Every Prime Ideal is Maximal if $a^n=a$ for any Element $a$ in the Commutative Ring
Problem 530
Let $R$ be a commutative ring with identity $1\neq 0$. Suppose that for each element $a\in R$, there exists an integer $n > 1$ depending on $a$.
Then prove that every prime ideal is a maximal ideal.
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Hint.
Let $R$ be a commutative ring with $1$ and $I$ be an ideal of $R$.
Recall the following facts:
- $I$ is a prime ideal if and only if $R/I$ is an integral domain.
- $I$ is a maximal ideal if and only if $R/I$ is a field.
Proof.
Let $I$ be a prime ideal of the ring $R$. To prove that $I$ is a maximal ideal, it suffices to show that the quotient $R/I$ is a field.
Let $\bar{a}=a+I$ be a nonzero element of $R/I$, where $a\in R$.
It follows from the assumption that there exists an integer $n > 1$ such that $a^n=a$.
Then we have
\[\bar{a}^n=a^n+I=a+I=\bar{a}.\]
Thus we have
\[\bar{a}(\bar{a}^{n-1}-1)=0\]
in $R/I$.
Note that $R/I$ is an integral domain since $I$ is a prime ideal.
Since $\bar{a}\neq 0$, the above equality yields that $\bar{a}^{n-1}-1=0$, and hence
\[\bar{a}\cdot \bar{a}^{n-2}=1.\]
It follows that $\bar{a}$ has a multiplicative inverse $\bar{a}^{n-2}$.
This proves that each nonzero element of $R/I$ is invertible, hence $R/I$ is a field.
We conclude that $I$ is a field.
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