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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..

MathMan
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2 Answers2

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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 :

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

  • A finite integral domain is a field. If $R/P$ is finite, then $P $will be maximal – MathMan 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 –  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... –  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? – MathMan Jul 13 '14 at 20:42
  • why would $r\cdot r^{n-1}=r$ imply $r^{n-1}=1$? –  Jul 13 '14 at 20:44
  • Alright :) . $r^n=r \implies r(r^{n-1}-1)=0$ . But, I know nothing about zero divisors? I might be able to conclude something from here :

    Hence, if $r^n=r$ then $(r+P) = r^n + P \implies r(r^{n-1}-1) \in P \implies (r^{n-1}-1) \in P\forallr \in P$ as $r$ can belong to $R - P$

     – MathMan Jul 13 '14 at 20:56
  • I do not know why you are making it so big deal... $r^n=r\Rightarrow r(r^{n-1}-1)=0$ and because we are in integral domain we have either $r=0$ or $r^{n-1}=1$.. Done!!! –  Jul 13 '14 at 21:01
  • But $R$ is not an integral domain, $R/P$ is an integral domain!? – MathMan Jul 13 '14 at 21:03
  • You can go modulo that ideal and you can assume $R$ to be integral domain... I am not going to write all those steps one by one.. please spend some time to understand what i have done.... Good luck! Good night!! –  Jul 13 '14 at 21:08
  • Uhm sorry about that. I don't understand how we can assume R to be an integral domain though – MathMan Jul 13 '14 at 21:56
  • @BenWest I got that $R/I$ must be an integral domain. But, in the answer above , it has been claimed that $R$ is also an integral domain and that the solution to the equation : $r(r^{n-1}-1) =0$ is that either $r=0$ or $r^{n-1}=1$. My question is, why we haven't accounted for the zero divisors of $R$. – MathMan Jul 13 '14 at 22:49
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    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
  • @PraphullaKoushik I got this. I know I should have got this a long time ago :( but think not sleeping enough has lately taking a toll on me. Gonna sleep now! – MathMan Jul 14 '14 at 00:01
  • @VHP : Happy that at last you have got it.. –  Jul 14 '14 at 07:37
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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.

D_S
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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. – MathMan Jul 13 '14 at 23:27
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    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 – MathMan 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
  • If i have to go without the homomorphisms : $R/P$ is an integral domain $\implies$ $R/P$ has no zero divisors. Hence, since the zero element in $R/P$ is $P$ means that in the equation $r + P = r^n + P $ or $r(r^{n-1}-1) \in P \implies r \in P or(r^{n-1}-1) \in P$. Where am I making a mistake? – MathMan Jul 13 '14 at 23:41
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    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
  • Let us continue this discussion in chat. – D_S Jul 13 '14 at 23:46
  • Right. From here, $(r + P).(r^{n-2} + P) = 1+P $ . Hence, $R/P$ is a field. Am i correct? – MathMan Jul 13 '14 at 23:46