Modern Algebra Lecture Notes: Rings and fields set 3
Kevin Broughan
University of Waikato, Hamilton, New Zealand
May 6, 2010
Polynomial Rings
Definition
LetRbe a commutative ring andx a “symbol” or “indeterminate”. The by R[x] we mean the set of formal sums
anxn+an−1xn−1+· · ·+a1x1+a0
where theai ∈R. We abbreviate these expressions byf(x),g(x)etc and call thempolynomials inx with coefficients inR.
Operations inR[x]
We makeR[x] into a ring by defining addition and multiplication as thoughx was a real variable and the expressionsf(x), g(x) were ordinary polynomial functions ofx:
f(x) := anxn+an−1xn−1+· · ·+a1x1+a0, g(x) := bnxn+bn−1xn−1+· · ·+b1x1+b0, f(x) +g(x) :=
n
X
j=0
(aj+bj)xj,
f(x).g(x) :=
n
X
j=0
cjxj wherecj=
j
X
i=0
aibj−i.
Examples
Notes
(1) These expressionsf(x) are not functions, but become so ifx is given values in a ring. Indeed , ifa∈R then the mapf →f(a) is a homomorphism R[x]→R.
(2) Theai orbi could be zero and that these operations emulate ordinary polynomial arithmetic.
(1) InZ[x],f(x) =x+ 1, g(x) =x2+ 3 =⇒ f(x) +g(x) = (0 + 1)x2+ (1 + 0)x+ (1 + 3) =x2+x+ 4.
(2) InZ[x],
f(x) =x+1,g(x) =x2+3 =⇒ f(x).g(x) = (x+1)(x2+3) =x3+x2+3x+3.
(3) We writecj=Pj
i=0aibj−i =a0bj+a1bj−1+· · ·+ajb0, theCauchy product. So in Example (2) the coefficient ofx is
c1=a0b1+a1b0= 1.0 + 1.3 = 3.
R[x] is a commutative ring
There are all of the axioms to check. The zero polynomialf(x) =a0= 0 is the additive identity and ifR has an identity 1, the identity ofR[x] isf(x) = 1, both constant polynomials.
R[x] is commutative because firstly, inf(x).g(x),
cj=
j
X
i=0
aibj−i =
j
X
i=0
bj−iai
using the commutativity ofR. Then re-index withk=j−i soi=j−kmaking
cj=
j
X
k=0
bkaj−k=
j
X
i=0
biaj−i
and this is just thej’th coefficient forg(x).f(x).
Thatf(x) +g(x) =g(x) +f(x), f(x) + 0 =f(x), 1.f(x) =f(x) are easy, but good practice for becoming familiar with the notation.
Leading coefficient and degree
Definitions
The termsan are calledcoefficientsand the coefficient which is non-zero with the highest value ofnis called theleading coefficient, with the value ofnbeing thedegreeof the polynomial. The zero polynomial has, by decree, no degree, and a constant non-zero polynomial degree 0. If the leading coefficient is 1, the polynomial is calledmonic.
Examples
(1)f(x) = 0x5+ 3x4+x3+x2−2x−3 inZ[x] has leading coefficient 3, degree 4 and is not monic.
(2)g(x) = 1 has degree zero,g(x)−g(x) has no degree.
(3) IfR=Z/4Z,f(x) = 8x5+ 5x4+ 1 has degree 4 and is monic.
(4) IfR=Q[y] andf(x)∈R[x] is defined as
f(x) = (y2+ 1/2)x3−yx+ (y4−3/4), thenf(x) has degree 3 and leading coefficienty2+ 1/2 so is not monic.
Theorem 13
IfR is an integral domain so isR[x]
Proof
We have already seen thatR[x] is a commutative ring with 1. So letam, bnbe the leading (non-zero) coefficients for the non-zero polynomialsf(x), g(x):
f(x) := amxm+am−1xm−1+· · ·+a1x1+a0, g(x) := bnxn+bn−1xn−1+· · ·+b1x1+b0.
Then the coefficient of the highest possible power ofx inf(x).g(x) is
cm+n=
m+n
X
i=0
ai.bm+n−i =am.bn,
which, sinceRis an integral domain, is non-zero. (Ifi <m, m+n−i>nso bm+n−i = 0. Ifi >m,ai = 0.) ThereforeR[x] is an integral domain.
Corollary
degf(x).g(x) =m+n= degf(x) + degg(x).
Theorem 14 The division identity
IfF[x] is the ring of polynomials with coefficients in a fieldF and
f(x), g(x)∈F[x] withg(x)6= 0 then there areunique polynomialsq(x), r(x) inF[x] with
f(x) =q(x).g(x) +r(x), r(x) = 0 or degr(x)<degg(x).
Note
This identity is vital for any treatment of polynomials in Modern Algebra. It represents the familiar division of one polynomial by another giving a remainder of lower degree than the divisor, when the division is not “exact”. The “exact”
case is wheng(x) “divides”f(x).
Example
Dividex3−x2+ 2x−3 byx−2:
x2 + x + 4
x−2 x3 − x2 + 2x − 3 x3 − 2x2
x2 + 2x − 3 x2 − 2x
4x − 3
4x − 8
5
Hence,x3−x2+ 2x−3 = (x−2)(x2+x+ 4) + 5.
Proof of Theorem 14
We will first consider the existence ofq(x) andr(x). Let S={f(x)−g(x)h(x) :h(x)∈F[x]}and assume that
g(x) =a0+a1x+· · ·+anxn
is a polynomial of degreen. This set is nonempty sincef(x)∈S.
Iff(x) is the zero polynomial, then
0 =f(x) = 0·g(x) + 0;
hence, bothqandrmay be chosen as the zero polynomial.
Now suppose that the zero polynomial is not inS. In this case the degree of every polynomial inSis nonnegative. Choose a polynomialr(x) of smallest degree inS; hence, there must exist aq(x)∈F[x] such that
r(x) =f(x)−g(x)q(x), or
f(x) =g(x)q(x) +r(x).
We need to show that the degree ofr(x) is less than the degree ofg(x).
Assume that degg(x)≤degr(x). Sayr(x) =b0+b1x+· · ·+bmxm and m≥n. Then
f(x)−g(x)[q(x) + (bm/an)xm−n] = f(x)−g(x)q(x)
−(bm/an)xm−ng(x)
= r(x)−(bm/an)xm−ng(x)
= r(x)−bmxm
+ terms of lower degree is inS.
This is a polynomial of lower degree thanr(x), which contradicts the fact that r(x) is a polynomial of smallest degree inS; hence, degr(x)<degg(x).
To show thatq(x) andr(x) are unique, suppose that there exist two other polynomialsq0(x) andr0(x) such thatf(x) =g(x)q0(x) +r0(x) and degr0(x)<degg(x) orr0(x) = 0.
Assumeq(x)6=q0(x). Then
f(x) =g(x)q(x) +r(x) =g(x)q0(x) +r0(x), and
g(x)[q(x)−q0(x)] =r0(x)−r(x).
Ifg is not the zero polynomial, then
degg(x)≤deg(g(x)[q(x)−q0(x)]) = deg(r0(x)−r(x)).
However, the degrees of bothr(x) andr0(x) are strictly less than the degree of g(x); therefore,q(x) =q0(x) and sor(x) =r0(x) also.
