Class Note for MATH 790 at KU
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Date Created: 02/06/15
Polynomial Rings Linear Algebra Notes Satya Mandal September 277 2005 1 Section 1 Basics De nition 11 A nonempty set R is said to be a ring if the follow ing are satis ed 1 R has two binary operations called addition and multipli cation 2 R has an abelian group structure with respect to addition 5 The additive identity is called zero and denoted by U 4 Distributiuity For 7112 6 R we have ay2 aya2 and y 2k ya 2a 5 We assume that there is a multiplicative identity denoted by 10 Note the multiplication need not be commutative So7 it is possible that my a ya Also note that not all non zero elements have an inverse For example Let R MmlF be the set of all n X n matrices n 2 2 Then R is a ring but multiplication is not commutative Following are few more de nitions De nition 12 Let R be a ring 1 We say R is commutative if my ya for all my 6 R 2 A commutative ring R is said to be an integral domain if xy0gta0 or y0 3 Let A be another ring A map f R a A is said to be a ring homomorphism if for all my 6 R we have f y f fy7fxy ffy and f11 4 For a ring R7 an Rialgebra is a ring A with together with a ring homomorphism f R a A Remark 11 Let l be a eld and A be an Fialgebra The textbook calls such an algebra as Linear Algebra Note that A has a natural vector space structure Exercise 11 Let l be a eld and f F a A be ring homomor phism Then f is 1 1 This means that ifA is an Fialgebm then lF Q A Proof It is enough to show that if f 0 then m 0 Are you sure that it is enough Assume m a 0 and f 0 We have f11So71 fx 1 ff 1So7f7 0 2 Polynomials We do not look at polynomials as functions Polynomilas are formal expressions and in algebra they are manipulated formally De nition 21 Let l be a eld and N 07 1727 be the set of non negative integers 1 Let f denote the set of all functions f N a F So7 f a07a17a27 12 E F is the set of all in nite sequences in F De ne addition and multiplication on f naturally see the book f is called the power series ring 2 Let X 07170707 6 f Then any element f a2 6 f can be written as f i aiXi 2 0 with apprpriate meaning of in nite sum attached 3 Notation Usual notation for the power series ring is EX f Elements in EX are called power series over F 4 Let EX f E EX f a0a1XanX7L7a E F Note that EX is a subring of We say that EX is the polynomial ring over F 5 Importantly two polynomials f7 9 are equal if and only if coe icients of Xi are same for both f and 9 Theorem 21 Let EX be the polynomial ring over over a eld F 1 Suppose f79791792 E EX and f is non zero Then fg0 g0 and f91f92 9192 2 f E EX has an inverse in EX if and only if f is a nonzero scalar 3 Section 4 Division and Ideals Theorem 31 Division Algorithm Let F is a eld and EX be a polynomial ring over F Let d a 0 be a polynomial and degD n Then for any f E EX7 there are polynomials q7r E EX such that fqdr r0 degrltn In fact q7r are UNIQUE for a given f 3 Proof Write a proof Corollary 31 Let l is a eld and EX be a polynomial ping over F Let f be a nonzero polynomial and c E F Then fc 0 if and only if X 7 c divides f in EX Further a polynomial f with degf n has atmost n roots in lF Proof Obvious By division algorithm7 we have f X 7 CQ R Where R762 6 EX and either R 0 or degR 0 We have 0 fc R0 R Therefore f X 7 CQ For the proof of the last assertion7 use induction on n 31 GCD De nition 31 Let l be a led and EX be the polynomial ring Let f177f7 E EX be polynomials7 not all zero An element d E EX is said to be a Greatest common divisor gcd if 1 dm v ll7r7 2 If there is an elment d E EX such that am v i1r then d d Lemma 31 Let l be a led and EX be the polynomial ping Let f177f7 E EX be polynomials not all zero Suppose d1 and d2 are two GCDs off177fT Then d1 udz for some unit it E F Further if we assume that both d17d2 are monie then d1 d2 That means monie GOD off17fT E EX is UNIQUE Proof By property 2 of the de nition7 d1 udg and d2 vdl for some u7v E FlX Hence d1 uvdl Since d1 a 07 we have uv 12397 so u is an unit Now7 if d17d2 are monic then comparing the coe icients of the top degree terms in the equation d1 udg it follows that u 1 and hence d1 12 This completes the proof Remarks 1 Note that Z has only two unit7 1 and 1 When you computed GCD of integers7 de nition assumes that the GOD is positive That