Theorem. Prime ideals of AA are also maximal [math-LRSC]

Let AA be a Boolean ring. Every prime ideal of AA is a maximal ideal.

Proof [local-0]

The facts we need here are

  1. For each ideal II of AA, the quotient ring A/IA / I is also a Boolean ring. Because for every i∈A/Ii \in A / I we have a ring morphism φ:A→A/I\varphi : A \to A / I and an element a∈Aa \in A such that φ(a)=i\varphi(a) = i To show A/IA / I is a Boolean ring, means for each i∈A/Ii \in A / I we have i∗i=ii * i = i. Now i∗i=φ(a)∗φ(a)=φ(a∗a)=φ(a)=ii * i = \varphi(a) * \varphi(a) = \varphi(a * a) = \varphi(a) = i as desired.
  2. If a Boolean ring RR is also an integral domain, it has only two elements. Because from a∈Ra \in R we have a∗a=aa * a = a rewrite it to a∗(a−1)=0a * (a - 1) = 0 Now because RR is an integral domain, hence a=0∨(a−1)=0a = 0 \lor (a - 1) = 0 for all a∈Ra \in R. Therefore, the only two elements are 00 and 11.

For each prime ideal xx, the quotient ring A/xA / x is an integral domain, and hence has finite elements, which implies it's also a field, hence xx is a maximal ideal.