Proof: Let $\mathfrak{p}$ be a prime ideal in $A$. Note that since $\mathfrak{p}$ is a prime ideal we have that $A / \mathfrak{p}$ is an integral domain.
We will show first that $A / \mathfrak{p}$ is a field. Choose $x + \mathfrak{p} \in A / \mathfrak{p}$ such that $x + p \neq 0_{A/ \mathfrak{p}}$. By hypothesis $x^n = x$ for some $n > 1$. Observe then that $$(x+\mathfrak{p})^n = x^n + \mathfrak{p} = x + \mathfrak{p}.$$
We then have that
\begin{align*}
(x+\mathfrak{p})^n = x + \mathfrak{p} &\implies (x+\mathfrak{p})^n - (x+\mathfrak{p}) = 0_{A/\mathfrak{p}} \\
&\implies (x + \mathfrak{p})\left((x+\mathfrak{p})^{n-1} - 1_{A/\mathfrak{p}}\right) = 0_{A/\mathfrak{p}} \\
&\implies x+ \mathfrak{p} = 0_{A/\mathfrak{p}} \ \ \ \text{or} \ \ \ (x+\mathfrak{p})^{n-1} = 1_{A/\mathfrak{p}}
\end{align*}
where the last implication follows from the fact that $A / \mathfrak{p}$ is an integral domain. Now since $x + \mathfrak{p} \neq 0_{A/\mathfrak{p}}$ we must have that $(x+\mathfrak{p})^{n-1} = 1_{A/\mathfrak{p}}$. But then we have $(x+ \mathfrak{p})(x+\mathfrak{p})^{n-2} = 1_{A/\mathfrak{p}}$ and so $(x+ \mathfrak{p})^{n-2}$ is an inverse of $x + \mathfrak{p}$.
Note that since $n > 1$ it follows that $(x+ \mathfrak{p})^{n-2}$ makes sense. Thus every non-zero element of $A/ \mathfrak{p}$ has an inverse and so $A / \mathfrak{p}$ is a field. Hence $\mathfrak{p}$ is maximal in $A$. $\square$
Claim: $A$ is a ring such that every primary ideal is prime if and
only if $A$ is absolutely flat.
Let us say a ring is a PP(primary is prime) ring if every primary ideal is
prime.
Suppose $A$ is absolutely flat, then $A$ is PP. (exercise 3 in page 55 of "Introduction to Commutative Algebra" by Atiyah & Macdonald)
We only need to show that if $A$ is PP then $A$ is absolutely flat.
Notice that every primary ideal of $S^{-1}A$ is of the form $S^{-1}\mathfrak{q}$
where $\mathfrak{q}$ is a $\mathfrak{p}$-primary ideal such that
$\mathfrak{p}\cap S=\emptyset$. And if $J/I$ (here $J\supset I$) is a
primary ideal of $A/I$ then every zerodivisor in $A/J=(A/I)/(J/I)$ is
nilpotent, so $J$ is is primary ideal of $A$.
Now it is easy to show that, if $A$ is PP, then $S^{-1}A$ and
$A/I$ is PP for a multiplicatively closed subset $S$ and an ideal
$I$ and $\mathfrak{m}^2=\mathfrak{m}$ for every maximal ideal
$\mathfrak{m}$.
Based on these properties we are able to prove that PP rings are absolutely
flat.
We first show every prime ideal in $A$ is maximal. Suppose not, there are
two distinct prime ideals $\mathfrak{p}\subset \mathfrak{m}$ where
$\mathfrak{m}$ is maximal. Now $B=(A/\mathfrak{p})_{\mathfrak{m}}$ is
a PP, local domain(but not a field) we also use $\mathfrak{m}$ to
denote the maximal ideal of $B$. Let $0\neq b\in \mathfrak{m}$,
suppose $\mathfrak{q}$ is a minimal prime containing $b$, then
$C=B_{\mathfrak{q}}$ is a PP, local domain(not a field) and the only maximal ideal of
$C$ is minimal over the ideal $(b)=bC$ so $(b)$ is primary and hence
maximal, $(bC)^2=bC$, thus $bC=0$. It is impossible. We have proved
that every prime ideal in $A$ is maximal.
Let $\mathfrak{m}$ be any prime ideal in $A$, then $A_{\mathfrak{m}}$
is PP. If $A_{\mathfrak{m}}$ is not a field, pick any nonezero $x\in
\mathfrak{m}A_{\mathfrak{m}}$, then $(x)$ is primary hence equals
$\mathfrak{m}A_{\mathfrak{m}}$, so $\mathfrak{m}A_m=0$,
contradiction.
Hence every prime ideal $\mathfrak{m}$ in $A$ is maximal and
$A_{\mathfrak{m}}$ is a field, so $A$ is absolutely flat. We are done.
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