# 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 correct. The notation $x_p$ is rather strange. Usually one denotes the residue classes by using bars, hats, etc. Commented Jun 17, 2019 at 7:00
• @user26857 I'll keep that in mind. Thanks. :) Commented Jun 17, 2019 at 21:44

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.