Boolean ring and algebraic numbers
Boolean ring
Let $B$ be a commutative ring with unity. We say that $B$ is a Boolean ring if $x^2=x$ for all $x \in B$. The name “Boolean” certainly rings a bell of the idea of bool values in programming, or in general, the Boolean algebra that is frequently used in logic, digital electronics and computer science.
In this post, we will examine Boolean rings on a level of commutative algebra, followed by an explicit example in algebraic number theory.
Basic properties of Boolean rings
Throughout, let $B$ be a Boolean ring.
Proposition 1. In the Boolean ring $B$, we have
- $2x=0$ for all $x \in B$.
- Every prime ideal $\mathfrak{p} \subset B$ is maximal, and $B/\mathfrak{p}$ is a field with two elements.
- Every finitely generated ideal in $A$ is principal.
Proof. For 1, notice that
For 2, it suffices to show that for every prime ideal $\mathfrak{p} \subset B$, we have $B/\mathfrak{p} \cong \mathbb{Z}/2\mathbb{Z}$.
Pick $x \in B \setminus \mathfrak{p}$. Then in $B/\mathfrak{p}$ we have $\overline{x}^2=\overline{x}$, where $\overline{x}=x+\mathfrak{p} \in B/\mathfrak{p}$. Therefore $\overline{x}(\overline{x}-\overline{1})=0$. However, since $B/\mathfrak{p}$ is entire, we see that we must have $\overline{x}=\overline{1}$ since $x \not\in \mathfrak{p}$. Therefore there are exactly two elements in $B/\mathfrak{p}$,...
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