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?
Best Answer
You could always just use the obvious elementary proof, if someone forced you to.
Suppose $ab\in M$ and $a\notin M$. Then $(a,M)=R$, so $1=ax+m$ for some $x\in R$, $m\in M$. ($(a,M)$ denotes the ideal generated by $a$ and $M$, which is equal to the ideal $aR+M$.)
This yields $b=abx+bm\in M$.