Quotients of the Gaussian Integers (8.3.6-7)

Dummit and Foote Abstract Algebra, section 8.3, exercises 6-7:

6. (a) Prove $\mathbb{Z}[i]/(1+i)$ is a field of order 2.
(b) Let $q∈\mathbb{Z}$ be a prime with $q ≡ 3 \mod{4}$. Prove $\mathbb{Z}[i]/(q)$ is a field of order $q^2$.
(c) Let $p∈\mathbb{Z}$ be a prime with $p ≡ 1 \mod{4}$ and let $p=\pi \overline{\pi}$ be its factorization in $\mathbb{Z}[i]$. Prove $\mathbb{Z}[i]/(p)≅\mathbb{Z}[i]/(\pi) \times \mathbb{Z}[i]/(\overline{\pi})$ as rings, $\mathbb{Z}[i]/(p)$ is of order $p^2$ and therefore $\mathbb{Z}[i]/(\pi)$ and $\mathbb{Z}[i]/(\overline{\pi})$ are fields of order $p$.
7. Let $\pi∈\mathbb{Z}[i]$ be irreducible.
(a) For $n \geq 0$ an integer, show that $(\pi^{n+1}) \subseteq (\pi^n)$ and that multiplication by $\pi^n$ induces an isomorphism $\mathbb{Z}[i]/(\pi)≅(\pi^n)/(\pi^{n+1})$ as additive abelian groups.
(b) Prove $|\mathbb{Z}[i]/(\pi^n)|=|\mathbb{Z}[i]/(\pi)|^n$.
(c) Prove for $\alpha≠0$ that $|\mathbb{Z}[i]/(\alpha)|=N(\alpha)$.

Proof: (6) (a) By the division algorithm, every element in $\mathbb{Z}[i]/(1+i)$ can be represented by an element of norm $< 2$. We have $-i=-(1+i)+1$,$i=(1+i)-1$, and $-1=-(1+i)(1-i)+1$, so that the order is $≤2$. Since $1+i$ is not a unit, we have $(1+i) \subset \mathbb{Z}[i]$, so that the order must be 2. Since $\mathbb{Z}[i]/(1+i)$ is now a finite integral domain by the primality of $1+i$, we have it is a field.

(b) Observe arbitrary $a+bi∈\mathbb{Z}[i]$. Let $a=a'+v_aq$ and $b=b'+v_bq$ with $0≤ a',b' < q$. We have $a+bi=(a'+v_aq)+(b'+v_bq)i=a'+b'i+q(v_a+v_bi)$, so that in this sense $a+bi$ can be reduced modulo $q$. This allows a maximum order of $q^2$. Now, assume reduced $\overline{a+bi} = \overline{c+di}$ in $\mathbb{Z}[i]/(q)$, so that $a+bi=c+di+qr$ for some $r∈\mathbb{Z}[i]$. We therefore have $(a-c)+(b-d)i=qr$, i.e. $q$ divides $(a-c)$ and $(b-d)$. Since $a,b,c,d < q$, we have $a=c$ and $b=d$, i.e. $a+bi=c+di$. Therefore there are precisely $q^2$ elements. This order counting holds when $q$ is an arbitrary integer.

Once again let $\overline{a+bi}$ be a typical (nonzero) element. Since $q$ is irreducible thus prime thus $\mathbb{Z}[i]/(q)$ is an integral domain, we have $\overline{(a+bi)(a-bi)}≠\overline{0}$. Letting $v^{-1}$ denote the inverse of $\overline{(a+bi)(a-bi)}$ modulo $q$, we have $\overline{(a+bi)(v^{-1}(a-bi))}=\overline{1}$, so that $\mathbb{Z}[i]/(q)$ is a field.

Alternatively, by a previous exercise $\mathbb{Z}[i]/I$ is finite for any nonzero ideal $I$, so these finite integral domains are clearly fields.

(c) Since $(\pi)$ is prime and thus maximal by a property of PIDs, and since $\pi$ is not associate to $\overline{\pi}$ i.e. $\pi r ≠ \overline{\pi}$ for any $r∈\mathbb{Z}[i]$, we have $(\pi,\overline{\pi})=\mathbb{Z}[i]$ so that the Chinese Remainder Theorem holds by $(\pi)(\overline{\pi})=(\pi\overline{\pi})=(p)$. Parallel to the proof above, $\mathbb{Z}[i]/(p)$ is of order $p^2$ and since $(\pi),(\overline{\pi})≠\mathbb{Z}[i]$ we have they are both finite integral domains (fields) of order $p$.

