Could anyone explain to me why maximal ideals are prime?

I’m approaching it like this, let R be a commutative ring with 1 and let A be a maximal ideal. Let a,b∈R:ab∈A.

I’m trying to construct an ideal B such that A⊂B≠A As this would be a contradiction. An alternative idea I had was to prove that R/A is an integral domain, but this reduces to the same problem.

EDIT: Ergh.. just realized that I’ve learnt a theorem that states is A is a maximal ideal then R/A is a field

**Answer**

Here’s a proof that doesn’t involve the quotient R/A.

Suppose that A is not prime; then there are a,b∈R∖A such that ab∈A. Let B be the ideal generated by A∪{a}; B={x+ar:x∈A and r∈R}. Clearly A⊊, so B = R, 1_R \in B, and hence 1_R = x + ar for some x\in A and r\in R. Then b = b1_R = b(x+ar) = bx + bar. But bx \in bA \subseteq RA = A, and bar \in Ar \subseteq AR = A, so b \in A. This contradiction shows that A is prime.

**Attribution***Source : Link , Question Author : Freeman , Answer Author : Brian M. Scott*