A Ring of Homomorphisms (7.1.30)

Dummit and Foote Abstract Algebra, section 7.1, exercise 30:

Let $A=\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} ...$ be the direct product of copies of $\mathbb{Z}$ indexed by positive integers and let $R$ be the ring of homomorphisms from $A$ to itself. Let $φ$ be the element of $R$ defined by $φ(a_1,a_2,a_3,...)=(a_2,a_3,...)$. Let $ψ$ be the element of $R$ defined by $ψ(a_1,a_2,a_3,...)=(0,a_1,a_2,a_3,...)$.
(a) Prove $φψ=1≠ψφ$.
(b) Exhibit infinitely many right inverses for $φ$.
(c) Find a nonzero $\pi∈R$ such that $φ\pi = 0 ≠ \piφ$.
(d) Prove $\nexists \lambda∈R~(\lambdaφ=0)$.

Proof: (a) $$(φψ)(a_1,a_2,a_3,...)=φ(ψ(a_1,a_2,a_3,...))=φ(0,a_1,a_2,...)=(a_1,a_2,a_3,...)$$ (b) Let $\pi$ be the homomorphism defined in part (c). We claim $\pi^n+ψ$ is a right inverse for any $n∈\mathbb{Z}^+$. We have $φ(\pi^n+ψ)=φ\pi\pi^{n-1}+φψ=φψ=1$, so now it suffices to prove $\pi^n+ψ≠\pi^m+ψ$ for any $m≠n$: $$\pi^n+ψ=\pi^m+ψ⇒\pi^n=\pi^m⇒$$ $$\pi^n(1,0,...)=\pi^m(1,0,...)⇒(2^n,0,...)=(2^m,0,...)⇒n=m$$ (c) Let $\pi$ be defined by $\pi(a_1,a_2,a_3,...)=(2a_1,0,0,...)$. Prove it is a homomorphism: $\pi((a_1,a_2,...)+(b_1,b_2,...))=\pi((a_1+b_1,a_2+b_2,...)=(2(a_1+b_1),0,...)=$$(2a_1,0,...)+(2b_1,0,...)=\pi(a_1,a_2,...)+\pi(b_1,b_2,...)$. We have $(φ\pi)(a_1,a_2,...)=φ(\pi(a_1,a_2,))=φ(2a_1,0,...)=(0,0,...)$, so that $φ\pi=0$. Now, $(\piφ)(0,1,0,0,...)=(2,0,0,...)$, so that $\piφ≠0$.
(d) By exercise 28(b), since $φ$ has $ψ$ a right inverse, $φ$ is not a right zero divisor.$~\square$