(7) (a) The containment should be evident, so we must demonstrate the group isomorphism by $φ(x+(\pi))=x\pi^n+(\pi^{n+1})$. Well-definedness:$$x+(\pi)=y+(\pi)⇒x=y+\pi r⇒\pi r = x-y ⇒ \pi^{n+1} r = x\pi^n-y\pi^n⇒$$$$x\pi^n+(\pi^{n+1})=y\pi^n+(\pi^{n+1})⇒φ(x+(\pi))=φ(y+(\pi))$$Homomorphism:$$φ(x+(\pi)+y+(\pi))=φ((x+y)+(\pi))=$$$$x\pi^n+y\pi^n+(\pi^{n+1})=φ(x+(\pi))+φ(y+(\pi))$$ Surjectivity: $$x∈(\pi^n)/(\pi^{n+1})⇒x=\pi^n r + (\pi^{n+1})⇒φ(r+(\pi))=x$$ Injectivity: $$φ(x+(\pi))=(\pi^{n+1})⇒x\pi^n∈(\pi^{n+1})⇒\pi \mid x⇒x+(\pi)=0+(\pi)$$ (b) Assume $\pi$ is not associate to $1+i$. Let $N(\pi)=p$. We claim $\mathbb{Z}[i]/(p^n)≅\mathbb{Z}[i]/(\pi^n) \times \mathbb{Z}[i]/(\overline{\pi^n})$. The greatest common divisor of $\pi^n$ and $\overline{\pi^n}$ according to proposition 13 is $1$, and since $\mathbb{Z}[i]$ is a Euclidean Domain it can written as a linear combination of $\pi^n$ and $\overline{\pi^n}$, implying $(\pi^n,\overline{\pi^n})=\mathbb{Z}[i]$. Since $(\pi^n)(\overline{\pi^n})=(p^n)$, the Chinese Remainder Theorem takes care of the claim.

Now, we have $\mathbb{Z}/(p^n)$ is of order $p^{2n}$, therefore $|\mathbb{Z}[i]/(\pi^n)| \cdot |\mathbb{Z}[i]/(\overline{\pi^n})| = p^{2n}$ and in particular these former two are orders of $p$ power. We are now in a position to prove our claim by induction; assuming the claim doesn't hold for some particular $n$, we have either $|\mathbb{Z}[i]/(\pi^n)| < p^n$ or $|\mathbb{Z}[i]/(\overline{\pi^n})| < p^n$. Without loss of generality we can assume the former, so that $|\mathbb{Z}[i]/(\pi^n)|=|\mathbb{Z}[i]/(\pi^m)|$ for some $m < n$ by induction. Write $\mathbb{Z}[i]/(\pi^m)≅(\mathbb{Z}[i]/(\pi^n))/((\pi^m)/(\pi^n))$ so that necessarily $(\pi^m) \subseteq (\pi^n)$, a contradiction by $\pi^m \not ∈ (\pi^n)$.

So assume $\pi=1+i$, which by the associativity of $1+i$ to $1-i$ implies this is the last case we have to consider. We see that this is the case when $n=2m$ is even, as $\mathbb{Z}[i]/(\pi^{2m})=\mathbb{Z}/((\pi^2)^m)=\mathbb{Z}/(2^m)$, which has $2^{2m}=2^n=|\mathbb{Z}[i]/(\pi)|^n$ elements as claimed.

So assume $n=2m+1$. We first show that $|\mathbb{Z}[i]/(\pi^{2m+1})|$ is of $2$ power; by way of contradiction, assume there is some prime $p$ dividing the order of this quotient. Since the quotient may be considered a group under addition, where exponentiation of addition is multiplication, by Cauchy's Theorem we have some $\overline{a+bi}∈\mathbb{Z}[i]/(\pi^{2m+1})$ such that $\overline{p}$ is the smallest positive integer for which $\overline{p(a+bi)}=\overline{0}$. In other words, $\pi^{2m+1}$ doesn't divide $a+bi$ so that there are less than $2m+1$ factors of $\pi$ in the decomposition of $a+bi$, yet since $\pi \not \mid p$ for any $p≠2$ we must also have $\pi^{2m+1} \not \mid p(a+bi)$, a contradiction by $\overline{p(a+bi)}=\overline{0}$.

