代数学讲稿3-6

1、1,Chapter III: Rings,2,Contents,1. Rings and Homomorphisms 2. Ideals 3. Factorization in Commutative Rings 4. Rings of Quotients and Localization 5. Rings of Polynomials and Formal Power Series 6. Factorization in Polynomial Rings,3,3.6 Factorization in Polynomial Rings,4,Main Points,Consider the to

2、pics introduced in Section 3 in the context of polynomial rings over a commutative ring. Begin with two basic tools: the concept of the degree of a polynomial and the division algorithm.,5,Main Points (Cont.),Factors of degree one of a polynomial are then studied; finding such factors is equivalent

3、to finding roots of the polynomial. Consider irreducible factors of higher degree: Eisensteins irreducibility criterion is proved and it is shown that the polynomial domain Dx1,xn is a unique factorization domain if D is.,6,Degree of A Monomial,Let R be a ring. The Degree of a nonzero monomial ax1k1

4、xnknRx1,xn is the nonnegative integer k1+kn. If f is a nonzero polynomial in Rx1,xn, then by Thm 5.4.,7,Degree of A Polynomial,The (total) degree of the polynomials f is the maximum of the degrees of the monomials s.t. ai0 (i=1,m). The (total) degree of f is denoted deg f.,8,Remarks,A nonzero polyno

5、mial f has degree zero iff f is a constant polynomial. A polynomial which is a sum of monomials, each of which has degree k, is said to be homogeneous of degree k.,9,Review,For each k (1kn), Rx1,xk-1,xk+1, xn is a subring of Rx1,xn. The degree of f in xk is the degree of f considered as a polynomial

6、 in one indeterminate xk over the ring Rx1,xk-1, xk+1,xn.,10,Example,The polynomial 3x12x22x32+3x1x34-6x23x3Zx has degree 2 in x1, degree 3 in x2, degree 4 in x3 and total degree 6.,11,Deg 0,Define the degree of the zero polynomial to be - and Adopt the following conventions about the symbol deg 0=-

7、: -n, (-)+n=-=n+(-) for every integer n; (-)+(-)=-.,12,Arithmetic of Degrees,Theorem 6.1 Let R be a ring and f, gRx1,xn. (i) deg(f+g) max (deg f, deg g). (ii) deg(fg) deg f + deg g. (iii) If R has no zero divisors, then deg(fg) = deg f + deg g. (iv) If n=1 and the leading coefficient of f or g is no

8、t a zero divisor in R (in particular, if it is a unit), then deg(fg) = deg f + deg g.,13,Remark,The theorem is also true if deg f is taken to mean “degree of f in xk.”,14,Sketch of Proof of 6.1,Since we shall apply this theorem primarily when n=1 we shall prove only that case. (i) is easy (ii) is tr

9、ivial if f=0 or g=0. If has degree n and has degree m, then fg = a0b0+(an-1bm+anbm-1)xn+m-1+anbmxm+n has degree at most m+n. Since an0bm, fg has degree m+n if one of an, bm is not a zero divisor. ,15,The Division Algorithm,Theorem 6.2 Let R be a ring with identity and f, gRx0 s.t. the leading coeffi

10、cient of g is a unit in R. Then | q, rRx s.t. f = qg+r, deg r deg f, let q=0 and r=f. If deg g deg f, then , , with an0, bm0, mn, and bm a unit in R. Proceed by induction on n=deg f. If n=0, then m=0, f=a0, g=b0 and b0 is a unit. Let q=a0b0-1 and r=0; then deg rdeg g and qg+r = (a0b0-1)b0 = a0 = f.,

11、17,Proof (Cont.),Assume that the existence part of the theorem is true for polynomials of degree less than n=deg f. A straightforward calculation shows that the polynomial (ambm-1xn-m)g has degree n and leading coefficient an. Hence (anbm-1xn-m) is a polynomial of degree less than n and leading coef

12、ficient an.,18,Proof (Cont.),Hence, is a polynomial of degree less than n. By the induction hypothesis there are polynomials q and r s.t.,19,Proof (Cont.),Therefore, if q=anbm-1xn-m+q, then (Uniqueness) Suppose f=q1g+r1, and f=q2g+r2 with deg r1deg g and deg r2deg g. Then q1g+r1=q2g+r2 implies (q1-q

13、2)g= deg(r1-r2). Since the leading coefficient of g is a unit, Thm 6.1 implies deg(q1-q2)+deg g= deg(q1-q2)g=deg(r1-r2).,20,Proof (Cont.),Since deg(r1-r2)max(deg r2,deg r1)1, whence f(c)=0. Conversely, if f(c)=0, then m1 (Thm 6.6). If m=1, then f(x)=g(x)+(x-c)g(x). Consequently, if f(c)=0, then 0=f(

14、c)=g(c) by Cor 5.6, which is a contradiction. Therefore, m1.,44,Proof (Cont.),(ii) By Cor 6.4 and Thm 3.11 kf+hf=1D for some k, hDx Ex. If c is a multiple root of f, then by Cor 5.6 and (i) 1D=k(c)f(c)+h(c)f(c)=0, which is a contradiction. Hence c is simple root. (iii) If f is irreducible and f0, th

