Q:

Prime ideal and maximal ideal

Let $R$ be a ring such that every $x$ satisfying $x^n = x$ for some $n > 1$. Show that every prime ideal in $R$ is maximal.

A:

Let $\mathfrak p\subseteq R$ be a prime ideal, and suppose that there exists an ideal $\mathfrak a\subseteq R$ such that $$\mathfrak p\subseteq\mathfrak a\subseteq R\tag{1}$$
Let $x\in\mathfrak a$, then there exists an integer $n>1$, such that $$x=x^n\iff x(1-x^{n-1})=0\tag{2}$$ 
$\mathfrak p$ is a prime ideal, so $0\in\mathfrak p$ gives that either $x$ or $1-x^{n-1}$ are also in $\mathfrak p$.

If $x\in\mathfrak p$ then $\mathfrak p\subseteq\mathfrak a$, and so $$\mathfrak p=\mathfrak a\tag{3}$$
If $1-x^{n-1}\in\mathfrak p$ then $1-x^{n-1}\in\mathfrak a$. Since both $x,1-x^{n-1}\in\mathfrak a$ we also have that $1-x^{n-1}+x^{n-1}=1\in\mathfrak a$, which gives $$\mathfrak a=R\tag{4}$$

This shows that $\mathfrak p$ is a maximal ideal.

