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Basic properties of the countable direct sum of the integers

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.1 Exercise 7.1.20

Let $R$ be the collection of sequences $(a_i)$ of integers, indexed by $\mathbb{N}$, such that all but finitely many $a_i$ are 0. (This is called the direct sum of countable copies of $\mathbb{Z}$.) Prove that $R$ is a ring under componentwise addition and multiplication which does not have an identity.

Solution: In Exercise 5.1.17, we saw that R is a subgroup of $\prod_\mathbb{N} \mathbb{Z}$, so that $R$ is an abelian group.

Now suppose$(a_i), (b_i) \in R$, where $a_i = 0$ when $i \notin A$ and $b_i = 0$ when $i \notin B$ for some finite sets $A$ and $B$. Then $(a_i)(b_i) = (a_ib_i)$, and $a_ib_i = 0$ when $i \notin A \cap B$. Since $A \cap B$ is finite, $(a_i)(b_i) \in R$. Thus $R$ is a subring of $\prod_\mathbb{N} \mathbb{Z}$.

Suppose now that $(e_i) \in R$ is an identity element. Then for all $(r_i) \in R$, $$(r_i) = (e_i)(r_i) = (e_ir_i),$$ so that $r_i = e_ir_i$; similarly, $r_i = r_ie_i$ for all $r_i \in \mathbb{Z}$. Thus $e_i = 1$ for all $i$. But then $(e_i) \notin R$.