Proof

Suppose M is a maximal ideal of R. Let a and b be elements of R such that ab M.

If neither a nor b belongs to M, then R = (a) + M and R = (b) + M, because of the maximality of M. This implies the existence of elements r, s R and m, n M such that

1 = ar + m ,     1= bs + n.

Multiplying left-hand sides and right-hand sides yields

1 = abrs + arn + bsm + mn.

As ab, m, n M, we find 1 M, a contradiction.

Hence a or b belongs to M, proving that M is a prime ideal.