Exercise 1.7

Question

Let $A$ be a ring in which every element $x$ satisfies $x^n = x$ for some $n > 1$ (depending on $x$). Show that every prime ideal in $A$ is maximal.

Amit Awasthi · Accepted Answer

\begin{proof}
  Let $\mathfrak{p} \subseteq A$ be a prime ideal, and consider $x \in A \setminus \mathfrak{p}$. Let $\overline{x}$ be the residue of $x$ in $A/\mathfrak{p}$. Then, $\overline{x}^n = \overline{x}$, so $\overline{x}(1-\overline{x}^{n-1}) = \overline{x} - \overline{x}^n = 0$. Since $\mathfrak{p}$ prime implies $A/\mathfrak{p}$ is a domain and $\overline{x} 
e 0$ by choice of $x$, we then have $\overline{x}^{n-1} = 1$. Thus, $\overline{x}\overline{x}^{n-2} = 1$, and so every $0 
e \overline{x} \in A/\mathfrak{p}$ is a unit, i.e., $A/\mathfrak{p}$ is a field. Thus, $\mathfrak{p}$ is maximal.
\end{proof}

Exercise 1.7

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Exercise 1.7

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