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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 $n$ greater than 1 such that $a^n=a$. Prove that every prime ideal of $R$ is a maximal ideal of $R$.

My attempt: Let $I$ be a prime ideal and assume $A$ is an ideal of $R$ that contains $I$. Let $a \in I$. Then $a^n \in I$ also and so $a + a^n = a( a^{n-1}+1) \in I$.

Since $a \in I$ and $I$ is a prime ideal, $a^{n-1} +1 \notin I$, but $a^{n-1}+1 \in A$. This implies $a^n \in A$ and $1 \in A$. Since $1 \in A$, we must have $A = R$. Since $I$ is properly contained in $A =R$, the prime ideal $I$ must be maximal.

blueskyscroll
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  • Why $a^{n-1}+1\in A$? – eranreches Dec 05 '17 at 20:21
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    Hint: $0=a^n-a=a(a^{n-1}-1)\in I$. – Steve D Dec 05 '17 at 20:24
  • Hm, since $I$ is contained in $A$, I'm thinking if $a+a^n = a(a^{n-1}+1) \in I \subset A$, then both $a$ and $a^{n-1}+1$ must be in $A$. Is this not true? If so, my argument is not valid. – blueskyscroll Dec 05 '17 at 20:25
  • It's nice that you've done your work, but if you want your solution evaluated, you should post it as a solution to existing copies of this question, like this one. There is no sense in posting a separate request to "check me" for every possible string of symbols that proves a problem. That is only one duplicate out of many, actually. – rschwieb Dec 05 '17 at 20:27
  • @rschwieb ah, I see. I didn't realize I could do that! Thank you for the link and advice on how to approach getting my work evaluated.. – blueskyscroll Dec 05 '17 at 20:32

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Your reasoning is flawed where you state that $a^{n-1}+1\in A$. For the proof, assume $I\subsetneq A$ and take $a\in A\setminus I$. There exists $n\in \mathbb{N}$ such that $a^{n}=a$. Thus

$$I\ni 0=a-a^{n}=a\left(1-a^{n-1}\right)$$

$I$ is prime and $a\notin I$, forcing $1-a^{n-1}\in I\subsetneq A$. But $a\in A$, and therefore $1\in A=R$ - which proves that $I$ is maximal.

eranreches
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