TITLE: Prove every maximal ideal of $A$ is a prime ideal (Hint: Use the fact that $J$ is a maximal ideal iff $A/J$ is a field.)
QUESTION [8 upvotes]: Let $A$ be a commutative ring with unity. Prove: Proof every maximal ideal of $A$ is a prime ideal (Hint: Use the fact that $J$ is a maximal ideal iff $A/J$ is a field.)

In the question before this I proved that $J$ is a prime ideal iff $A/J$ is an integral domain.  Now, I have what I think is a "pseudoproof" and as such, am not satisfied.
$\rightarrow$ Let $J$ denote an arbitrary maximal ideal of $A$.  Since $J$ is a maximal ideal of $A$, $A/J$ is a field.  Because every field is an integral domain, $A/J$ is an integral domain.  Since $A/J$ is an integral domain, $J$ is a prime ideal.  Thus, every maximal ideal $J$ of $A$ is a prime ideal.
Is there another way to prove this directly?

REPLY [3 votes]: An easy way for me:
Let $ab\in J$ where $a,b \in A$
Let $a\notin J $ To show $b\in J$.
Since $a\notin J,a+J\neq 0+J$ Hence $a+J $ has an inverse in $A/J$
So $\exists (c+J) \in A/J$ such that $(a+J)(c+J)=1+J\implies ac+J=1+J\implies ac-1\in J\implies bac-b\in J\implies abc-b\in J$
Again $ab\in J\implies abc \in J$
Thus $b=b-abc+abc \in J$