Proposition 3

(a) If the $A_\alpha$ are non-zero, A is non-zero. (b) If the $A_\alpha$ are integral domains, A is an integral domain. (c) If the $A_\alpha$ are fields, A is a field.

Let $0_\alpha, 1_\alpha$ be the zero and unit of $A_\alpha$ and 0, 1 the zero and unit of A. There exists $\alpha \in I$ such that $f_\alpha(0_\alpha) = 0, f_\alpha(1_\alpha) = 1$. If $0 = 1$, there exists $\beta \geq \alpha$ such that $f_{\beta\alpha}(0_\alpha) = f_{\beta\alpha}(1_\alpha)$, that is $0_\beta = 1_\beta$. This proves (a).

Suppose that $A_\alpha$ are integral domains. Then A is commutative and non-zero by (a). Let $x, y$ be elements of A such that $xy = 0$. There exists $\alpha \in I$ and $x_\alpha, y_\alpha \in A_\alpha$ such that $x = f_\alpha(x_\alpha), y = f_\alpha(y_\alpha)$. Then $f_\alpha(x_\alpha y_\alpha) = xy = 0 = f_\alpha(0_\alpha)$. Hence there exists $\beta \geq \alpha$ such that $f_{\beta\alpha}(x_\alpha y_\alpha) = f_{\beta\alpha}(0_\alpha)$. As $A_\beta$ is an integral domain, it follows that $f_{\beta\alpha}(x_\alpha) = 0_\beta$ or $f_{\beta\alpha}(y_\alpha) = 0_\beta$, hence $x = 0$ or $y = 0$. This proves (b).

Suppose that the $A_\alpha$ are fields. Then $A \neq \{0\}$ by (a). Let $x$ be a non-zero element of A. There exist $\alpha \in I$ and $x_\alpha \in A_\alpha$ such that $x = f_\alpha(x_\alpha)$. Then $x_\alpha \neq 0$ and $f_\alpha(x_\alpha^{-1})$ is the inverse of $x$ in A. This proves (c).

Let $\mathfrak{E} = (E_\alpha, f_{\beta\alpha})$ and $\mathfrak{E}' = (E'_\alpha, f'_{\beta\alpha})$ be two direct systems of magmas (resp. monoids, groups, rings). A homomorphism of $\mathfrak{E}$ into $\mathfrak{E}'$ is a direct system $(u_\alpha)_{\alpha \in I}$ of mappings $u_\alpha : E_\alpha \to E'_\alpha$ such that each $u_\alpha$ is a homomorphism. Under these conditions, the mapping $u = \lim u_\alpha$ from $E = \lim E_\alpha$ to $E' = \lim E'_\alpha$ is a homomorphism (cf. Set Theory, III, § 7, no. 6).