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

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?

Answer 1

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 $b=abx+bm\in M$.

Answer 2

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 $b=b-abc+abc \in J$