We have $(\pi^{2m+2}) \subset (\pi^{2m+1}) \subset (\pi^{2m})$, so that $$|(\mathbb{Z}[i]/(\pi^{2m+2}))/((\pi^{2m})/(\pi^{2m+2}))| <$$$$|(\mathbb{Z}[i]/(\pi^{2m+2}))/((\pi^{2m+1})/(\pi^{2m+2}))| <$$$$|(\mathbb{Z}[i]/(\pi^{2m+2}))/((\pi^{2m+2})/(\pi^{2m+2}))|$$implying $$|\mathbb{Z}[i]/(\pi^{2m})| = 2^{2m} < |\mathbb{Z}[i]/(\pi^{2m+1})| < |\mathbb{Z}[i]/(\pi^{2m+2})| = 2^{2m+2}$$Since $|\mathbb{Z}[i]/(\pi^{2m+1})|$ is of $2$ power, we must have $|\mathbb{Z}[i]/(\pi^{2m+1})|=2^{2m+1}=|\mathbb{Z}[i]/(\pi)|^{2m+1}$.

(c) Let $\alpha = \pi_1^{a_1}...\pi_n^{a_n}$ be its decomposition, so that $|\mathbb{Z}[i]/(\alpha)|=|\mathbb{Z}[i]/((\pi_1^{a_1})...(\pi_n^{a_n}))|$, the latter of which we claim is equivalent to $|\mathbb{Z}[i]/(\pi_1^{a_1})| ... |\mathbb{Z}[i]/(\pi_n^{a_n})|$ by the Chinese Remainder Theorem; taking any $(\pi_x^{a_x})$ and $(\pi_y^{a_y})$ with $x≠y$ we can see $(\pi_x^{a_x},\pi_y^{a_y})=\mathbb{Z}[i]$ since the greatest common divisor of the pair is $1$. The original claim now follows by$$|\mathbb{Z}[i]/(\alpha)|=|\mathbb{Z}[i]/(\pi_1^{a_1})...(\pi_n^{a_n})|=|\mathbb{Z}[i]/(\pi_1^{a_1})| ... |\mathbb{Z}[i]/(\pi_n^{a_n})| =$$$$|\mathbb{Z}[i]/(\pi_1)|^{a_1} ... |\mathbb{Z}[i]/(\pi_n)|^{a_n}=N(\pi_1)^{a_1} ... N(\pi_n)^{a_n} = N(\alpha)~\square$$

Rings of Fractions of Principle Ideal Domains (8.2.8)

Dummit and Foote Abstract Algebra, section 8.2, exercise 8:

Let $R$ be a principle ideal domain, and let $D$ be a multiplicatively closed subset of $R$. Prove that $D^{-1}R$ is a principle ideal domain.

Proof: Let $I$ be a typical ideal of $D^{-1}R$, and $J$ be the subset of elements of $R$ appearing as numerators of elements of $I$. We claim $J$ is an ideal: For any $a,c∈J$, we have some $\dfrac{a}{b},\dfrac{c}{d}∈I$ by definition, so $\dfrac{a}{b} \cdot \dfrac{b^2}{b} = \dfrac{ae}{e}∈I$ for any $e∈D$ and similarly $\dfrac{ce}{e}∈I$. Now $(\dfrac{ae}{e}-\dfrac{ce}{e}) \cdot \dfrac{e}{e^2}=\dfrac{(a-c)e}{e} \cdot \dfrac{e}{e^2}=\dfrac{a-c}{e}∈I$ so that $a-c∈J$ and $J$ is closed under subtraction. For any $r∈R$, we have $\dfrac{re}{e} \cdot \dfrac{a}{b}=\dfrac{ra}{b}∈I$ so that $ra∈J$ and $J$ is an ideal. Due to the hypothesis, $J=(\alpha)$ for some $\alpha∈R$.

