§ 3. Rational fractions

1. Definition of rational fractions

Definition 1

Let K be a commutative field and I a set. The field of fractions (I, p. 116) of the integral domain $K[(X_i)_{i \in I}]$ is denoted by $K((X_i)_{i \in I})$ or $K(X_i)_{i \in I}$ . Its elements are called rational fractions in the indeterminates $X_i$ with coefficients in K.

For $I = \{1, 2, \ldots, n\}$ we write $K(X_1, X_2, \ldots, X_n)$ in place of $K((X_i)_{i \in I})$ .

Let A be an integral domain and K its field of fractions. The ring $A[(X_i)_{i \in I}]$ may be identified with a subring of $K[(X_i)_{i \in I}]$ , hence also of $K((X_i)_{i \in I})$ . For each $f \in K[(X_i)_{i \in I}]$ there exists a non-zero element $\alpha$ of A such that $\alpha f \in A[(X_i)_{i \in I}]$ . Hence every element of $K((X_i)_{i \in I})$ can be written as $u/v$ where $u, v \in A[(X_i)_{i \in I}], v \neq 0$ . Thus $K((X_i)_{i \in I})$ may be identified with the field of fractions of $A[(X_i)_{i \in I}]$ .

Now let K be a commutative field, I a set and J ⊂ I. Put B = K[(X_i)_{i ∈ J}], then K[(X_i)_{i ∈ I}] = B[(X_i)_{i ∈ I - J}], and by what has been said, K((X_i)_{i ∈ I}) may be identified with K'((X_i)_{i ∈ I - J}), where K' = K((X_i)_{i ∈ J}).

2. Degrees

Let K be a commutative field. For every element r of K((X_i)_{i ∈ I}) there exist u, v ∈ K[(X_i)_{i ∈ I}] such that v ≠ 0 and r = $\frac{u}{v}$ . The relation $\frac{u}{v} = \frac{u_1}{v_1}$ , where v ≠ 0, v_1 ≠ 0 is equivalent to uv_1 = vu_1; if r ≠ 0, we have u ≠ 0 and u_1 ≠ 0, and so deg u + deg v_1 = deg v + deg u_1 (IV, p. 9), or also deg u − deg v = deg u_1 − deg v_1. The rational integer deg u − deg v thus depends only on r; we call it the degree, or the total degree of r, and denote it by deg r. We shall agree to write deg 0 = −∞. If J ⊂ I, we can likewise define the degree with respect to the X_j of index j ∈ J. When r is a polynomial, these notions coincide with those defined in IV, p. 2.

Proposition 1

— Let r, s be two rational fractions.

(i) If deg r ≠ deg s, we have

r + s \neq 0 \quad \text{and} \quad \deg(r + s) = \sup(\deg r, \deg s)

If deg r = deg s, we have deg(r + s) ≤ deg r.

(ii) We have deg(rs) = deg r + deg s.

We can limit ourselves to the case where r and s are non-zero.

Let us write r = $\frac{u}{v}$ , s = $\frac{w}{z}$ , where u, v, w, z are non-zero polynomials. We have rs = $\frac{uw}{vz}$ , and so

\deg(rs) = \deg(uw) - \deg(vz) = \deg u - \deg v + \deg w - \deg z = = \deg r + \deg s.

On the other hand, we have r + s = $\frac{uz + vw}{vz}$ . Suppose that deg r ≠ deg s, in other words deg u + deg z ≠ deg w + deg v. Then uz + vw ≠ 0, and

\begin{align*} \deg(r + s) &= \deg(uz + vw) - \deg(vz) \\ &= \sup(\deg(uz), \deg(vw)) - \deg(vz) \\ &= \sup(\deg(uz) - \deg(vz), \deg(wv) - \deg(vz)) \\ &= \sup(\deg r, \deg s). \end{align*}

