(This question originates from Pinter's Abstract Algebra Chapter 18 Exercise J6.)
- Let $A$ be a commutative ring.
- For each element $a$ in $A$, the function $\pi_a$ defined by $\pi_a(x)=ax$ satisfies the identity $\pi_a(x+y)=\pi_a(x)+\pi_a(y)$.
- Let $\mathscr{A}$ denote the set $\{\pi_a: a\in A\}$ with the two operations: \begin{align*} [\pi_a + \pi_b](x) &= \pi_a(x) + \pi_b(x) \quad \text{and}\quad \pi_a\pi_b = \pi_a\circ\pi_b \end{align*}
- Let $\phi: A\rightarrow\mathscr{A}$ such that $\phi(a)=\pi_a$.
Prove that if $A$ has a unity, then $\phi$ is an isomorphism.
It can be verified that $\mathscr{A}$ is a ring, and that $\phi$ is a homomorphism.
Question:
If we construct the inverse function $\phi^{-1}: \mathscr{A}\rightarrow A$ such that $\phi^{-1}(\pi_a)=a$, wouldn't it be sufficient to show the bijection between $\mathscr{A}$ and $A$, and therefore $\phi$ an isomorphism? What is the significance of having unity in this question?
If the ring does not have unity, the map $\phi$ may fail to be injective, i.e., there may exist nonzero $a \in A$ such that $ax = 0$ for all $x \in A$. Your supposed inverse is not well-defined in this case. For example, for each prime $p$ one can define a ring of order $p$ with the additive structure of $\mathbb{Z}/p\mathbb{Z}$ but where the product of any two elements is $0$. In this case, the map $\phi$ is trivial.
If the ring does have unity, then if $a \in ker\phi$ in particular implies $$a = a \cdot 1 = 0,$$ thus $\phi$ is injective.