Let R be a commutative ring with an identity such that for all $r\in$ R, there exists some $n>1$ such that $r^n = r$. Show that any prime ideal is maximal. (Atiyah and MacDonald, Introduction to Commutative Algebra, Chapter 1, Exercise 7.)

Any hints?

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    $\begingroup$ You are probably assuming the ring is commutative: if so, please edit the question and add that hypothesis. In any case, your hypothesis actually implies that the ring must be commutative, by a theorem of Jacobson, but the proof of this is quite not as simple as one might want. $\endgroup$ – Mariano Suárez-Álvarez Oct 11 '12 at 7:42
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    $\begingroup$ Yes, you are correct, I apologize. $\endgroup$ – user41916 Oct 11 '12 at 8:04
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    $\begingroup$ Oh, there is no need to apologize! Just keeping things precise is good for the universe :-) $\endgroup$ – Mariano Suárez-Álvarez Oct 11 '12 at 8:06

Hint: Reduce to the case that $R$ is an integral domain satisfying $\forall r \exists n (r^n=r)$, and show that $R$ is a field.

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    $\begingroup$ Slightly more verbosely: in order to prove that a prime ideal $\mathfrak p\subseteq R$ is maximal, it is enough to show that the quotient $R/\mathfrak p$ is a field. Now the quotient is an integral domain and it has the same property that you supposed $R$ itself has. &c $\endgroup$ – Mariano Suárez-Álvarez Oct 11 '12 at 7:41

Let $\,D\,$ be any integer domain and let $\,d\in D\,$ be s.t. $\,d^n=d\,\,,\,\,1<n\in\Bbb N\,$ , then:

$$d^n=d\Longrightarrow d(d^{n-1}-1)=0\Longleftrightarrow d=0\,\,\,\text{or}\,\,\,d^{n-1}=1,$$

so if $\,d\,$ is not zero then it must be a unit.


In our case: let $\,I\leq R\,$ be a prime ideal and let $\, r\in R\setminus I\,$, then:

$$\exists\,n\in\Bbb N\,\,s.t.\,\,r^n=r\Longrightarrow \left(r+I\right)^n=r^n+I=r+I\in R/I$$

Now use the first part with $\,D:=R/I\,\,\,,\,\,d=r+I\,$ and deduce $\,R/I\,$ is actually a field...


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 $A$. $\square$


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