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Suppose that R is a commutative ring with unity such that for each $a$ in $R$ there is a positive integer $n > 1$ such that $a^n =a$. Prove that every prime ideal of $R$ is a maximal ideal of R.

Attempt: Suppose $I$ is a prime ideal of $R$. Let $I'$ be an ideal of $R$ such that $I \subseteq I' \subseteq R$.

Given that $\forall~~a \in R, \exists~~ n>1$ such that $a^n=a$.

$\implies $ if $b \notin I$, then, for any positive integer $i,b^i \notin I$

Now, since $I$ is a prime ideal $\implies$ if $a,b \in R, ab \in I \implies a \in I$ or $b \in I$

Also, this means, for any $b \notin I , ~r \notin I, br \notin I$ and also that $br,(br)^2,(br)^3,..... \notin I$

By the definition of ideal, we know that for any $a \in I, ar \in I,~~~\forall~~r \in R$

I need to prove that either $I = I'$ or $I'=R$

If I want to prove the first one, I must show that $I \subseteq I'$ and $I' \subseteq I$

If I want to prove the second one, then I need to prove that $1 \in I'$. If I somehow know that there exists an element $c \notin I, c \in I'$ whose inverse $d$ exists anywhere in $R$, then, $I'$ being an ideal contains $cd=1$

I know I haven't been able to use the existence of prime ideals nor the existence of prime ideal satisfactorily enough. How do I proceed further. Thank you for your help..

Edit: $R/I$ will be an integral domain but to prove it to be a field, we must prove that every element is invertible with respect to multiplication. $r = r^n \implies r(r^{n-1}-1)=0 \implies ... $ Can I deduce something from here? Thanks..

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2 Answers 2

up vote 2 down vote accepted

I'll explain Praphulla Koushik's argument in more detail so that you can see what he is getting at. Let $R$ be a ring, and $\ast$ be the property you mentioned: namely that for any $a \in R$ there exists an $n > 1$ such that $a^n = a$.

Lemma 1: Let $S$ be an integral domain. If $S$ satisfies property $\ast$, then $S$ is a field.

Proof: $S$ is contained in a field $K$ (for example, its quotient field), in which every nonzero element of $S$ has an inverse. If $0 \neq a \in S$, we need to show that in fact $a^{-1} \in S$. By hypothesis $a^n = a$ for some $n > 1$, so $a(a^{n-1} - 1) = 0$. Since $a \neq 0$ and $S$ is an integral domain, we have that $a^{n-1} - 1 = 0$, or $a^{n-1} = 1$. Thus $aa^{n-2} = 1$. Since $n > 1$, you have $n - 2 \geq 0$, so $a^{-1} = a^{n-2}$ is in fact in $S$.

Lemma 2: If $\phi: R \rightarrow T$ is a surjective homomorphism of rings, and $R$ satisfies property $\ast$, then so does $T$.

Proof: Every element of $T$ takes the form $\phi(a)$ for some $a \in R$. By hypothesis every such $a$ is equal to $a^n$ for some $n > 1$. But then $\phi(a) = \phi(a^n) = \phi(a)^n$.

Now for the proof of the theorem. Let $R$ be a commutative ring with identity which satisfies property $\ast$. If $P$ is a prime ideal of $R$, then $R/P$ is a ring. The map $R \rightarrow R/P$ given by $a \mapsto a + P$ is a surjective homomorphism, so by Lemma 2, $R/P$ satisfies property $\ast$.

Also $R/P$ is an integral domain, and since it satisfies property $\ast$, it is a field by Lemma 1. Therefore $P$ is in fact a maximal ideal.

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Thank you. It seems Gallian has introduced this question before ring homomorphisms. Took away so much of time duh. But, thanks a lot for the answer. –  Wanderer Jul 13 '14 at 23:27
You're welcome. Of course you can avoid ring homomorphisms and Lemma 2 by showing directly that $R/P$ satisfies property $\ast$. –  D_S Jul 13 '14 at 23:28
I am just still a little curious . I agree that $(r + P) = (r+P)^ n \implies (r+P) = (r^n+P)$ . Does not this $\implies r(1-r^{n-1}) \in P?$ and Not $r(1-r^{n-1})=0$ ? Thanks –  Wanderer Jul 13 '14 at 23:32
You are confusing the zero element of $R$ with the zero element of $R/P$. The elements of $R/P$ are SUBSETS of $R$ (in fact, cosets, i.e. sets of the form $a + P = \{a + p : p \in P\}$). In $R/P$, the zero element is $0 + P$, which is literally the set $P$. So saying that $r(1- r^{n-1}) \in P$ is the same as saying that $r(1-r^{n-1}) + P = 0 + P$. –  D_S Jul 13 '14 at 23:36
You didn't make a mistake with what you wrote in your last comment. You're right that $r(r^{n-1}-1) \in P$ implies that either $r$ or $r^{n-1} - 1$ is in $P$, since $P$ is a prime ideal. If $r+P$ is chosen to be a nonzero element of $R/P$, then $r + P \neq 0 + P$, i.e. $r \not\in P$. Therefore $r^{n-1}-1 \in P$. –  D_S Jul 13 '14 at 23:44

Hint :

Enough to prove $R/P$ is a field..

See that $R/P$ is already an integral domain and then see what would $r^n=r$ imply?

I repeat $R/P$ is integral domain...

Spoiler :

$r^n=r\Rightarrow r^n-r=0\Rightarrow r(r^n-1)=0\Rightarrow???$

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A finite integral domain is a field. If $R/P$ is finite, then $P $will be maximal –  Wanderer Jul 13 '14 at 20:33
@VHP : why would $R/P$ a finite integral domain? I am not saying that integral domain would be finite.. all i am saying is that every element in $R/P$ is invertible –  Praphulla Koushik Jul 13 '14 at 20:33
there may exists such case to which i do not want to pay attention... what i have said is actually much simpler and straightforward argument... –  Praphulla Koushik Jul 13 '14 at 20:38
$r^n =r \implies r.r^{n-1}=r \implies r^{n-1}=1 \implies r .r^{n-2}=1 \implies~~ \forall ~~r \in R, r^{-1}$ exists as $n>1$. Hence, every element in $R/P$ is invertible. Am i correct? –  Wanderer Jul 13 '14 at 20:42
No one is claiming that the $R$ you started with is an integral domain. He is saying that any integral domain $R$ which satisfies the given property (namely for any $a \in R$ there exists an $n > 1$ such that $a^n = a$) must be a field. He also claims you should be able to figure out the rest of the proof from there. –  D_S Jul 13 '14 at 23:18

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