Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2)
Kevin Broughan
University of Waikato, Hamilton, New Zealand
May 13, 2010
Remainder and Factor Theorem 15
Definition of factor
Iff(x) =g(x).h(x) inR[x] we sayg(x) is afactoroff(x) inR[x].
Examples
(1) The polynomialx+ 2 is a factor ofx2+ 5x+ 6 inZ[x].
(2) The polynomialx2+ 1 = (x+i)(x−i) inZ[i], but has no factors inZ[x].
(3) Polys likef(x) =x8+ 3x6−6x5+ 3x4−12x3+ 10x2−6x+ 9 are difficult to factor in any ring.
InZ[x] we havef(x) = x2+ 1
x3+x−32
. How is this done ?
Theorem 15
LetF be a field,a∈F an element andf(x)∈F[x] a polynomial. Then f(x) = (x−a)q(x) +f(a), i.e. f(a) is the remainder when we dividef(x) by x−a. Iff(a) = 0 then (x−a) is a factor off(x).
Proof of Theorem 15
By the division identity inF[x] there exist polynomialsq(x), r(x) such that f(x) = (x−a)q(x) +r(x) withr(x) = 0 or degr(x)<degx−a= 1.
Thereforer(x) =constant=r(a).
Butf(a) = (a−a)q(a) +r(a) =r(a) sof(a) =r(a) and we can write f(x) = (x−a)q(x) +f(a).
Iff(a) = 0 we havef(x) = (x−a)q(x) + 0 = (x−a)q(x) sox−ais a factor off(x).
Theorem 16: A poly over a field of degree n has at most n zeros in the field
Proof
The polynomial must be non-zero. Ifnis zero the polynomial has no zeros.
Assume every polynomial of degree up to and includingnhas at mostnzeros and letf(x) be a polynomial of degreen+ 1.
Iff(x) has no zeros we are done. Ifais a zero off(x), by Theorem 15 we can writef(x) = (x−a)kg(x), by Theorem 15, wherek≥1 andg(a)6= 0, degf(x) =k+ degg(x), so the degree ofg(x) is≤n.
By the inductive hypothesis it has at mostnzeros,a1, . . . ,amsay withm≤n.
Every zero ofg(x) is a zero off(x). Ifbis a zero off(x) other thanathen 0 =f(b) = (b−a)kg(b) sobis a zero ofg(x).
Hence the zeros off(x) are{b,a1, . . .am}so are in number less than or equal ton+ 1, completing the proof by induction.
Notes
This is not true in general for polynomials over a ring: x2+x has 4 zeros in Z/6Z.
IfF ⊂K andK is also a field, a so-calledextension field, then we have actually shownf(x) has at most degf(x) zeros inK.
Examples
(1)x2+ 1 has no zeros inQbut a full set inQ(i) and inC. (2)x2−2 has no zeros inQbut a full set inQ(√
2).
Definition of a principal ideal domain
Aprincipal ideal domainor PID, is an integral domainR where every ideal has the formhaifor somea∈R.
Examples:
(1)Z: IfA⊂Zis an ideal leta>0 be the smallest positive element ofA.
ThenA=hai.
(2) IfF is a field thenF[x]isa principal ideal domain. This is Theorem 17 proved below.
(3) The ringZ[√
5]isa principal ideal domain, butZ[√
−5] isnot. So being a PID is a big issue and quite subtle.
Proof of Theorem 17
LetA⊂F[x] be an ideal. IfA={0}we haveA=h0i.
IfA6={0}letg(x) be a polynomial inAof minimum degree. We can assume g(x) is monic. Then we claimA=hg(x)i.
To see this letf(x)∈A. Then, by the division identity there are polynomials q(x), r(x) such thatf(x) =q(x).g(x) +r(x) where eitherr(x) = 0 or degr(x)<degg(x).
Ifr(x) = 0 thenf(x) =q(x).g(x)∈ hg(x)i.
Otherwiser(x) =f(x)−q(x).g(x)∈A. But degr(x)<degg(x) so this case is impossible. ThereforeA=hg(x)i.
Factoring polynomials
Definitions
IfF is a field we say a polynomialf(x)∈F[x] isirreducible overF if it cannot be expressed as the product of two polynomials overF with strictly lower degrees than that off(x).
IfRis an integral domain we say a polynomialf(x)∈R[x] isirreducible overR if whenever we writef(x) =g(x).h(x) we must have eitherg(x) orh(x) a unit inR[x].
Examples
(1)x+ 1 is irreducible overQ[x] andZ[x].
(2) A unit inR[x] is a constant polynomialf(x) =u whereuis a unit inR.
(3) 3x+ 6 is irreducible overQ[x] but factors non-trivially as 3.(x+ 2) inZ[x].
(4)x2+ 1 is irreducible inR[x] but factors as (x+i)(x−i) inC[x].
(5)x2+x+ 1 is irreducible inQ[x] but factors as (x+ 2)(x+ 2) in (Z/3Z)[x].
