# 36 Fields of Rational Functions

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1 36 Fields of Rational Functions The reader might have missed the familiar quotient rule (\f(f,g)) = f g fg in Lemma It was missing because f is not a polynomial. g We now introduce these quotients f g. 36. Definition: Let D be an integral domain and x, x indeterminates over D. Then D[x] and D[x ] are integral domains (Lemma 33.6, Lemma 33.0).. An element in the the field of fractions of D[x] is called a rational function (in x) over D.. The field of fractions of D[x] will be called the field of rational functions over D (in x). and will be denoted by D.. An element in the the field of fractions of D[x ] is called a rational function (in x ) over D. The field of fractions of D[x ] will be called the field of rational functions over D (in x ) and will be denoted by D(x ). Thus a rational function over D is a fraction f g of two polynomials over D, with g 0. Two rational functions f and f are equal if and only if the polynomials f and f are equal. Two rational functions f and f are added and multiplied according to the rules f + f = f + f, f f = f f. Here and are distinct from the zero polynomial over D. This terminology is unfortunate and misleading, because a rational function is not a function in the sense of Definition 3.. A rational function is not a function of the 'rational' kind, whatever that might mean. The technical term we defined is rational function, a term consisting of two words "rational" and "function". The meaning of the 430

2 words "rational" and "function" do not play any role in Definition 36.. A rational function is a fraction of polynomials over D. The reader should exercise caution about this point. One should not conclude that x x and x + in are different rational functions, on grounds that that their domains are different, since the domain of the first one does not contain, whereas is in the domain of the second one. Neither of them has a domain, for neither of them is a function. And these rational functions are equal because the polynomials (x ) and (x )(x + ) in [x] are equal. 36. Lemma: Let D be an integral domain and F the field of fractions of D. Let x be an indeterminate over D. Then D = F. Proof: F consists of the fractions a b, where a,b D and b 0; and D consists of the fractions a n x n + a n x n a x + a 0 b m x m + b m x m b x + b 0, where a n,a n,...,a,a 0,b m,b m,...,b,b 0 D and the denominator is distinct from the zero polynomial in D[x]. Finally, F consists of the fractions c n x n + c n x n c x + c 0 d m x m + d m x m d x + d 0, where c n,c n,...,c,c 0,d m,d m,...,d,d 0 from the zero polynomial in F[x]. F and the denominator is distinct An element of D is identified with the fraction a in F (Theorem 3.5), whence D F. Thus D[x] F[x] as sets. Note that two elements f and g p of D are equal in D if and only if fq = gp in D[x], and q this holds if and only if fq = gp in F[x], so if and only if f g 43

3 and p are equal in F. Thus every element of D is in F and q equality in D coincides with equality in F. So D F. Next we show F n Then p = i=0 a i D. Let p q m b i x i, q = j=0 F, with p, q F[x], q 0. c j d j x j, where a i,b i,c j,d j D, b i 0, d j 0 for all i,j and not all of c j are equal to 0 D. We put b = b 0 b... b n b n and d = d 0 d... d m d m. Then dbp and dbq are polynomials in D[x], and hence p q = dbp dbq D. So F D. This proves D = F. As an illustration of Lemma 36., observe that 3 x 7 x x + 3 x 5(56x is equal to the rational function x + ) 4(x in [x]. + 0x 5) 36.3 Remark: Let D be an integral domain and F the field of fractions of D. Then D(x ) = field of fractions of D[x ] = field of fractions of D[x ][x n ] = D[x ](x n ) = D(x )(x n ) by Lemma 36., with D[x ], D(x ), x n in place of D,F,x, respectively. Also, we have D(x ) = F(x ), for this is true when n = (Lemma 36.) and, when it is true for n = k, so that D(x ) = F(x ), it is also true for n = k + : D(x,x k+ ) = D(x )(x k+ ) = F(x )(x k+ ) = F(x,x k+ ), the last equation by the remark above, with F in place of D and k + in place of n. 43

