rings lect 7

8/6/2019 Rings Lect 7

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7 Factorisation in Q[X ]

In this section we deal with some very concrete matters concerning the factorisation of poly-nomials with rational coefficients. Any such polynomial is a rational multiple of a polynomial

with integer coefficients, and so we may concentrate on the integral associate. For such poly-nomials we have various techniques to help us, in particular we can work ‘modulo p’ forvarious primes; or we can use other arithmetic tricks. We describe some of these, and weprove the important theorem which reconciles factorisation of polynomials over Q and overZ.

7.1 Factorisation in Z[X ] and Z p[X ]

We begin with a general lemma.

Lemma 7.1.1. Let R and S be rings, and let f : R → S be a homomorphism, f : a → a.

Then π : R[X ] → S [X ] given by π :N

k=0 akX k →N

k=0 akX k is a homomorphism with

image f (R)[X ] and kernel (ker f )[X ].

Proof. In view of how we add and multiply polynomials all we need to check is that

ak + bk = ak + bk and

r+s=n

arbs =

r+s=n

arbs.

The remarks on kernel and image are clear.

We can use what we have learned about rings and homomorphisms to give a usefulpractical test for irreducibility1 in the ring Z[X ].

Proposition 7.1.2 (Eisenstein’s criterion). Let f (X ) =N

k=0ck

X k be a polynomial in Z[X ],and let p ∈ Z be a prime. Suppose that p|ck for all k = 0, 1, . . . , N − 1, p |cN and p2 |c0.

Then f (X ) has no factor of smaller degree in Z[X ].

Proof. Suppose that we had that f (X ) = g(X )h(X ), with g(X ) =N 1

k=0 akX k and h(X ) =N 2k=0 bkX k and N 1, N 2 > 1. Extending the natural homomorphism Z → Z p to the poly-

nomial rings we would then have that π(g(X )π(h(X )) = π(f (X )) = cN X N = 0, so thatπ(g(X )) = aX N 1 and h(X ) = bX N 2 by the unique factorisation in the Euclidean domainZ p[X ]. Hence we have that a0 = b0 = 0; that is p|a0 and p|b0, yielding p2|a0b0 = c0 contraryto hypothesis.

There are many examples of this. For instance, (X 2+125)2(X 3+25)4+5 has only factorsof degree 16 in Z[X ] and so, being monic, is irreducible.

More famous are the cyclotomic polynomials or prime order. Let Φ p(X ) := Xp−1X−1

; thenΦ p has no factor of degree k with 1 < k < ( p − 1). We can’t use Eisenstein directly; butinstead we apply the result to Φ p(X + 1); the details are an exercise.

1Well, almost irreducibility: see later.

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7.2 Factorization in Z[X ] and Q[X ]

Although we often work with polynomials with integral coefficients, we are really interestedin whether these factorise in the ring Q[X ]. It is clear that in a trivial way a polynomialmay be a product of two irreducibles in Z[X ] but of only one irreducible in Q[X ]—take for

example f (X ) = 2X . We can easily check that in the large ring Q[X ] this is irreducible; butin the small ring it is the product of two irreducibles 2 and X . What we want to see is thatessentially nothing more complicated than this is possible.

Theorem 7.2.1. Let 0 = h ∈ Z[X ]. Suppose that h = f · g with f, g ∈ Q[X ]. Then there

exist f , g ∈ Z[X ] with deg f = deg f and deg g = deg g, such that h = f · g.

Proof. Write f =M

k=0 akX k, g =N

k=0 bkX k and let ck :=

r+s=k arbs be the coefficientsof h.

