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## Counterexamples regarding one-sided zero divisors and inverses

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.1 Exercise 7.1.30

Let $A = \prod_\mathbb{N} \mathbb{Z}$ be the direct product of countably many copies of $\mathbb{Z}$. Recall from Exercise 7.1.19 that $A$ is a ring under componentwise addition and multiplication. Let $R$ be the ring of all additive group homomorphisms $A \rightarrow A$ as described in Exercise 7.1.29 (i.e. with pointwise addition and composition). Let $\varphi$ be the element of $R$ defined by ($\prod a_i) \mapsto (\prod b_i)$, where $b_i = a_{i+1}$, and let $\psi$ be the element of $R$ defined by $(\prod a_i) \mapsto (\prod b_i)$, where $b_0 = 0$ and $b_{i+1} = a_i$.

(1) Prove that $\varphi\psi = 1$, but that $\psi\varphi \neq 1$. (That is, $\psi$ is a right inverse of $\varphi$ but not a left inverse.)
(2) Exhibit infinitely many right inverses for $\varphi$.
(3) Find a nonzero element $\pi \in R$ such that $\varphi\pi = 0$ but $\pi\varphi \neq 0$.
(4) Prove that there is no nonzero element $\lambda \in R$ such that $\lambda\varphi = 0$ (i.e. $\varphi$ is a left zero divisor but not a right zero divisor.)

Solution:

(1) Let $(\prod a_i) \in A$. Note that $$(\varphi \circ \psi)(\prod a_i) = \varphi(\prod b_i),$$ where $b_0 = 0$ and $b_{i+1} = a_i$. Now $\varphi(\prod b_i) = \prod c_i$, where $c_i = b_{i+1} = a_i$ for all $i \in \mathbb{N}$; thus $(\varphi \circ \psi)(\prod a_i) = \prod a_i$, and we have $\varphi \circ \psi = 1$. On the other hand, if $a_0 \neq 0$, the 0-th coordinate of $(\prod a_i)$ is nonzero while the 0-th coordinate of $(\psi \circ \varphi)(\prod a_i)$ is zero. Thus $\psi \circ \varphi \neq 1$.

(2) For each $k \in \mathbb{N}$, define $\psi_k : A \rightarrow A$ as follows: $\psi_k(\prod a_i) = \prod b_i$, where $b_0 = a_k$ and $b_{i+1} = a_i$. First we show that $\psi_k$ is a group homomorphism. If $(\prod a_i), (\prod b_i) \in A$, then $$\psi_k((\prod a_i) + (\prod b_i)) = \psi_k(\prod a_i+b_i) = \prod c_i,$$ where $c_0 = a_k + b_k$ and $c_{i+1} = a_i + b_i$. Then $(\prod c_i) = (\prod d_i) + (\prod e_i)$, where $d_0 = a_k$ and $d_{i+1} = a_i$ and $e_0 = b_k$ and $e_{i+1} = b_i$. Thus $$\psi_k((\prod a_i) + (\prod b_i)) = \psi_k(\prod a_i) + \psi_k(\prod b_i).$$ Hence $\psi_k$ is in fact a group homomorphism. Now we show that $\varphi \circ \psi_k = 1$. To that end, note that $(\varphi \circ \psi_k)(\prod a_i) = \varphi(\prod b_i)$ where $b_0 = a_k$ and $b_{i+1} = a_i$. Then $\varphi(\prod b_i) = \prod c_i$, where $c_i = b_{i+1} = a_i$. Thus $(\varphi \circ \psi_k)(\prod a_i) = \prod a_i$ for all $\prod a_i$, and we have $\varphi \circ \psi_k = 1$. Thus we have exhibited infinitely many right inverses for $\varphi$.

(3) Define $\pi : A \rightarrow A$ by $\pi(\prod a_i) = \prod b_i$, with $b_0 = a_0$ and $b_{i+1} = 0$. First we verify that $\pi$ is a group homomorphism; $$\pi((\prod a_i) + (\prod b_i)) = \pi(\prod a_i + b_i) = \prod c_i,$$ where $c_0 = a_0 + b_0$ and $c_{i+1} = 0$. Thus $\prod c_i = (\prod d_i) + (\prod e_i)$, where $d_0 = a_0$ and $d_{i+1} = 0$ and $e_0 = b_0$ and $e_{i+1} = 0$; now $$\pi((\prod a_i) + (\prod b_i)) = \pi(\prod a_i) + \pi(\prod b_i).$$ Thus $\pi$ is a group homomorphism. Moreover, $(\varphi \circ \pi)(\prod a_i) = \varphi(\prod b_i)$, where $b_0 = a_0$ and $b_{i+1} = 0$, and $\varphi(\prod b_i) = \prod c_i$, where $c_i = b_{i+1} = 0$. Thus $(\varphi \circ \pi)(\prod a_i) = \prod 0 = 0$ for all $\prod a_i$, and we have $\varphi \circ \pi = 0$. However, consider $\prod a_i \in A$ defined by $a_i = 1$. Now $(\pi \circ \varphi)(\prod a_i) = \pi(\prod b_i)$, where $b_i = a_{i+1} = 1$, so that the 0-th coordinate of $(\pi \circ \varphi)(\prod a_i)$ is 1. Thus $\pi \circ \varphi \neq 0$.

(4) Suppose $\lambda \circ \varphi = 0$. Recall that $\psi_k$ (defined above) is a right inverse for $\varphi$; then $$\lambda \circ \varphi \circ \psi_k = 0 \circ \psi_k,$$ so that $\lambda = 0$. In particular, $\varphi$ is not a right zero divisor.