If $A$ is absolutely flat then every primary ideal is maximal

Question

Given the ring $A$ is commutative with an identity element.

If $A$ is absolutely flat (i.e. each $A$-module is flat) then every primary ideal is maximal.

This exercise 4.3 comes from the classical text of Atiyah-Macdonald : introduction to the commutative algebra.

Attempt:

Suppose $\mathfrak{q}$ is a primary ideal in $A$ and fix an $x\in A-\mathfrak{q}.$ Then $\bar{x}\in A/\mathfrak{q}$ is non-zero since $x\not\in\mathfrak{q}$

Recall that $A$ is absolutely flat $\Longleftrightarrow$ every principal ideal is idempotent.

Then as $x\in\langle x\rangle=\langle x^{2}\rangle$, we have $x=ax^{2}$ for some $a\in A$ and hence $x(ax-1)=0\in\mathfrak{q}$ and we thus have $(ax-1)^{n}\in\mathfrak{q}$ for some integer $n>0$ since $x\not\in\mathfrak{q}.$

Therefore, we get that $(ax-1)^{n}=\bar{0}$ in $A/\mathfrak{q}$. It follows that $\bar{a}\bar{x}-\bar{1}\in A/\mathfrak{q}$ is nilpotent and thus $\bar{a}\bar{x}=(\bar{a}\bar{x}-\bar{1})+\bar{1}\in A/\mathfrak{q}$ is unit which implies $\bar{x}\in A/\mathfrak{q}$ is unit. Therefore, $A/\mathfrak{q}$ is a field. It follows that $\mathfrak{q}$ must be a maximal ideal in $A$.

I am not very sure if my proof is valid. Any suggestion or comment I will be grateful.

Answer 1

This is a perfect proof, and very neatly written. Obviously, you are on a level where you no longer need this kind of supervision. You can determine if a proof is correct or not.

Or is there a particular point in the proof that you are uncertain about?

Answer 2

A short proof of primary ideal of an absolutely flat ring is maximal:

Let $A$ be an absolutely flat ring and $x\in A$. From Exercise 2.27 of Atiyah-Macdonald, $<x>=<x^2>=....=<x^n>$ for every positive integer $n$.

Suppose $q$ is a primary ideal of $A$ and $m$ a maximal ideal containing it. Take any $x\in m$.

Since $x=bx^2$ for some $b\in A$, $x(1-bx)= 0$ and $1-bx\notin m$ (otherwise 1 is in $m$). Thus $x^n\in q$ for some positive integer $n$. But from $<x>=<x^n>$ implies $x= ax^n$ for some $a\in A$. Thus $x\in q$ and we conclude.