Suppose that deg r = deg s, that is deg u + deg z = deg w + deg v. If r + s ≠ 0, then we have

\begin{align*} \deg(r + s) &= \deg(uz + vw) - \deg(vz) \\ &\leq \deg(uz) - \deg(vz) = \deg r \end{align*}

\* The mapping r ↦ −deg r is thus a discrete valuation on the field K((X_i)_{i ∈ I}). \*

3. Substitutions

Let K be a commutative field, E a unital associative K-algebra, $x = (x_i)_{i \in I}$ a family of pairwise permutable elements of E. Let $B = K[(X_i)_{i \in I}]$ and $S_x$ the set of all non-zero $v \in B$ such that $v(x)$ is invertible in E. Let $u \in B,\ v \in S_x$ and $f = \frac{u}{v} \in K((X_i)_{i \in I})$ . The element $u(x)\ v(x)^{-1} = v(x)^{-1}u(x)$ is defined in E; moreover, if $u_1,\ v_1$ are two polynomials such that $f = \frac{u_1}{v_1}$ and $v_1 \in S_x$ , then $uv_1 = vu_1$ , hence $u(x)v_1(x) = v(x)u_1(x)$ and so

u(x)\ v(x)^{-1} = u_1(x)\ v_1(x)^{-1}.

Let $f \in K((X_i)_{i \in I})$ . If there exists at least one couple $(u, v)$ such that $f = \frac{u}{v}$ and $v \in S_x$ , we shall say that x is substitutable in $f$ ; the element $u(x)\ v(x)^{-1}$ which only depends on $f$ and $x$ is then denoted by $f(x)$ or $f((x_i))$ or $f((x_i)_{i \in I})$ .

Proposition 2

Let K be a commutative field, E a unital associative K-algebra and $x = (x_i)_{i \in I}$ a family of pairwise permutable elements of E. The set $S_x^{-1}B$ of $f \in K((X_i)_{i \in I})$ such that x is substitutable in $f$ is a K-subalgebra of $K((X_i)_{i \in I})$ . The mapping $f \mapsto f(x)$ is a unital homomorphism $\varphi$ of $S_x^{-1}B$ into E. The image $\varphi(S_x^{-1}B)$ is the set of all $yz^{-1}$ , where y runs over the unital subalgebra $K[x]_E$ of E generated by the family x and where z runs over the set of all invertible elements of $K[x]_E$ .

Let $f_1 = \frac{u_1}{v_1},\ f_2 = \frac{u_2}{v_2}$ be two elements of $K((X_i)_{i \in I})$ such that $v_1,\ v_2 \in S_x$ . We have $f_1 + f_2 = \frac{u_1v_2 + u_2v_1}{v_1v_2},\ f_1f_2 = \frac{u_1u_2}{v_1v_2}$ and $v_1,\ v_2 \in S_x$ . Hence $S_x^{-1}B$ is a K-subalgebra of $K((X_i)_{i \in I})$ . The rest of the proposition is clear.

Corollary

Let L be a commutative field, K a subfield of L, $x = (x_i)_{i \in I}$ a family of elements of L, M the set consisting of the $x_i$ , U the set of all $f \in K((X_i)_{i \in I})$ such that x is substitutable in $f$ and $\varphi$ the homomorphism $f \mapsto f(x)$ of U into L. Then $\varphi(U)$ is the subfield of L generated by $KUM$ .

Let L' be the subfield of L generated by $KUM$ . We have

K \cup M \subset \varphi(U) \subset L'

and

\varphi(U)

is a subring of L. Now Prop. 2 implies that

\varphi(U)

is a subfield of L, whence

\varphi(U) = L'

Let $f \in K((X_i)_{i \in I})$ and let $(g_i)_{i \in I}$ be a family of elements of $K((Y_l)_{l \in L})$ . If (g_i) is substitutable in f, then f((g_i)) is an element of K((Y_l)_{l \in L}). In particular, (X_i)_{i \in I} is substitutable in f and f = f((X_i)_{i \in I}).

