Show that any prime ideal from such a ring is maximal.

Solution 1:

Proof: Let $\mathfrak{p}$ be a prime ideal in $A$. Note that since $\mathfrak{p}$ is a prime ideal we have that $A / \mathfrak{p}$ is an integral domain.

We will show first that $A / \mathfrak{p}$ is a field. Choose $x + \mathfrak{p} \in A / \mathfrak{p}$ such that $x + p \neq 0_{A/ \mathfrak{p}}$. By hypothesis $x^n = x$ for some $n > 1$. Observe then that $$(x+\mathfrak{p})^n = x^n + \mathfrak{p} = x + \mathfrak{p}.$$ We then have that \begin{align*} (x+\mathfrak{p})^n = x + \mathfrak{p} &\implies (x+\mathfrak{p})^n - (x+\mathfrak{p}) = 0_{A/\mathfrak{p}} \\ &\implies (x + \mathfrak{p})\left((x+\mathfrak{p})^{n-1} - 1_{A/\mathfrak{p}}\right) = 0_{A/\mathfrak{p}} \\ &\implies x+ \mathfrak{p} = 0_{A/\mathfrak{p}} \ \ \ \text{or} \ \ \ (x+\mathfrak{p})^{n-1} = 1_{A/\mathfrak{p}} \end{align*} where the last implication follows from the fact that $A / \mathfrak{p}$ is an integral domain. Now since $x + \mathfrak{p} \neq 0_{A/\mathfrak{p}}$ we must have that $(x+\mathfrak{p})^{n-1} = 1_{A/\mathfrak{p}}$. But then we have $(x+ \mathfrak{p})(x+\mathfrak{p})^{n-2} = 1_{A/\mathfrak{p}}$ and so $(x+ \mathfrak{p})^{n-2}$ is an inverse of $x + \mathfrak{p}$.

Note that since $n > 1$ it follows that $(x+ \mathfrak{p})^{n-2}$ makes sense. Thus every non-zero element of $A/ \mathfrak{p}$ has an inverse and so $A / \mathfrak{p}$ is a field. Hence $\mathfrak{p}$ is maximal in $A$. $\square$