Prime ideals are maximal if all $r$ satisfy $\,r^{n_r} = r$ for some $n_r > 1$

Question

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.

Answer 1

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.