We now claim $I=(\dfrac{\alpha e}{e})$. Since $\alpha∈J$, as before we have $\dfrac{\alpha e}{e}∈I$ so that $(\dfrac{\alpha e}{e}) \subseteq I$. Now, observe arbitrary $\dfrac{x}{y}∈I$, and write $x=z\alpha$. We have $\dfrac{\alpha e}{e} \cdot \dfrac{z}{b} = \dfrac{a}{b}$, so that $I \subseteq (\dfrac{\alpha e}{e})$ and now $I=(\dfrac{\alpha e}{e})$ is a principle ideal, and $D^{-1}R$ is a principle ideal domain.$~\square$

(p-1)th Roots of the p-Adic Integers (7.6.11e)

Dummit and Foote Abstract Algebra, section 7.6, exercise 11e:

Show that if $a_1 \not ≡ 0 \mod{p}$, then there is an element $a=(a_i)$ in the inverse limit $\mathbb{Z}_p$ satisfying $a_j^{p-1} ≡ 1 \mod{p^j}$ and $\mu_{j1}(a_j)=a_1$ for all $j$. Deduce that $\mathbb{Z}_p$ contains $p-1$ distinct $(p-1)$th roots of 1.

Proof: We claim that for any $a_1 \not ≡ 0 \mod{p}$, the coordinates of $a$ are uniquely determined and satisfy the given criteria. We proceed by induction on the $n$th coordinate: The case holds true for $n=1$, so observe the $n+1$th coordinate. $a_{n+1}^{p-1} ≡ 1 \mod{p^{n+1}}$ if and only if its associated polynomial representation $(b_0+b_1p+...+b_np^n)^{p-1} ≡ 1 \mod{p^{n+1}}$ holds true, so this is equivalent to solving for $b_n$ since $a_n=b_0+b_1p+...+b_{n-1}p^{n-1}$ has been uniquely determined.$$(b_0+b_1p+...+b_np^n)^{p-1} ≡ 1 \mod{p^{n+1}}⇔$$$$((b_0+b_1p+...+b_{n-1}p^{n-1})+(b_np^n))^{p-1} ≡ 1 \mod{p^{n+1}}⇔$$$$\sum_{k=0}^{p-1} {p-1\choose k}(b_0+b_1p+...+b_{n-1}p^{n-1})^k(b_np^{n})^{p-1-k}≡1 \mod{p^{n+1}}⇔$$$$(b_0+b_1p+...+b_{n-1}p^{n-1})^{p-1}+$$$$(p-1)(b_0+b_1p+...+b_{n-1}p^{n-1})^{p-2}(b_np^n) ≡ 1 \mod{p^{n+1}}$$Since $(b_0+b_1p+...+b_{n-1}p^{n-1})^{p-1}≡ 1 \mod{p^n}$ has been uniquely determined, we have $(b_0+b_1p+...+b_{n-1}p^{n-1})^{p-1}=1+vp^n$ for some integer $v$.$$1+vp^n+(p-1)(b_0+b_1p+...+b_{n-1}p^{n-1})^{p-2}(b_np^n) ≡ 1 \mod{p^{n+1}}⇔$$$$1+vp^n-b_0^{p-2}b_np^n ≡ 1 \mod{p^{n+1}}⇔$$$$1+(v-a_1^{p-2}b_n)p^n ≡ 1 \mod{p^{n+1}}⇔$$$$v-a_1^{p-2}b_n ≡ 0 \mod{p}⇔$$$$b_n≡a_1^{2-p}v \mod{p}~\square$$

Direct Limits (7.6.8)

Dummit and Foote Abstract Algebra, section 7.6, exercise 8-9:

