Exercise 1.11

Question

A ring $A$ is \emph{Boolean} if $x^2 = x$ for all $x \in A$. In a Boolean ring $A$, show that
  \begin{enumerate}[label=\roman*)]
    \item $2x = 0$ for all $x \in A$;
    \item every prime ideal $\mathfrak{p}$ is maximal, and $A/\mathfrak{p}$ is a field with two elements;
    \item every finitely generated ideal in $A$ is principal.
  \end{enumerate}

Amit Awasthi · Accepted Answer

\begin{proof}[Proof of $i)$]
  $x+x = (x+x)^2 = x^2 + 2x + x^2 = x+x+2x$, so $2x = 0$.
\end{proof}
\begin{proof}[Proof of $ii)$]
  Every prime ideal $\mathfrak{p}$ is maximal by Exercise 7, so $A/\mathfrak{p}$ is a field. Now consider $x \in A \setminus \mathfrak{p}$, and let $\overline{x}$ be the residue of $x$ in $A/\mathfrak{p}$. Then, $x^2 = x$ implies $\overline{x}(\overline{x}-1) = 0$. But $A/\mathfrak{p}$ is a field, hence $\overline{x} = 0$ or $1$, i.e., $A/\mathfrak{p} = \{0,1\}$.
\end{proof}
\begin{proof}[Proof of $iii)$]
  We induce on the number of generators $n$ of an ideal $\mathfrak{a} \subseteq A$. Suppose $n=2$, and $\mathfrak{a} = (x,y)$. Then clearly $(x+y-xy) \subseteq \mathfrak{a}$, and $\mathfrak{a} \subseteq (x+y-xy)$ since $x(x+y-xy) = x^2+xy-x^2y = x$ and similarly $y(x+y-xy) = xy+y^2-xy^2 = y$. If $n>2$, let $\mathfrak{a} = (x,\mathfrak{a}')$; by inductive hypothesis, $\mathfrak{a}' = (y)$ for some $y$, and so $\mathfrak{a} = (x,y) = (x+y-xy)$ by the above.
\end{proof}

Exercise 1.11

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Exercise 1.11

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