4 In the remainder of this paragraph, we discuss partial fraction expansions of ratinonal functions Lemma: Let K be a field and let f be a nonzero polynomial in K[x]. Let q, r be two nonzero, relatively prime polynomials of positive degree in K[x]. Suppose deg f deg qr and suppose that f is relatively prime to qr. Then there are uniquely determined nonzero polynomials a, b in K[x] such that ar + bq = f, deg a deg q, deg b deg r. Proof: We first prove the existence of a and b. Since q, r are relatively prime, there are polynomials h, k in K[x] with hr + kq =. Multiplying both sides of this equation by f and putting A = fh, B = fk, we obtain Ar + Bq = f. We now divide A by q and B by r: A = sq + a, a = 0 or deg a B = ur + b, b = 0 or deg b deg q, deg r. Thus ar + bq = (A sq)r + (B ur)q = (Ar + Bq) (s + u)qr = f (s + u)qr. We claim s + u is the zero polynomial in K[x]. Otherwise, we would have deg (s + u) 0, deg (s + u)qr deg qr, and since by hypothesis deg f deg qr, deg f (s + u)qr deg qr, so that ar + bq 0; in particular, both a and b cannot be zero. Assume, without loss of generality, that a 0 in case one of a, b is zero and that deg ar deg bq in case neither of them is zero. Then we get the contradiction deg [f (s + u)qr] = deg (ar + bq) 433

5 deg ar = deg a + deg r deg q + deg r = deg qr. Thus s + u, and consequently (s + u)qr is the zero polynomial in K[x]. This gives ar + bq = f. It remains to show that a and b are distinct from the the zero polynomial in K[x]. Both of them cannot be 0, for then f would be also 0, which it is not by hypothesis. If one of them is 0, say if a = 0, then b 0 and f = bq would not be relatively prime to qr (because q is of positive degree, so not a unit in K[x]), against the hypothesis. This proves the existence of a, b. It remains to show the uniqueness of a and b. If we have also. a r + b q = f, deg a deg q, deg b deg r, we obtain 0 = f f = (a r + b q) (ar + bq) = (a a )r (b b)q, so (a a )r = (b b)q. (*) Hence r (b b)q in K[x] r b b in K[x] as r and are q relatively prime. Now b b implies b b 0 and this gives deg r deg (b b) max{deg b, deg b} deg r, a contradiction. Thus b = b and we get then a = a from (*). So a and b are uniquely determined Lemma: Let K be a field and let f be a nonzero rational function g in K,with deg f deg g. Suppose that f and g are both monic and that f is relatively prime to g. Assume g = qr, where qr are two relatively prime polynomials of positive degree in K[x]. Then there are uniquely determined nonzero polynomials a, b in K[x] such that f g = f qr = a q + b r and deg a deg q, deg b deg r. 434

6 Proof: If f is a nonzero rational function in K, then f is a nonzero g polynomial in K[x], and f is relatively prime to g = qr. As f and g are monic, these conditions determine f and g uniquely. The polynomials q, r are relatively prime and deg f is smaller than deg qr. So the hypotheses of Lemma 36.4 are satisfied and therefore there are uniquely determined nonzero polynomials a,b in K[x] such that f = ar + bq, and deg a deg q, deg b deg r. Dividing both sides of the equation above by g = qr, we see that there are uniquely determined nonzero polynomials a,b in K[x] such that f g = f qr = a q + b r and deg a deg q, deg b deg r. By induction on m, we obtain the following lemma Lemma: Let K be a field and let f be a nonzero rational g function in K,with deg f deg g. Suppose that f and g are both monic and that f is relatively prime to g. Assume g = q q... q m, where q, q,...,q m. are pairwise relatively prime monic polynomials of positive degree in K[x].. Then there are uniquely determined nonzero polynomials a, a,...,a m in K[x] such that f g = f q q...q m = a q + a q a m q m and deg a i deg q i for all i =,,...,m Lemma: Let K be a field and x an indeterminate over K.. Let g be a polynomial in K[x] of degree. Then, for any f K[x],. there are uniquely determined polynomials r 0, r, r,...,r n such that 435