8. Let $I$ be an indexing set with partial order $≤$, with $A_i$ an additive abelian group for all $i∈I$. Furthermore, $I$ is a directed set: For any $i,j∈I$, there exists $k∈I$ such that $i,j≤k$. Suppose that for every pair of indices $i,j$ with $i≤j$ there exists a map $\rho_{ij} : A_i → A_j$ such that $\rho_{jk} \circ \rho_{ij}$ for all $i≤j≤k$ and $\rho_{ii} = 1$. Letting $B$ be the disjoint union of all the $A_i$, define a relation $\sim$ on $B$ by$$a \sim b ⇔ ∃k∈I (i,j≤k \land \rho_{ik}(a)=\rho_{jk}(b))$$for $a∈A_i$ and $b∈A_j$.
(a) Show that $\sim$ is an equivalence relation on $B$. The set of equivalence classes is called the direct limit of the direct system $\{A_i\}$ and is denoted $\varinjlim A_i$. For the rest of the exercise let $A = \varinjlim A_i$.
(b) Let $\overline{x}$ denote the class of $x$ and define $\rho_i : A_i → A$ by $\rho_i(a)=\overline{a}$. Show that if $\rho_{ij}$ is injective for all $i,j$ then $\rho_i$ is injective for all $i$ (so that each $A_i$ can be viewed as a subset of $A$).
(c) Assume $\rho_{ij}$ are all group homomorphisms. For $a∈A_i$,$b∈A_j$ show that the operation$$\overline{a}+\overline{b}=\overline{\rho_{ik}(a)+\rho_{jk}(b)}$$ is well defined, for any $k$ such that $i,j≤k$. Show that this makes $A$ into an abelian group, and that $\rho_i$ in part (b) are group homomorphisms.
(d) Show that if $A_i$ are commutative rings with 1 and $\rho_{ij}$ are ring homomorphisms that send 1 to 1, then $A$ may likewise be given the structure of a commutative ring with 1 such that $\rho_i$ are all ring homomorphisms.
(e) Under the hypotheses in (c), prove that the direct limit has the following universal property: if $C$ is any abelian group such that for each $i∈I$ there is a homomorphism $φ_i : A_i → C$ with $φ_i = φ_j \circ \rho_{ij}$ whenever $i≤j$, then there is a unique homomorphism $φ : A → C$ such that $φ \circ \rho_i = φ_i$ for all $i$.

Proof: (a) Reflexivity: $\rho_{ii}(a)=a=\rho_{ii}(a)⇒a \sim a$. Symmetry: $a \sim b$$⇒ \rho_{ik}(a)=\rho_{jk}(b) ⇒ \rho_{jk}(b)=\rho_{ik}(a) ⇒ b \sim a$ for some $k$. Transitivity: Let $c∈A_l$. $a \sim b \land b \sim c ⇒ \rho_{ik}(a)=\rho_{jk}(b) \land \rho_{jm}(b)=\rho_{lm}(c)$ for some $k$ and $m$ with $i,j≤k$ and $j,l≤m$. Let $k,m≤n$. We have $$\rho_{in}(a)=\rho_{kn} \circ \rho_{ik}(a)=\rho_{kn} \circ \rho_{jk}(b)=\rho_{jn}(b)=$$$$\rho_{mn} \circ \rho_{jm}(b)=\rho_{mn} \circ \rho_{lm}(c)=\rho_{ln}(c)$$ so that $a \sim c$.
(b) Assume $a,b∈A_i$ and $\rho_{i}(a)=\rho_{i}(b)$, so that $\overline{a}=\overline{b} ⇒ a \sim b$. Then $\rho_{ij}(a)=\rho_{ij}(b)$ for some $j$, so that $a=b$.
(c) Start with well-definedness. Let $\overline{a'}=\overline{a}$ by $\rho_{i'n}(a')=\rho_{in}(a)$ for some $i,i'≤n$, and denote $a'∈A_{i'}$, and likewise for $b'$ with $m$ in place of $n$. It suffices to show $\rho_{i'k'}(a')+\rho_{j'k'}(b') \sim \rho_{ik}(a) + \rho_{jk}(b)$ for arbitrary applicable $k'$. Choose $k^o$ such that $n,k,k',m≤k^o$ (by choosing a value greater than or equal to $n$ and $k$, and another greater than or equal to $m$ and $k'$, and letting $k^o$ be a value greater than or equal to both). We have$$\rho_{k'k^o}(\rho_{i'k'}(a')+\rho_{j'k'}(b'))=\rho_{k'k^o} \circ \rho_{i'k'}(a')+\rho_{k'k^o} \circ \rho_{j'k'}(b')=\rho_{i'k^o}(a')+\rho_{j'k^o}(b')=$$$$\rho_{nk^o} \circ \rho_{i'n}(a')+\rho_{mk^o} \circ \rho_{j'm}(b')=\rho_{nk^o} \circ \rho_{in}(a) + \rho_{mk^o} \circ \rho_{jm}(b)=$$$$\rho_{ik^o}(a)+\rho_{jk^o}(b)=\rho_{kk^o} \circ \rho_{ik}(a)+\rho_{kk^o} \circ \rho_{jk}(b)=\rho_{kk^o}(\rho_{ik}(a)+\rho_{jk}(b))$$Due to the homomorpicity of $\rho_{ij}$, we have $\overline{e}=0$ in $A$ for any $e∈A_i$ a zero and $-\overline{a}=\overline{-a}$. Associativity, closure, and abelianness are assured by their demonstration in $A_i$. Furthermore, for $a,b∈A_i$, we have $\rho_{i}(a)+\rho_{i}(b)=\overline{a}+\overline{b}=\overline{\rho_{ii}(a)+\rho_{ii}(b)}=\rho_{i}(a+b)$.
(d) This is simply part (c) with the operation of addition replaced by multiplication, together with the evident fact that $\overline{e}=1$ in $A$ for any $e∈A_i$ an identity.
(e) Any such homomorphism $φ$ is uniquely defined by $φ(\overline{a})=φ \circ \rho_{i}(a)=φ_i(a)$, so now well definedness and homomorphicity must be demonstrated: $φ(\overline{a'})=φ_{i'}(a)=φ_n \circ \rho_{i'n}(a')=φ_n \circ \rho_{in}(a)=φ_i(a)=φ(\overline{a})$, and $φ(\overline{a}+\overline{b})=φ(\overline{\rho_{ik}(a)+\rho_{jk}(b)})=φ_k(\rho_{ik}(a)+\rho_{jk}(b))=φ_i(a)+φ_j(b)=φ(\overline{a})+φ(\overline{b})$where addition is interchangeable with multiplication.$~\square$

