Prime ideal in a ring with some property is maximal

Question

Let $R$ be a ring where for every element $a \in R$, there exists a positive integer $n_a \gt2$ such that $a^{n_a}=a$. Prove that every prime ideal in $R$ is maximal.

I think that I would want to prove this by contradiction, assuming that a prime ideal $P$ is not maximal. Then, there exists $M \subset R$ (properly). This would imply that there exists $a \in M\setminus P$. Then $a^{n_a}=a \in M\setminus P$. Then $M\setminus P$ is prime. And here I am stuck. I don't see how this helps.

Would I rather want to use the fact that then $R/P$ is an integral domain?

Answer 1

Suppose $\;P\le R\;$ is prime $\;\iff R/P\;$ is an integral domain, and let $\;0\neq a+P\in R/P\;$ .

We're given that

$$\exists\,\,2<n_a\in\Bbb N\;\;s.t.\;\;a^{n_a}=a\implies (a+P)^{n_a}=a^{n_a}+P=a+P\implies$$

$$a(a^{n_a-1}-1)+P=P(=\overline 0)$$

and as $\;R/P\;$ is an integral domain we're done.

Answer 2

Hint $\ $ More generally, assume every $\,r\in R\,$ is a root of a reverse-$\rm\color{#c00}{monic}$ polynomial $\,f\in R[x],\,$ $\,f = \color{#c00}u x^n + x^{n+1} g(x),\,$ with $\,\color{#c00}u\,$ a $\rm\color{#c00}{unit}$ (invertible). In the domain $\,R/P,$ every $\,r\,$ is a root of some $\,f = x^n(\color{#0a0}{u+x g(x)})\,$ so either $\,r^n = 0\,$ so $\, r=0,\,$ or $\,\color{#0a0}{-rg(r) = u}\,$ unit, so $\,r\mid u\,$ unit $\,\Rightarrow\,r\,$ unit. Therefore $\,R/P$ is a field, so $\,P\,$ is maximal.