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)$\displaystyle M^a+N^b=A (a,b\ge1)$.

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- Mar 15th 2010, 08:28 PMKaKacomaximal(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 PMaliceinwonderland
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 $\displaystyle M^a+N^b=A ,a,b\ge1$.