Thread: relatively prime ideals

Suppose $\displaystyle R$ is a ring and $\displaystyle I, J $ are relatively prime ideals of $\displaystyle R$; that is, $\displaystyle I + J = R$. Prove that for any integers $\displaystyle n, m \leq 1$, $\displaystyle I^n+J^m=R$.

Originally Posted by Erdos32212

Suppose $\displaystyle R$ is a ring and $\displaystyle I, J $ are relatively prime ideals of $\displaystyle R$; that is, $\displaystyle I + J = R$. Prove that for any integers $\displaystyle n, m \leq 1$, $\displaystyle I^n+J^m=R$.

first of all, that should be $\displaystyle m,n \geq 1,$ and also i assume that your ring is commutative with unity. so you only need to show that $\displaystyle 1 \in I^n +J^m.$ this is very easy:

since $\displaystyle I+J=R,$ there exist $\displaystyle r \in I, \ s \in J$ such that $\displaystyle r+s=1.$ thus: $\displaystyle 1=(r+s)^{m+n}=r^n \sum_{i=0}^m \binom{n+m}{i}r^{m-i}s^i \ + \ s^m \sum_{i=m+1}^{n+m} \binom{n+m}{i}r^{n+m-i}s^{i-m} \in I^n + J^m. \ \Box$