Proposition 3

— Let E be an algebra over K which is associative, commutative, unital and non-zero. Let f \in K((X_i)_{i \in I}) and for each i \in I, let g_i \in K((Y_l)_{l \in L}). Given a family y = (y_l)_{l \in L} of elements of E, suppose that y is substitutable into each g_i and (g_i(y))_{i \in I} is substitutable in f. Then: (i) (g_i)_{i \in I} is substitutable in f; (ii) if we denote by h the element f((g_i)) of K((Y_l)_{l \in L}), then y is substitutable in h and h(y) = f((g_i(y))).

We may take I to be finite. By hypothesis, for each $i \in I$, $g_i$ can be put in the form $p_i/q_i$ where $p_i, q_i \in K[(Y_l)_{l \in L}]$ and $q_i(y)$ is invertible in E. Likewise f can be written as $u/v$, where $u, v \in K[(X_i)_{i \in I}]$ and $v((g_i(y)))$ is invertible. Let $m = \sup(\deg u, \deg v)$, and let $w = \prod_{i \in I} q_i \in K[(Y_l)_{l \in L}]$, $u_1 = u((g_i)) w^m$, $v_1 = v((g_i)) w^m$. The polynomial u is a K-linear combination of monomials $\prod_{i \in I} X_i^{\nu_i}$ such that $\sum_{i \in I} \nu_i \leq m$. We have $w^m \prod_{i \in I} g_i^{\nu_i} = w^m \left( \prod_{i \in I} p_i^{\nu_i} \right) \left( \prod_{i \in I} q_i^{\nu_i} \right)^{-1} \in K[(Y_l)_{l \in L}]$ by the choice of m. Hence $u_1 \in K[(Y_l)_{l \in L}]$ and similarly $v_1 \in K[(Y_l)_{l \in L}]$. Moreover, $v_1(y) = (w(y))^m v((g_i(y)))$ is invertible. Hence $v_1 \neq 0$, because $E \neq 0$, and so $v((g_i)) \neq 0$. The family $(g_i)$ is thus substitutable in f. Besides we have $f((g_i)) = u_1/v_1$, hence y is substitutable in $h = f((g_i))$, and $h(y) = u_1(y)/v_1(y) = u((g_i(y)))/v((g_i(y))) = f((g_i(y)))$.

Let K be a commutative field, E a commutative associative and unital K-algebra, and let f \in K((X_i)_{i \in I}). Let T_f be the set of all x = (x_i)_{i \in I} \in E^I which are substitutable in f. The mapping x \mapsto f(x) of T_f into E is called the rational function associated with f (and E); we sometimes denote it by \tilde{f}. If g \in K((X_i)_{i \in I}) we have T_f \cap T_g \subset T_{f+g}, T_f \cap T_g \subset T_{fg}, hence the rational function associated with f + g (resp. fg) is defined on T_f \cap T_g and has the same value on this set as \tilde{f} + \tilde{g} (resp. \tilde{f}\tilde{g}). Let T'_f be the set of x \in T_f such that f(x) is invertible; if x \in T'_f, x is substitutable in 1/f and the rational function associated with 1/f takes at x the value f(x)^{-1}.

If K is an infinite commutative field, f \in K((X_i)_{i \in I}), g \in K((X_i)_{i \in I}) and \tilde{f}, \tilde{g} are the rational functions associated with f, g (and K), and if \tilde{f}(x) = \tilde{g}(x) for all x \in T_f \cap T_g then f = g. For if f = u/v and g = u_1/v_1, where u, v, u_1, v_1 are polynomials, we have u(x)v_1(x) = u_1(x)v(x) for all x such that v(x)v_1(x) \neq 0, hence uv_1 = u_1v (IV, p. 18, Th. 2). Therefore the mapping f \mapsto \tilde{f} is injective and we shall often identify f and \tilde{f}.

\* Using the factoriality of $K[(X_i)_{i \in I}]$ (Comm. Alg., VII, § 3, No. 2 p. 502 and Cor. of Th. 2 p. 506), one easily shows the following: for every $f \in K((X_i)_{i \in I})$ there exist $u, v \in K[(X_i)_{i \in I}]$ such that: 1) $f = u/v$; 2) for $x \in K^I$ to be substitutable in $f$ it is necessary and sufficient that $v(x) \neq 0.$