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.

  • 1
    $\begingroup$ The proof is correct. The notation $x_p$ is rather strange. Usually one denotes the residue classes by using bars, hats, etc. $\endgroup$
    – user26857
    Commented Jun 17, 2019 at 7:00
  • $\begingroup$ @user26857 I'll keep that in mind. Thanks. :) $\endgroup$ Commented Jun 17, 2019 at 21:44

1 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.

  • 1
    $\begingroup$ Oh! You are right! This is what I was looking for. I'll leave the question open until tomorrow morning in case something else comes up. Thanks a lot. :) $\endgroup$ Commented Jun 17, 2019 at 7:33

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