APP. B] ALGEBRAIC SYSTEMS 461
Supposepdividesfgbut notf. Sincepis irreducible, the polynomialsfandpmust then be relatively prime.
Thus there exist polynomialsm, n∈K[t]such thatmf+np=1. Multiplying this equation byg, we obtain
mf g+npg =g. Butpdividesfgand sopdividesmf g. Also,pdividesnpg. Therefore,pdivides the sum
g=mf g+npg.
Now supposepdividesf 1 f 2 ···fn.Ifpdividesf 1 , then we are through. If not, then by the above resultp
divides the productf 2 ···fn. By induction onn,pdivides one of the polynomials in the productf 2 ···fn. Thus the
lemma is proved.B.37.Prove Theorem B.24 (Unique Factorization Theorem): Letfbe a nonzero polynomial inK[t]. Thenfcan
be written uniquely (except for order) as a productf=kp 1 p 2 ···pnwherek∈Kand thep’s are monic
irreducible polynomials inK[t].
We prove the existence of such a product first. Iffis irreducible or iff∈K, then such a product clearly exists.
On the other hand, supposef=ghwheregandhare nonscalars. Thengandhhave degrees less than that off.By
induction, we can assumeg=k 1 g 1 g 2 ···grandh=k 2 h 1 h 2 ···hswherek 1 ,k 2 ∈Kand thegiandhjare monic
irreducible polynomials. Accordingly, our desired representation follows:
f=(k 1 k 2 )g 1 g 2 ···grh 1 h 2 ···hs
We next prove uniqueness (except for order) of such a product forf. Suppose
f=kp 1 p 2 ...pn=k′q 1 q 2 ...qm where k, k′∈K
and thep 1 ,...,pn,q 1 ,...,qmare monic irreducible polynomials. Nowp 1 dividesk′q 1 ...qm. Sincep 1 is irreducible
it must divide one of theq’s by Lemma B.23. Sayp 1 dividesq 1. Sincep 1 andq 1 are both irreducible and monic,
p 1 =q 1. Accordingly,kp 2 ...pn=k′q 2 ...qm. By induction, we have thatn=mandp 2 =q 2 ,...,pn=qmfor
some rearrangement of theq’s. We also have thatk=k′. Thus the theorem is proved.
B.38.Prove Theorem B.25: Supposef(t)is a polynomial over the real fieldR, and suppose the complex number
z=a+bi, b=0, is a root off(t). Then the complex conjugatez ̄=a−biis also a root off(t). Hence
the following is a factor off(t):
c(t)=(t−z)(t− ̄z)=t^2 − 2 at+a^2 +b^2
Dividingf(t)byc(t)where deg(c)=2, there existq(t)and real numbersMandNsuch that
f(t)=c(t)q(t)+Mt+N( 1 )
Sincez=a+biis a root off(t)andc(t), we have, by substitutingt=a+biin (1),
f(z)=c(z)q(z)+M(z)+n or 0= 0 q(z)+M(z)+N or M(a+bi)+N= 0
ThusMa+N=0 andMb=0. Sinceb=0, we must haveM=0. Then 0+N=0orN=0. Accordingly,
f(t)=c(t)q(t)andz ̄=a−biis a root off(t).SupplementaryProblems
OPERATIONS AND SEMIGROUPS
B.39. Consider the setNof positive integers, and let∗denote least common multiple (lcm) operation onN.
(a) Find 4∗6, 3∗5, 9∗18, 1∗6.
(b) Is (N,∗) a semigroup? Is it commutative?
(c) Find the identity element of∗.
(d) Which elements inN, if any, have inverses and what are they?
B.40. Let∗be the operation on the setRof real numbers defined bya∗b=a+b+ 2 ab.
(a) Find 2∗3, 3∗(−5), and 7∗(1/2). (b) Is (R,∗) a semigroup? Is it commutative?
(c) Find the identity element of∗. (d) Which elements have inverses and what are they?