# relatively prime ideals

• November 8th 2008, 10:00 PM
Erdos32212
relatively prime ideals
Suppose $R$ is a ring and $I, J$ are relatively prime ideals of $R$; that is, $I + J = R$. Prove that for any integers $n, m \leq 1$, $I^n+J^m=R$.
• November 8th 2008, 11:38 PM
NonCommAlg
Quote:

Originally Posted by Erdos32212
Suppose $R$ is a ring and $I, J$ are relatively prime ideals of $R$; that is, $I + J = R$. Prove that for any integers $n, m \leq 1$, $I^n+J^m=R$.

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

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