Collapsing Units Between Modular Rings (7.6.7)

Dummit and Foote Abstract Algebra, section 7.6, exercise 7:

Let $n \mid m$, and prove that the natural projection map $(\mathbb{Z}/m\mathbb{Z})^{\times} → (\mathbb{Z}/n\mathbb{Z})^{\times}$ is surjective.

Proof: Firstly, if $u$ is a unit of $\mathbb{Z}/m\mathbb{Z}$, then $(u,m)=1$, so that $(u,n)=1$, so that $u$ is a unit of $\mathbb{Z}/n\mathbb{Z}$, proving the mapping is as claimed. To show surjectivity, let $p_i$ be a prime not dividing $n$ but dividing $m$ for all applicable $i$. For a typical unit $x∈(\mathbb{Z}/n\mathbb{Z})^{\times}$, solve the simultaneous system of equations$$y ≡ x \mod{n}$$$$y ≡ 1 \mod{\prod p_i}$$which is always possible given that $n$ and $\prod p_i$ are coprime. For any prime $q$ dividing $m$, we have $q$ divides either $n$ or $\prod p_i$. In the former case, since $y$ is coprime to $n$, we have $y$ coprime to $q$. In the latter case, since $y$ is coprime to $\prod p_i$, we have $y$ coprime to $q$. Since $y$ is coprime to all the prime divisors of $m$, it is coprime to $m$ itself, and is thus a unit of $(\mathbb{Z}/m\mathbb{Z})^{\times}$ which naturally projects to $x$ in $(\mathbb{Z}/n\mathbb{Z})^{\times}$.$~\square$

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 $n∈\mathbb{Z}^+$.

Zorn's Lemma on the Real Numbers (7.5.6)

Dummit and Foote Abstract Algebra, section 7.5, exercise 6:

Prove that $\mathbb{R}$ contains a subring $A$ with identity and maximal under inclusion such that $\dfrac{1}{2} \not ∈ A$.

Proof: Let $\mathcal{F}$ be the set of all subrings of $\mathbb{R}$ not containing $\dfrac{1}{2}$, and let $\mathcal{C}$ be a typical chain of subrings $R_0 \subseteq R_1 \subseteq ...$. Admit$$R=\bigcup_{n∈N}R_n$$and prove it is an upper bound of $\mathcal{C}$. For any $a,b∈R$, we have $a∈R_x$ and $b∈R_y$ for some $x,y$ implying $a-b∈R$ and $ab∈R$ so that $R$ is a subring. Furthermore, by definition of the union, $1∈R$ and $\dfrac{1}{2} \not ∈ R$. By Zorn's Lemma, $\mathcal{F}$ has a maximal element.$~\square$