The quadratic and cubic formulas
Quadratic
Iff(x) =ax2+bx+c,a,b,c ∈Canda6= 0 then inC[x] f(x) =a(x−α)(x−β) whereα, β=−b±
√
b2−4ac 2a
Cubic
Iff(x) =x3+ax+bthen f(x) = (x−α)(x−β)(x−γ) where
α =
p√3
3√
4a3+ 27b2−9b
√3
232/3 −
q3 2 3a p√3
3√
4a3+ 27b2−9b
β =
1−i√
3
a 22/3√3
3p√3 3√
4a3+ 27b2−9b
−
1 +i√ 3
p√3 3√
4a3+ 27b2−9b 2√3
232/3
γ =
1 +i√
3 a 22/3√3
3p√3 3√
4a3+ 27b2−9b
−
1−i√ 3p√3
3√
4a3+ 27b2−9b 2√3
232/3
Factors for low degree polynomials
Degree 1
If a non-zero polynomial over a field has degree 0 it is irreducible. For degree 1, f(x) =ax+b, thenf(−b/a) = 0 andf(x) is irreducible.
Degree 2 or 3
Iff(x)∈F[x] has a zero or root,f(a) = 0 thenf(x) is reducible since f(x) = (x−a)g(x). Iff(x) has degree 2 or 3 then it is reducible if and only if it has a factor of degree 1 if and only if it has a zero/root.
Degree 4 or more
OverQ[x] (x2+ 1)(x2+ 2) =x4+ 3x2+ 2 so the right hand side is reducible but has no root inQ.
Polynomials over finite fields
Finding roots is easy, just test each element of the field and see if the polynomial vanishes, e.g.f(x) =x4+ 3x+ 1 overZ/5Z hasf(1) = 0 so x−1 =x+ 4 divides exactly with quotient 4 +x+x2+x3. This does not vanish atx = 0,1,2,3,4 modulo 5, therefore it is irreducible (overZ/5Z).
Gauss’ Lemma
Definition
Letf(x) =anxn+· · ·+a0∈Z[x]. Thecontentoff(x), denoted c(f(x)), is the gcd of the coefficients, i.e. gcd(a0, . . . ,an). Ifc(f(x)) = 1 we sayf(x) is primitive.
Example
(1)f(x) = 2x2−4x+ 12 =⇒ c(f(x)) = 2 f(x) = 2x3−3x+ 6 =⇒ c(f(x)) = 1.
Statement of the Lemma
Letf(x), g(x)∈Z[x] be primitive. Thenf(x).g(x) is also primitive, i.e. the product of two primitive polynomials is primitive.
Reduction Modulop
A useful operation on polynomials with integer coefficients is reduction modulo pwherepis a fixed prime. This is a homomorphism of polynomial rings:
f(x)∈Z[x]→θ(f(x))∈Zp[x] : anxn+· · ·a0→[an]xn+· · ·+ [a0].
Proof of Gauss’ Lemma
Letf(x), g(x) be primitive and supposef(x).g(x) is not primitive. Let the primepdivide the content, i.e. p divides every coefficient off(x).g(x).
Then inZp[x],θ(f(x).g(x)) = 0. But this is an integral domain (becauseZpis an integral domain) and, becauseθis a homomorphism, 0 =θ(f(x)).θ(g(x)).
Thereforeθ(f(x)) = 0 orθ(g(x)) = 0. Suppose the former is true. But this meansp divides each of the coefficients off(x).
But this is impossible, sincef(x) is primitive.
Factoring over Q equals factoring over Z
Example
6x4+ 8x3+ 3x2+ 7x+ 4 = (2x+83)(3x3+32x+32) = (3x+ 4)(2x3+x+ 1) Content fact
Iff(x)∈Z[x] then we can writef(x) =c(f(x)).f1(x) wherec(f1(x)) = 1.
Example
f(x) = 12x4−6x3+ 3x+ 18 = 3(4x4−2x3+x+ 6) =c(f(x)).f1(x) where f1(x) has content 1, i.e. is primitive.
Theorem 18
Statement
Iff(x)∈Z[x] andf(x) =g(x).h(x) factors inQ(x) thenf(x) =g1(x).h1(x) in Z[x] and the degrees of degg1(x) = degg(x), degh1(x) = degh(x).
Proof
Makef(x) primitive by dividing bothf(x) andg(x) byc(f(x)).
Letabe the least common multiple of the denominators of the coefficients of g(x) andb the LCM of the denominators ofh(x) soag(x)∈Z[x] and bh(x)∈Z[x].
Letag(x) =c(ag(x)).g1(x) andbh(x) =c(bh(x)).h1(x), sog1(x), h1(x) are primitive and inZ[x].
Then, sincef(x) is primitive,c(abf(x) =ab. Butabf(x) =ag(x)bh(x) = c(ag(x).g1(x)c(bh(x)).h1(x) =c(ag(x))c(bh(x)g1(x).h1(x).
Since, by Gauss,g1(x).h1(x) is primitive the content of the RHS is c(ag(x)).c(bh(x)) which must be the content of the LHS, so they cancel leading tof(x) =g1(x).h1(x) the factorization overZ[x].