7 and f = r 0 + r g + r r n g n r i = 0 or deg r i deg g for all i =,,...,n. Proof: From deg, we know that g 0.. So we may divide f by g and obtain f = q 0 g + r 0, where q 0, r 0 K[x], with r 0 = 0 or deg r 0 deg g. Here q 0 and r 0 are uniquely determined by f and g (Theorem 34.4) and we have f = r 0 + q 0 g. If q 0 = 0, we are done (with n = 0). Otherwise, since f = q 0 g + r 0, deg and r 0 = 0 or deg r 0 deg g, we have deg q 0 deg f (Lemma 33.3). We now divide q 0 by g and obtain q 0 = q g + r, where q, r K[x], with r = 0 or deg r deg g. Here q and r are uniquely determined by q 0 and g (hence by f and g) and f = r 0 + r g + q. If q = 0, we are done. Otherwise, deg q deg q 0. We then divide q by g and obtain q = q g + r, where q, r K[x], with r = 0 ordeg r deg g. Here q and r are uniquely determined by q and g (hence by f and g) and f = r 0 + r g + r + q g 3. If q = 0, we are done. Otherwise, we have deg q deg q. We continue this process. As the degrees of q 0, q, q,... get smaller and smaller, this process cannot go on indefinitely.. Sooner or later, we will meet a q n equal to 0 K[x]. Then, with uniquely determined r 0,r,r,... r n, we have f = r 0 + r g + r r n g n, where r i = 0 or deg r i deg g for all i =,,...,n. In the situation of Lemma 36.7, the unique expression f = r 0 + r g + r r n g n of f, where r i = 0 or deg r i g-adic expansion of f. deg g for all i =,,...,n, is called the 36.8 Theorem: Let K be a field and p q a nonzero rational function in K, where p,q K[x] are relatively prime in K[x]. Let u be the leading coefficient of q and let q = u m m... g t m t be the decomposition of q into polynomials irreducible over K, where g i are monic. Then there are uniquely determined polynomials G, 436

8 a (), a (),...,a m (),a (), a (),...,a m (),...,a (t), a (t),...,a mt (t) in K[x] such that p q = G + a () + a () a m () m + a () a (t) t + a () + a (t) t a m () m a mt (t) g t m t and deg a i (k) deg g k or a i (k) = 0 for all i and k. Proof: We divide p by q and find unique polynomials G, H in K[x] with p = qg + H, deg H deg q or H = 0. In the latter case, everything is proved (a (k) i = 0 for all i and k). If H 0, let v be the leading coefficient of H and put c = v/u. Then H and q are relatively prime (since p and q are). We have H = vh, where h is monic, relatively prime to q and p q = G + c h q with deg h deg q.. We may use Lemma 36.6 and get uniquely determined nonzero polynomials b, b,...,b t in K[x] such that h q = b m + b m b t g t m t and deg b k deg g k m k for all k =,,...,t. We put f k = cb k. Then p q = G + f m + f m f t g t m t and, since c is uniquely determined by p and q,. the polynomials f k are also uniquely determined. Since in the g k -adic expansion deg f k = deg b k deg g k m k, f k = r 0 + r g k + r g k r n g k n 437

9 of f k, the polynomials r s = 0 for s m k. So let f k = a (k) g k m k + a (k) g k m k a (k) mk g k + a (k) mk be the g k -adic expansion of f k. The polynomials a (k),a (k) (k),...,a mk (k) in K[x] are uniquely determined and deg a i deg g k or a (k) i = 0 for all i =,,...,m k. Hence, for all k =,,...,t, there holds f k a (k) g k m = k k + a (k) k a mk (k) g k m k and this completes the proof. The equation p q = G + a () + a () a m () m + a () a (t) t + a () + a (t) t in Theorem 36.8 is known as the expansion of p q a m () m a mt (t) g t m t in partial fractions. Exercises. Let K be a field. For any nonzero rational function f g in K, we define the degree of f g, denoted by deg f g, by deg f g = deg f deg g. Prove that the degree of a rational function is well defined. Can you extend the degree assertions in Lemma 33.3 to rational functions?. Let K be a field. For any rational function f g in K, we define the derivative of f g, denoted by (f g ), by declaring 438

10 ( f g ) = f g fg. Prove that differentiation is well defined, i.e., prove that f g = a b implies ( f g ) = (a b ). 3. Extend Lemma 35.5 and Lemma 35.6 to derivatives of rational functions in one indeterminate over a field. 4. Expand x 3 + 3x + 8x + 6 (x 3 + 3x + 3)(x + x + 3) 4x 3 + 3x + x + x 5 + 4x 4 + 4x 3 + x + 5 and in partial fractions. 5. Let K be a field and let a,a,...,a m be pairwise distinct elements in K. Put g = (x a )(x a )... (x a m ) and let f be a nonzero polynomial in K[x] with deg f m. Show that f g m f(a = i )/g (a i ). x a i= i 439

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