Modern Algebra Lecture Notes: Rings and fields set 5 (Revision 4)
Kevin Broughan
University of Waikato, Hamilton, New Zealand
May 17, 2010
Theorem 19: the Mod p irreducibility test
Statement
Letpbe a prime andf(x)∈Z[x] is such thatθp(f(x)) has the same degree and is irreducible inZp[x]. Thenf(x) is irreducible overZ.
Notes
(1) There is a good fast method for factoring a polynomial over anyZp: Berlekamp’s algorithm.
(2) This can be used as an irreducibility test of general utility: try factoring for a number of primes until there is only one factor.
(3) Berlekamp can be used to factor overZeither by choosing a very large prime, or by “lifting” a factorization modp to higher powers ofp, until the integer coefficients “appear”.
Examples
(1) InZ[x] letf(x) = x2+ 1
x3+x+ 32
. Then this same factorization works modulopfor every primep>3 but overZ3,f(x) =x2(x2+ 1)3. (2)x4+ 2 is irreducible overZandZ5, but overZ7we have
f(x) = (x2+x+ 4)(x2+ 6x+ 4) and overZ3,f(x) = (x+ 1)(x+ 2)(x2+ 1), where each of the factors is irreducible.
(3) 2 +x3+ 2x7+x10is reducible but changingx3 to 2x3makes it irreducible.
Statement of Theorem 19:
Letpbe a prime andf(x)∈Z[x] is such thatθp(f(x)) has the same degree and is irreducible inZp[x]. Thenf(x) is irreducible overZ.
Proof
Assume, to get a contradiction,f(x) =g(x).h(x) overZwhere each ofg(x) andh(x) has degree less than that off(x) and greater than 0.
Thenθp(f(x)) =θp(g(x)).θp(h(x)). Then degθp(g(x)) = degg(x) and degθp(g(x)) = degg(x), giving a non-trivial factorization ofθp(f(x)), which is impossible.
Thereforef(x) is irreducible.
Theorem 20: Eisenstein’s irreducibility test
Statement
Letf(x) =a0+a1x+· · ·+anxnhave positive degree and suppose some prime p|a0,p|a1, . . . ,p|an−1butp-an andp2-a0. Thenf(x) is irreducible over Z.
Proof
To get a contradiction letf(x) =g(x).h(x) withg(x) =b0+· · ·+bmxmand h(x) =c0+· · ·+clxl. Thena0=b0c0andan=bmcl.
Sincep|a0we havep|b0orp|c0. Sincep2-a0pdivides one of these and not the other. Supposep|b0andp-c0.
Sincep-an we havep-bmso we may assumep|b0, b1, b2, . . . ,bt−1but p-bt. Butat =btc0+bt−1c1+· · ·+b0ct and we are givenp|at.
Thereforep|btc0sop|bt, which is false. Thereforef(x) is irreducible.
Example
12 + 4x+ 21x2−6x3+ 3x4+x7is irreducible overZsince we can takep= 3.
Theorem 21: f (x) irreducible implies F [x]/hf (x)i is a field
Proof
LetAbe the idealhf(x)iandB and ideal withA⊂B⊂F[x].
SinceF[x] is a principal ideal domain we must haveB=hg(x)ifor some polynomialg(x)∈F[x]. Butf(x)∈B so there is a polynomialh(x) such that f(x) =h(x)g(x).
But we are givenf(x) is irreducible, and we may assume non-zero. Hence one ofh(x) org(x) must be a non-zero constant polynomial (a unit ofF[x]). If both are constant,B=F[x]
If both are constant,B=F[x]. If only one is constant,B=A. ThereforeAis a maximal ideal.
HenceF[x]/A=F[x]/hf(x)i, by Theorem 5 (set 1), is a field.
Example:Find the inverse of [x+ 2] inQ[x]/hx2+ 1i: leta,b∈Qbe such that [x+ 2][ax+b] = [1]. ThenLHS= [ax2+x(2a+b) + 2b] or
[a(x2+ 1)−a+x(2a+b) + 2b] which is [x(2a+b) + 2b−a] so we want 2a+b= 0, 2b−a= 1 ora=−1/5, b= 2/5, giving the inverse [(−x+ 2)/5].
Theorem 22: irreducible polynomials over a field are like prime numbers
Statement
LetF be a field and letf(x) inF[x] be irreducible. Then iff(x)|g(x).h(x) in F[x] we must havef(x)|g(x) orf(x)|h(x).
Proof
Sincef(x) is irreducible,F[x]/hf(x)iis a field, hence an integral domain.
Sincef(x)|g(x).h(x) we haveg(x).h(x)∈ hf(x)i. Thus [g(x)][h(x)] = [g(x).h(x)] = [0] inF[x]/hf(x)i.
Hence [g(x)] = [0] or [h(x)] = [0]. But this meansg(x)∈ hf(x)ior h(x)∈ hf(x)i.
In other wordsf(x)|g(x) orf(x)|h(x).
Theorem 23: Unique factorization holds in Z[x]
Statement
Iff(x)∈Z[x] has positive degree then it can be factored, up to order, uniquely into the product of an integer and a set of polynomials of positive degree, irreducible overZ.
Proof
The integer in the statement is the content off(x) so first take it out as a factor,c(f(x)).f1(x) wheref1(x) is primitive and has the same degree asf(x).
Letn= degf1(x). Ifn= 1 the polynomial is irreducible and we are done. So assume the result is true for every primitive polynomial of degree up to some given natural numbern>1.
Iff1(x) is irreducible we are also done. So assumef1(x) has degreenand f1(x) =g(x).h(x) is a non-trivial factorization. Of courseg(x), h(x) must be primitive, by Gauss’ Lemma. Since their degrees are less thannthey will have a factorization into irreducible polynomials by the inductive assumption.
Combining these factorizations gives a factorization off1(x) into irreducibles and hence off(x).
Theorem 23: Uniqueness of the factors
The leading integer is unique because it is the content off(x), so we can assumef(x) is primitive.
Suppose it factors into irreducibles asf(x) =g1(x).g2(x)· · ·gn(x) and f(x) =h1(x)· · ·hm(x).
Then eachhj(x)|f(x) which implies, using Theorem 22 many times, for some i,hj(x)|gi(x).
But this meanshj(x).k(x) =gi(x) for somek(x)∈Z[x] andgi(x) is irreducible. Thusk(x) =±1, a unit inZ.
So for eachj there is ani withhj(x) =±gi(x). Similarly for eachi there is a correspondingj. If each of thegi(x)’s andhj(x) are distinct respectively, renumber to make thei0s andj0s identical som=nand
f(x) =g1(x)· · ·gn(x) =±h1(x)· · ·hn(x), gi(x) =±hi(x), 1≤i≤m.
Uniqueness of the factors continued
If some factors have multiplicity we can write
f(x) =g1(x)α1· · ·gm(x)αm=±g1(x)β1· · ·gm(x)βm for someαi, βi ∈N, and where thegi(x) are now all distinct.
But ifα16=β1, sayα1> β1, we can cancel all theβ1copies ofg1(x) from the right hand side leavingg1(x)|g2(x)β2· · ·gm(x)βm, sog1(x)|gj(x) for some j>1, which is impossible, since thegi(x) are all irreducible.
Henceα1=β1and we can cancel all of the powers ofg1(x) from each side.
Then attend tog2(x) in the same manner leading toα2=β2and so on, getting eventuallyαi=βi for alli, 1≤i≤m. Hence the factorization is unique up to±1.