15、en f and f are relatively prime since deg fdeg f. Therefore, f has no multiple roots in E by (ii). Conversely, suppose f has no multiple roots in E and b is a root of f in E. If f=0, then b is a multiple root by (i), which is a contradiction. Hence f0. ,45,More General Question,To determine the unit

16、s and irreducible elements in Dx, where D is an integral domain.,46,The Units in Dx,The units in Dx are precisely the constant polynomials that are units in D see the proof of Cor 6.4.,47,Irreducible Elements in Dx,If c is irreducible in D, then the constant polynomial c is irreducible in Dx. Every

17、first degree polynomial whose leading coefficient is a unit in D is irreducible in Dx. In particular, every first degree polynomial over a field is irreducible. Suppose D is a subring of an integral domain E and fDx Ex. Then f may be irreducible in Ex but not in Dx and vice versa.,48,Example,2x+2 is

18、 irreducible in Qx. However, 2x+2=2(x+1) and neither 2 nor x+1 is a unit in Zx, whence 2x+2 is reducible in Zx. X2+1 is irreducible over the real field, but factors over the complex field as (x+i)(x-i). Since x+i and x-i are not units in Cx, x2+1 is reducible in Cx.,49,Our Goal,To obtain what few ge

19、neral results there are in this area. The rest of the discussion will be restricted to polynomials over a unique factorization domain D. We shall prove that Dx1,xn is also a unique factorization domain. The proof requires some preliminaries, which will also provide a criterion for irreducibility in

20、Dx.,50,Content of Polynomials,Let D be a unique factorization domain and f=0inaixi a nonzero polynomial in Dx. A greatest common divisor of the coefficients is called a content of f and is denoted C(f). Note. The notation C(f) is ambiguous since greatest common divisors are not unique. But any two c

21、ontents of f are associates and any associate of a content of f is also a content of f.,51,An Equivalence Relation,Write bc whenever b and c are associates in D. is an equivalence relation on D. Since D is an integer domain, bc iff b=cu for some unit uD by Thm 3.2. If aD and fDx, then C(af)=aC(f) (E

22、x 4).,52,Primitive Polynomials,If fDx and C(f) is a unit in D, then f is said to be primitive. Clearly for any polynomial gDx, g=C(g)g1 with g1 primitive.,53,Gauss Lemma,Lemma 6.11 (Gauss) If D is a unique factorization domain and f, gDx, then C(fg)=C(f)C(g). In particular, the product of primitive

23、polynomials is primitive.,54,Proof,f=C(f)f1 and g=C(g)g1 with f1, g1 primitive. Consequently, C(fg)=C(C(f)f1C(g)g1)=C(f)C(g)C(f1g1). Hence it suffices to prove that f1g1 is primitive (that is, C(f1g1) is a unit). If f1=0inaixi and g1=0jnbjxj, then f1g1=0km+nckxk with ck=i+j=kaibj.,55,Proof (Cont.),I

24、f f1g1 is not primitive, then there exists an irreducible element p in R s.t. p|ck for all k. Since C(f1) is a unit and p can not divide C(f1), whence there is a least integer s s.t. p|ai for is and p can not divide as.,56,Proof (Cont.),Similarly there is a least integer t s.t. p|bj for jt and p can

25、 not divide bt. Since p divides cs+t=a0bs+t+asbt+ as+1bt-1+as+tb0, p must divides asbt. Since every irreducible element in D is prime, p|as or p|bt. This is a contradiction. Therefore f1g1 is primitive. ,57,Property of Associates,Lemma 6.12 Let D be a unique factorization domain with quotient field

26、F and let f and g be primitive polynomials in Dx. Then f and g are associates in Dx iff they are associates in Fx.,58,Sketch of Proof,If f and g are associates in the integral domain Fx, then f=gu for some unit u Fx (Thm 3.2(vi). By Cor 6.4 uF, whence u=b/c with b, cD and c0. Therefore, cf=bg. Since

27、 C(f) and C(g) are units in D, c=cC(f)=C(cf)=C(bg)=bC(g)=b.,59,Sketch of Proof (Cont.),Therefore, b=cv for some unit vD and cf=bg=vcg. Consequently, f=vg (Since c0), whence f and g are associates in Dx. The converse is trivial. ,60,Property of Irreducible Polynomials,Lemma 6.13 Let D be a unique fac

28、torization domain with quotient field F and f a primitive polynomial of positive degree in Dx. Then f is irreducible in Dx iff f is irreducible in Fx.,61,Sketch of Proof,Suppose f is irreducible in Dx and f =gh with g, hFx and deg g1, deg h1. Then g=0in(ai/bi)xi and h=0jm(cj/dj)xj with ai, bi, cj, d

