Lemma 2

Let E be a monoid and x an element of E. (i) There exists a unique homomorphism f of $\mathbf{N}$ into E with $f(1) = x$ and $f(n) = \frac{n}{1} x$ for all $n \in \mathbf{N}$. (ii) If x is invertible, there exists a unique homomorphism g of $\mathbf{Z}$ into E such that $g(1) = x$ and g coincides with f on $\mathbf{N}$.

Writing $f(n) = \frac{n}{1} x$ for all $n \in \mathbf{N}$, the formulae

(no. 1) express the fact that f is a homomorphism of $\mathbf{N}$ into E and obviously $f(1) = x$. If $f'$ is a homomorphism of $\mathbf{N}$ into E such that $f'(1) = x$, then $f = f'$, by § 1, no. 4, Proposition 1, (iv).

Suppose now that x is invertible. By no. 3, Corollary 2 to Proposition 4, $f(n) = \frac{n}{1} x$ is invertible for every integer $n \geq 0$. By construction, $\mathbf{Z}$ is the group of differences of $\mathbf{N}$ and hence (no. 4, Theorem 1) f extends uniquely to a homomorphism g of $\mathbf{Z}$ into E. If $g'$ is a homomorphism of $\mathbf{Z}$ into E with $g'(1) = x$, the restriction $f'$ of $g'$ to $\mathbf{N}$ is a homomorphism of $\mathbf{N}$ into E with $f'(1) = x$. Hence $f' = f$, whence $g' = g$.

We shall apply Lemma 2 to the case where the monoid E is $\mathbf{Z}$; for every integer $m \in \mathbf{Z}$ there therefore exists an endomorphism $f_m$ of $\mathbf{Z}$ characterized by $f_m(1) = m$. If m is in $\mathbf{N}$, the mapping $n \mapsto mn$ of $\mathbf{N}$ into $\mathbf{N}$ is an endomorphism of the magma $\mathbf{N}$ (Set Theory, III, § 3, no. 3, Corollary to Proposition 5); hence $f_m(n) = mn$ for all $m, n$ in $\mathbf{N}$.

Multiplication on $\mathbf{N}$ can therefore be extended to multiplication on $\mathbf{Z}$ by the formula $mn = f_m(n)$ for $m, n$ in $\mathbf{Z}$. We shall establish the formulae:

(2)

(3) (4) (5) (6) (7) (8)

for $x, y, z$ in $\mathbf{Z}$. (In other words, $\mathbf{Z}$ is a commutative ring.) The formulae $x(y+z) = xy + xz$ and $x.0 = 0$ express the fact that $f_x$ is an endomorphism of the additive monoid $\mathbf{Z}$ and $f_x(1) = x$ may be written $x.1 = x$. The endomorphism $f_x \circ f_y$ of $\mathbf{Z}$ maps 1 to $xy$ and hence is equal to $f_{xy}$, whence (3). Now $f_x(-y) = -f_x(y)$, that is $x(-y) = -xy$; similarly, the endomorphism $y \mapsto -xy$ of $\mathbf{Z}$ maps 1 to $-x$, whence $(-x).y = -xy$ and therefore

For $m, n$ in $\mathbf{N}$, $mn = nm$ (Set Theory, III, § 3, no. 3, Corollary to Proposition 5), whence $(-m).n = n(-m)$ and $(-m)(-n) = (-n)(-m)$; as $\mathbf{Z} = \mathbf{N} \cup (-\mathbf{N})$, $xy = yx$ for $x, y$ in $\mathbf{Z}$; and this formula means that (5) follows from (4) and completes the proof of formula (6) to (8).