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 $b=abx+bm\in M$.

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    $\begingroup$ What is that ordered pair, $(a,M)$? Why is it equal to $R$? $\endgroup$ – Al Jebr Dec 2 '14 at 3:26
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    $\begingroup$ $(a, M)$ is the ideal generated by $a$ and $M$ and it is equal to $R$ because it is larger than the maximal ideal. $\endgroup$ – user157227 Dec 2 '14 at 3:29
  • $\begingroup$ 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? $\endgroup$ – router Jan 6 '19 at 17:28
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    $\begingroup$ @router ok, I added a line to that effect. $\endgroup$ – 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 $b=b-abc+abc \in J$

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

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