is why GCD of integers is unique De nition 32 Let R be a commutative ring A nonempty subset I of R is said to be an ideal of R if 1 7yEI yEI 2 ER7 yEI yEI Example 31 Let R be a commutative ring Let f17 7ft 6 R Let IceRcglf1yf for 961 Then I is an ideal of R This ideal is sometime denoted by f17 7 Also IRf1RfT Theorem 32 Let l be a led and lle be the polynomial ring Let I be a non zero ideal 0leX Then 1 Hde for some d E EX ln fact7 for any non zero d E I with degd least7 we have I FXd Proof Let k mlndegf f 6 I7 f a 0 Pick d E I such that d a 0 and degd k Question Why such a d exists Now clairn IEXM Clearly7 I Q EXd Now7 let f E I By division f qd r with r00rdegrltkNoterf7quIWeprover0 lfr7 07 then degr lt k would contradicts the rninirnality of k So7 r 0 and f qd E EXd This completes the proof Theorem 33 Let l be a led and EX be the polynomial ring Let f7 fT E EX be polynomials not all zero 1 Then f177fT has a GOD In fact a GOD d off17fT is given by dq1f1quot39qrfr for some qi E EX 2 Two GODs dz er by a unit multiple 5 A mom39c GOD is unique Proof Write 1FWUHHM By above thorern7 I EXd for some d E EX We calirn that d is a GCD of fh 7 First note7 dq1f1quot39qrfr for some qi E EX Since fi 6 I7 what have df Now let 1 E I be such that d f 7 for l 17 77 We need to prove that d d This follows from the above equation This completes that proof that GCD exist We have alrady seen 2 and 3 before 4 Prime Factorization De nition 41 Let EX be a the polynomial ring over a eld F 1 An element f E EX is said to be a an reducible over l if f gh for some non unit g h E EX equivalently7 degg gt 0 and degh gt 0 2 f E EX is said to be irredubible over l if it is not reducible 3 A non scalar irreducible element f E EX over l is called a prime in EX Lemma 41 Let R be an integral domain For non zero g 6 R7 Rf Ry if and only if f ug for some unit in R Proof Easy Lemma 42 Let R EX be the polynomial ring over a eld F Let p E R be a prime element and f E R Then Rf Rp R ltgt p does not divide f Proof i We prove by contradiction Assume that p f Then f dp for some d E R Hence RfRp deRp Rp a R So7 this part of the proof is complete Assume p does not divide f By Theorem 327 we have Rf Rp Rd for some d E R Therefore f ud and p ed for some u7v E R Claim d is a unit If not7 since p is prime7 U is an unit Hence f ud uv lp That means7 p f This Will be a contradiction Therefore the claim is proved and d is a unit Hence Rf Rp Rd R The proof is complete Theorem 41 Let R EX be the polynomial ring over a eld F Let p E R be a prime element and y 6 R Then pfg either pf or py Proof Assume p 1 f9 and p does not divide f We will prove that p l 9 We have f9 pm for some it E R Also by above lemma 42 Rf Rp R Therefore 1 pf yp for some 11 E R Hence 9 zfg yp mop yp This completes the proof Corollary 41 Let R 1le be the polynomial ring over a eld 1 Let p E R be a prime element and f1f2 f E R Then Plf1f2quot39fr gt Plfi forsomei1r Proof Use induction and the above thoerern Theorem 42 Unique Factorization Let R 1le be the poly nomial ring over a eld 1 Let f E R be a nonzero element Then fuP1P2quot39pk where u E 1 is a unit and p1 pk are monie prime elements In fact this factorizton is unique except for order Proof First we prove that factorization as above of f is possible Let degf n we will use induction on n Case n 0 If n 0 then f is an unit and we are done Case n 1 lnthis case f uXv withuv E Fandu a 0 Write p X vu The p is prime and f up Case n gt 1 If f is prime then write f uX an1X 1 a1X 10 with up E 1 and u a 0 Write p fu The p is rnonic prime and f up Now if f is not a prime then f gh with 1699 lt n and degh lt n By induction 9 and h have factorization as desired The product of these two factorizations will give a desired factorization of f So the proof of existance of the factorization is complete Now we will prove the uniqueness of the factorization Suppose fup1p2pkvq1q2qm where uv are units and p qj are monic primes Assume degf 71 By comparing coef cients of X we get u 1 Therefore we have 9P1P2quot39PkqiltI2quot39qm where g fu is monic Now p1 l q1q2qm By Corollary 41 p1 l qj for some j we may assume j 1 and p1 l ql Since both p1q1 are monic primes we have p1 ql Hence it follows p2pkq2qm Therefore by induction k m and pi q upto order This com pletes the proof
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