Classification of Finite Boolean Rings with Identity (7.6.1-2)

Dummit and Foote Abstract Algebra, section 7.6, exercises 1-2:

1. An element $e∈R$ is idempotent if $e^2=e$. Assume $e$ is idempotent and within the center of $R$. Prove $Re$ and $R(1-e)$ are two-sided ideals of $R$ and $R≅Re \times R(1-e)$. Show $e$ and $1-e$ are identities for $Re$ and $R(1-e)$ respectively.
2. Let $R$ be a finite Boolean ring with identity. Prove $R≅\mathbb{Z}/2\mathbb{Z} \times ... \times \mathbb{Z}/2\mathbb{Z}$.

Proof: (1) $Re$ and $R(1-e)$ are already left ideals. For any $r_1e∈Re$ and $r∈R$, we have $r_1er=(r_1r)e∈Re$ so that $Re$ is a two-sided ideal. As well, we have $(1-e)^2=1-2e+e^2=1-2e+e=1-e$ and $r(1-e)=r-re=r-er=(1-e)r$ so that $(1-e)$ is idempotent and central and similarly $R(1-e)$ is a two-sided ideal.

Define a mapping $φ : R → Re \times R(1-e)$ by $r \mapsto (re,r(1-e))$. This is clearly a homomorphism of rings, and we can prove surjectivity by observing $φ(e)=(e^2,e(1-e))=(e,0)$ and $φ(1-e)=((1-e)e,(1-e)^2)=(0,1-e)$ and now for any $(r_1e,r_2(1-e)∈Re \times R(1-e)$ we have $φ(r_1e+r_2(1-e))=φ(r_1)(e,0)+φ(r_2)(0,1-e)=(r_1e,r_2(1-e))$. Finally, this mapping is injective since $$φ(r)=(0,0)⇒re=0 \land r(1-e)=0 ⇒$$$$r-re=r-0=0 ⇒ r=0$$ As we've already seen, $e$ and $(1-e)$ are central and idempotent, so that their identity qualities are deduced straightforwardly.$~\square$

(2) Lemma 1: Let $φ_i : A_i → B_i$ be an isomorphism of rings for all applicable $i$. We have$$A_1 \times A_2 \times ... ≅ B_1 \times B_2 \times ...$$Proof: Define a function that maps the $i$th coordinate of $A_1 \times A_2 \times ...$ to the $i$th coordinate of $B_1 \times B_2 \times ...$ as its image under $φ_i$. Due to the nature of the isomorphisms $φ_i$, we have this map is an isomorphism.$~\square$

Definition 1: Let $R$ be a ring with $r∈R$. Define the rank of $R$ from $r$ as the order of the smallest subset $A \subseteq R$ such that $r∈A$ and $A$ generates $R$.

We will proceed by induction on the rank of $R$ from $1$, where $R$ is a finite Boolean ring with identity. When the rank from $1$ is 1, since $R$ is of characteristic 2 we have $R ≅ \mathbb{Z}/2\mathbb{Z}$. Assume that for some integer $n$ we can deduce that a finite Boolean ring with rank from $1$ of $m≤n$ is necessarily isomorphic to a direct product of copies of $\mathbb{Z}/2\mathbb{Z}$, and now observe $R$ with rank from $1$ of $n+1$. Let $A=\{1,a,r_3,...,r_{n+1}\}$ be its fulfilling generating subset, with $1≠a$. Since $a$ is necessarily central and idempotent, we have $R ≅ Ra \times R(1-a)$ by the first exercise.

We claim that $\{1 \cdot a,a \cdot a,r_3 \cdot a,...,r_{n+1} \cdot a\}=\{a,r_3a,...,r_{n+1}a\}$ is a generating subset of $Ra$ containing the identity and of order $≤n$ so that $Ra$ has a rank from $a$ (the identity of $Ra$) of $≤n$ and so is isomorphic to some number of direct products of $\mathbb{Z}/2\mathbb{Z}$ by induction. This is immediate because any series of operations taking elements from this subset is divisible by $a$, and for any $ra∈Ra$ we have $ra=Na$ for some series of operations $N$ from $A$ and evidently $Na$ is replicable by a series of operations from the subset. Similarly the case holds for $R(1-a)$ where a potential generating subset containing its identity is $\{1 \cdot (1-a),a \cdot (1-a),r_3 \cdot (1-a),...,r_{n+1} \cdot (1-a)\}=$$\{1-a,0,r_3-r_3a,...,r_{n+1}-r_{n+1}a\}$. We may remove the superfluous zero from this generating subset. Therefore by induction and the lemma$$R≅Ra \times R(1-a) ≅(\mathbb{Z}/2\mathbb{Z} \times ... \times \mathbb{Z}/2\mathbb{Z}) \times (\mathbb{Z}/2\mathbb{Z} \times ... \times \mathbb{Z}/2\mathbb{Z}) ≅$$$$\mathbb{Z}/2\mathbb{Z} \times ... \times \mathbb{Z}/2\mathbb{Z}~\square$$ In addition, this provides a unique construction of the finite Boolean ring $R$ with identity of rank $n$ for any $n∈\mathbb{Z}^+$.