Exercise 1.10

Question

Let $A$ be a ring, $\mathfrak{N}$ its nilradical. Show that the following are equivalent:
  \begin{enumerate}[label=\roman*)]
    \item $A$ has exactly one prime ideal;
    \item every element of $A$ is either a unit or nilpotent;
    \item $A/\mathfrak{N}$ is a field.
  \end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}
  $i) \Rightarrow ii)$. $\mathfrak{p} = \mathfrak{N}$ by Prop.~1.8, so $x^n = 0$ for all $x \in \mathfrak{p}$. Now if $y 
otin \mathfrak{p}$, it is a unit by the contrapositive of Cor.~1.5.
  \par $ii) \Rightarrow iii)$. If $x \in A/\mathfrak{N}$, it can be lifted to $	ilde{x} \in A \setminus \mathfrak{N}$, and so $x\overline{	ilde{x}^{-1}} = \overline{	ilde{x}	ilde{x}^{-1}} = 1$.
  \par $iii) \Rightarrow i)$. $\mathfrak{N}$ is a maximal ideal since $A/\mathfrak{N}$ is a field. By Prop.~8, $\mathfrak{N} \subseteq \mathfrak{p}$ for any prime ideal $\mathfrak{p} \subseteq A$. By maximality, $\mathfrak{N} = \mathfrak{p}$, so $\mathfrak{N}$ is the unique prime ideal in $A$.
\end{proof}

Exercise 1.10

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Exercise 1.10

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