# 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.)

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?

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 $$b=abx+bm\in M$$.

• What is that ordered pair, $(a,M)$? Why is it equal to $R$? – Al Jebr Dec 2 '14 at 3:26
• $(a, M)$ is the ideal generated by $a$ and $M$ and it is equal to $R$ because it is larger than the maximal ideal. – user157227 Dec 2 '14 at 3:29
• That's a petite proof :D Just so other people don't get confused with notations. (a, M) represents the ideal M + <a> where <a> is ideal general by a and since sum of two ideals is an ideal we have M + <a> as another ideal. @rschweib can you please add a little more explanation so that newbies like me get it in the first glance? – router Jan 6 '19 at 17:28
• @router ok, I added a line to that effect. – rschwieb Jan 6 '19 at 18:21

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

• How did we get from $ac-1 \in J$ to $bac-b \in J$? – Al Jebr Dec 2 '14 at 3:47
• J is an ideal of $A$ so $b\in A$ and $ac-1\in J\implies bac-b\in J$ Got it – Learnmore Dec 2 '14 at 3:51
• Since $J$ absorbs products in $A$, we can multiply both sides on the left by $b$? – Al Jebr Dec 2 '14 at 3:54
• This is a sort of hybrid of the proof in the OP and the brute force proof I suggested. – rschwieb Dec 2 '14 at 4:06
• yes you are correct @JohannFranklin – Learnmore Dec 2 '14 at 4:44