Proof critique - Atiyah and MacDonald's Commutative algebra - Q 1.7 I'm trying to solve the exercises in Atiyah and MacDonald's Intro to comm. algebra. Here is my attempted solution to exercise 1.7.
Exercise: 
Let $A:\text{CRing}$ such that for every $x\in A$ there is an $n>1$ such that $x^n=x$. Show that every prime ideal of $A$ is maximal.
Proof:
Let $\mathfrak{p}$ be a prime ideal of $A$, and denote the projection $A\xrightarrow{\pi}A/\mathfrak{p}$ by $x\mapsto x_\mathfrak{p}$. We show that the integral domain $A/\mathfrak{p}$ is a field by showing that any nonzero element is a unit.
Take $x\in A$ with $x\not\in\mathfrak{p}$; then $x_{\mathfrak{p}}\neq 0$. By hypothesis there is $n>1$ such that $x^n=x$. Then $0=x^n-x=x(x^{n-1}-1)$, hence $x_\mathfrak{p}(x^{n-1}-1)_\mathfrak{p}=0$. Since $A/\mathfrak{p}$ is an integral domain and $x_\mathfrak{p}\neq 0$ this implies $(x^{n-1}-1)_\mathfrak{p}= 0$, i.e.: $x^{n-1}_\mathfrak{p}=1$. If $n=2$, this means $x_\mathfrak{p}=1$, if $n>2$ then $x_\mathfrak{p}x^{n-2}_\mathfrak{p}=1$. In each case, $x_\mathfrak{p}$ is a unit. $\square$
I'm looking for possible errors and suggestions. Thanks in advance.

The proof is fine. You can make it shorter by remarking that the given condition is preserved by homomorphisms.
Thus it’s sufficient to prove that if $R$ is an integral domain where, for every $x\in R$ there is $n>1$ with $x^n=x$, then $R$ is a field. 
Now your argument goes through. 
