# comaximal(coprime)

• Mar 15th 2010, 08:28 PM
KaKa
comaximal(coprime)
Let \$\displaystyle A\$ be a commutative ring with unity.
If \$\displaystyle M,N\$ are distinct maximal ideals of \$\displaystyle A\$, then
(1)\$\displaystyle M+N=A\$.
(2)\$\displaystyle M^a+N^b=A (a,b\ge1)\$.
• Mar 15th 2010, 10:45 PM
aliceinwonderland
Quote:

Originally Posted by KaKa
Let \$\displaystyle A\$ be a commutative ring with unity.
If \$\displaystyle M,N\$ are distinct maximal ideals of \$\displaystyle A\$, then
(1)\$\displaystyle M+N=A\$.
(2)\$\displaystyle M^a+N^b=A (a,b\ge1)\$.

For (1), the sum of ideals is again an ideal (link). Thus, M+N is an ideal containing M. By hypthesis, M+N should properly contain an maximal ideal M. Thus M+N=A.

For (2), every proper ideal in A is contained in a maximal ideal in A and note that A contains the unity (link).
Assume \$\displaystyle M^a+N^b , a,b\ge1\$ is a proper ideal in A. Then, \$\displaystyle M^a+N^b ,a,b\ge1\$ should be contained in a maximal ideal. It follows that \$\displaystyle M^a+N^b , a,b\ge1\$ should be contained in either M or N (check their intersection). Contradiction !
Thus, \$\displaystyle M^a+N^b , a,b\ge1\$ is an ideal in A which is not a proper ideal in A. We conclude that \$\displaystyle M^a+N^b=A ,a,b\ge1\$.