $k \geqslant n$, the term $i^{k}j^{m+n-k} \in I^{n}$. If $k < n$, then $m+n-k > m$, so $i^{k}j^{m+n-k} \in J^{m}$. Hence, $1 = (i+j)^{m+n} \in I^{n} + J^{m}$. For your particular question, take $I = (a), J = (b)$.
