Vtyjkyuk

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 $mathfrakq$ is a primary ideal in $A$ and fix an $xin A-mathfrakq.$ Then $barxin A/mathfrakq$ is non-zero since $xnotinmathfrakq$

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

Then as $xinlangle xrangle=langle x^2rangle$, we have $x=ax^2$ for some $ain A$ and hence $x(ax-1)=0inmathfrakq$ and we thus have $(ax-1)^ninmathfrakq$ for some integer $n>0$ since $xnotinmathfrakq.$

Therefore, we get that $(ax-1)^n=bar0$ in $A/mathfrakq$. It follows that $barabarx-bar1in A/mathfrakq$ is nilpotent and thus $barabarx=(barabarx-bar1)+bar1in A/mathfrakq$ is unit which implies $barxin A/mathfrakq$ is unit. Therefore, $A/mathfrakq$ is a field. It follows that $mathfrakq$ 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.

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?