Suppose that R is a commutative ring with unity such that for each $a$ in $R$ there is an integer $n > 1\mid a^n =a$. Every prime ideal is maximal? 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..
 A: 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.  
A: 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 : 

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