29、jD and bi0, dj0. Let b=b0b1bn and for each i let b*i=b0bi-1bi+1bn. If g1= 0in(aib*i)xiDx, then g1=ag2 with a=C(g1), g2Dx and g2 primitive.,62,Sketch of Proof (Cont.),Verify that g=(1D/b)g1=(a/b)g2 and deg g= deg g2. Similarly h=(c/d)h2 with c, dD, h2Dx, h2 primitive and deg h=deg h2. Consequently, f

30、=gh=(a/b)(c/d)g2h2, whence . bdf=acg2h2. Since f is primitive by hypothesis and g2h2 is primitive by Lem 6.11, bd=bdC(f)=C(bdf)=C(acg2h2)=acC(g2h2)=ac.,63,Sketch of Proof (Cont.),As in the proof of Lem 6.12, bd and ac associates in D imply that f and g2h2 are associates in Dx. Consequently, f is red

31、ucible in Dx, which is a contradiction. Therefore f is irreducible in Fx.,64,Sketch of Proof (Cont.),Conversely if f is irreducible in Fx and f=gh with g, hDx, then one of g, h (say g) is a constant by Cor 6.4. Thus C(f)=gC(f). Since f is primitive, g must be a unit in D and hence in Dx. Therefore,

32、f is irreducible in Dx. ,65,Property of Dx1,xn,Theorem 6.14 If D is a unique factorization domain, then so is the polynomial ring Dx1,xn.,66,Remark,Since a field F is trivially a unique factorization domain, Fx1,xn is a unique factorization domain.,67,Sketch of Proof of 6.14,Only prove Dx is a uniqu

33、e factorization domain. Since Dx1,xn=Dx1,xn-1xn by Cor 5.7, a routine inductive argument then completes the proof. If fDx has positive degree, then f=C(f)f1 with f1 a primitive polynomial in Dx of positive degree. Since D is a unique factorization domain, either C(f) is a unit or C(f) =c1cm with eac

34、h ci irreducible in D and hence in Dx.,68,Sketch of Proof of 6.14 (Cont.),Let F be the quotient field of D. Since Fx is a unique factorization domain (Cor 6.4) which contains Dx, f1=p*1p*n with each p*i an irreducible polynomial in Fx. The proof of Lem 6.13 shows that for each i, p*i=(ai/bi)pi with

35、ai, biD, bi0, ai/biF, piDx and pi primitive.,69,Sketch of Proof of 6.14 (Cont.),Clearly each pi is irreducible in Fx, whence each pi is irreducible in Dx by Lem 6.13. If a=a1an and b=b1bn, then f1=(a/b)p1pn. Consequently, bf1=ap1pn. Since f1 and p1pn are primitive (Lem 6.11), it follows (as in the p

36、roof of Lem 6.12) that a and b are associates in D. Thus a/b=u with u a unit in D.,70,Sketch of Proof of 6.14 (Cont.),Therefore, if C(f) is a nonunit, f=C(f)f1=c1cm(up1)p2pn with each ci, pi, and up1 irreducible in Dx. Similarly, if C(f) is a unit, f is a product of irreducible elements in Dx. (Uniq

37、ueness) Suppose f is a non-primitive polynomial in Dx of positive degree.,71,Sketch of Proof of 6.14 (Cont.),Verify that any factorization of f as a product of irreducible elements may be written f=c1cmp1pn with each ci irreducible in D, C(f)=c1cm and each pi irreducible (and hence primitive) in Dx

38、of positive degree. Suppose f=d1drq1qs with each dj irreducible in D, C(f)=d1dr and each qj irreducible primitive in Dx of positive degree. Then c1cn and d1dr are associates in D.,72,Sketch of Proof of 6.14 (Cont.),Unique factorization in D implies that n=r, and (after reindexing) each ci is an asso

39、ciate of di. Consequently, p1pn and q1qs are associates in Dx and hence in Fx. Since each pi resp. qj is irreducible in Fx by Lem 6.13, unique factorization in Fx (Cor 6.4) implies that n=s and (after reindexing ) each pi is an associate of qi in Fx.,73,Sketch of Proof of 6.14 (Cont.),By Lem 6.12 ea

40、ch pi is an associate of qi in Dx. ,74,Eisenstein Criterion,Theorem 6.15 (Eisenstein Criterion) Let D be a unique factorization domain with quotient field F. If f=0inaixi, deg f 1 and p is an irreducible element of D s.t. p can not divide an, p|ai for i= 0,1,n-1; p2 can not divide a0, then f is irre

41、ducible in Fx. If f is primitive, then f is irreducible in Dx.,75,Proof,f=C(f)f1 with f1 primitive in Dx and C(f)D; (in particular f1=f if f is primitive). Since C(f) is a unit in F (Cor 6.4), it suffices to show that f1 is irreducible in Fx. By Lem 6.13 we need only prove that f1 is irreducible in Dx. Suppose on the contrary that f1=gh with g=brxr+b0Dx, deg g=r1; and h=csxs+c0Dx, deg h=s1.,76,Proof (Cont.),Now p doe