Exercise 4.3, Atiyah Macdonald

Exercise 2.27 (Atiyah-Macdonald p.35) says $A$ is absolute flat iff every principal ideal is idempotent.

The problem can be proved using only this fact.

Let $q$ be any primary ideal of $R$. We will show $A/q$ is a field. Lex $x$ be any elements of $A$ which is not contained in $q$(This means $x$ is non-zero element in $A/q$). Then $(x)=(x^2)$. So for some $a$, $ax^2=x$,hence $x(ax-1)=0$. Since $x \notin q$ and $q$ is a primary ideal, there is some $n>0$ s.t. $(ax-1)^n \in q$ . Expanding $(ax-1)^n$ we know $1 -bx \in q$ for some $b \in A$. This means $x$ is invertible in $A/q$. i.e. $A/q$ is a field so $q$ is a maximal ideal.

sorry for my poor English...