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 $v(x) \neq 0.$

4. Differentials and derivations

Let $K$ be a commutative field. By III, p. 558, Prop. 5, every derivation $D$ of $K[(X_i)_{i \in I}]$ extends in a unique fashion to a derivation $\bar{D}$ of $K((X_i)_{i \in I})$ . If $D, D'$ are permutable derivations of $K[(X_i)_{i \in I}]$ , then the bracket $[D, D'] = DD' - D'D$ is zero, hence $[\bar{D}, \bar{D}']$ which is a derivation of $K((X_i)_{i \in I})$ extending $[D, D']$ is zero; in other words, $D$ and $D'$ are permutable. In particular the derivations $D_i$ (IV, p. 6) extend to derivations of $K((X_i)_{i \in I})$ again denoted by $D_i$ and which are pairwise permutable. If $f \in K((X_i)_{i \in I}), D_i f$ is also written $D_{x_i} f$ or $\frac{\partial f}{\partial x_i}$ or $f'_{x_i}$ . When there is only a single indeterminate $X$ one uses the notation $Df, \frac{df}{dX}, f'$ .

Let $B = K[(X_i)_{i \in I}], C = K((X_i)_{i \in I})$ . By III, p. 574, Prop. 23, the canonical mapping

\Omega_K(B) \otimes_B C \to \Omega_K(C)

is an isomorphism of vector $C$ -spaces. Bearing in mind III, p. 570, we see that the vector $C$ -space $\Omega_K(C)$ admits as a basis the family $(dX_i)_{i \in I}$ of the differentials of the $X_i$ . Let $\partial_i$ be the coordinate form of index $i$ on $\Omega_K(C)$ relative to that basis. Then the mapping $u \mapsto (\partial_i, du)$ of $C$ into itself is a derivation of $C$ which maps $X_i$ to 1 and $X_j$ to 0 for $j \neq i$ , and so is equal to $D_i$ ; in other words, we have

du = \sum_{i \in I} (D_i u) dX_i

for every $u \in C$ . If $I$ is finite, $(D_i)_{i \in I}$ is a basis of the vector $C$ -space of derivations of $C$ .

Proposition 4

— *Let $E$ be an associative, commutative and unital $K$ -algebra, $x = (x_i)_{i, I}$ a family of elements of $E$ and $f \in K((X_i)_{i, I})$ . Suppose that $x$ is substitutable in $f$ and $y = f(x)$ . (i) For every derivation $A$ of $K((X_i)_{i, I})$ which maps $K[(X_i)_{i, I}]$ into itself, $x$ is substitutable in $Af$ . (ii) For every derivation $D$ of $E$ into an $E$ -module we have

Dy = \sum_{i \in I} (D_i f)(x) \cdot Dx_i .

Let $f = \frac{u}{v}$ with $u, v \in K[(X_i)_{i \in I}]$ and $v(x)$ invertible in E. Let A be a derivation of $K((X_i)_{i \in I})$ mapping $K[(X_i)_{i \in I}]$ into itself. We have

\Delta f = \frac{(\Delta u)\ v - u(\Delta v)}{v^2}

and

v^2(x)

is invertible, hence x is substitutable in

Af

. Secondly put

r = u(x)

s = v(x)

; we have

y = s^{-1}r

, hence for every derivation D of E into an E-module we have

\begin{align*} Dy &= s^{-2}(s(Dr) - r(Ds)) \\ &= s^{-2}\left( s \sum_{i \in I} (D_iu)(x) \cdot Dx_i - r \sum_{i \in I} (D_iv)(x) \cdot Dx_i \right) \end{align*}

by Prop. 4 of IV, p. 6. Thus

Dy = \sum_{i \in I} w_i \cdot Dx_i

with

w_i = v(x)^{-2}(v(x)(D_iu)(x) - u(x)(D_iv)(x)) = (D_if)(x).

Exercises

See the exercises for § 3.