We may multiply f by the least common multiple of the denominators of its coefficients,and then divide the resulting polynomial by the highest common factor of its (integral!)

coefficients. In that way we get f = α

˜f , where α ∈

Qand

˜f is a polynomial of the samedegree as f , and has integral coefficients whose highest common factor is 1. We can find

corresponding β ∈ Q and g ∈ Z[X ] with g = β g.We can find corresponding γ ∈ Z—there are no denominators to clear so γ is integral—

and h ∈ Z[X ] with h = γ h.Then we have, after re-arranging the rationals,

n · h(X ) = m · f (X ) · g(X ), for some m, n ∈ Z.

(The X ’s are inserted into the polynomials for clarity.)How are m and n related? Well n is the highest common factor of the coefficients of the

polynomial on the left hand side. But the highest common factor of the coefficients of the

polynomial on the right hand side is m times the highest common factor of the coefficientsof f (X ) · g(X ). In a moment we will prove that the latter highest common factor is 1. So

m = n and we have thath(X ) = f (X ) · g(X ),

which we multiply by γ ∈ Z to get

h(X ) = γ · h(X ) = γ · f (X ) · g(X ).

Define f := γ · f and g := g and we are done.

The missing lemma is so important it deserves a separate subsection.

7.3 Gauss’s Lemma

We begin with some pieces of notation.

Definition 1. Let 0 = f ∈ Z[X ]. Then by the content of f , denoted by c(f ), we mean

the highest common factor of the coefficients of f . If the content of f is 1 then we say f is

primitive.

Proposition 7.3.1 (Gauss’s Lemma). Let f, g ∈ Z[X ] and suppose that c(f ) = c(g) = 1.

Then c(f · g) = 1.

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Proof. If false let the prime p divide all the coefficients of the product. Extend the naturalhomomorphism Z → Z p to a homomorphism π of the polynomial rings in the standard way.Then

0 = π(f · g) = π(f ) · π(g)

and so, since Z p[X ] is an integral domain, either π(f ) = 0 or π(g) = 0. But that implies pdivides every coefficient of f or every coefficient of g, a contradiction.

Corollary 7.3.2. Let f, g ∈ Z[X ]; then c(f · g) = c(f )c(g).

Proof. Trivial.

7.4 Z[X ] is a UFD

We can use Gauss’s Lemma and the techniques of the previous sections to prove that in thering Z[X ] every polynomial factorises uniquely into a product of primes in Z times a product

of primitive polynomials each irreducible in Q[X ].

7.5 An algorithm for factorisation in Q[X ]

How can we factorise a polynomial f ∈ Q[X ]? As we’ve seen we may assume that f ∈ Z[X ]and carry out our factorisations over Z.

Here are a couple of trivial starting points:

Lemma 7.5.1. Suppose that n ∈ Z. If f (n) = 0 then X − n|f .

Proof. Remainder Theorem and Gauss’s Lemma.

Lemma 7.5.2. Suppose that g|f in Z[X ]; and that n ∈ Z. Then g(n)|f (n).

Proof. Trivial.

With these in mind we search for factors of f in Z[X ] of degree d = 1, 2, . . . , deg f − 1.Of course whenever we find a factor we divide it out and then look recursively at two factorswe have found.

The following lemma controls the possible values of a factor of degree d may take.

Lemma 7.5.3. Let a0, a1, a2, . . . , ad ∈ Z be distinct integers, and suppose that f (ak) = 0 for

each k. Then there are only finitely many (d + 1)-tuples (b0, b1, . . . , bd) such that bk|f (ak) for

each k.

Proof. This is essentially the fact that Z is a UFD.

The following lemma allows us to reconstruct a polynomial of degree d from its values at(d + 1) places.

Lemma 7.5.4. Let a0, a1, a2, . . . , ad ∈ Z be distinct integers, and let b0, b1, . . . , bd ∈ Q. There

is exactly one polynomial g ∈ Q[X ] of degree at most d such that g(ak) = bk for all k.

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Proof. This is just the Chinese Remainder Theorem. To be absolutely specific we could, asLagrange did, take

g(X ) =d

k=0

bk

j=k

X − a j

ak − a j.

By the previous two lemmas we can now construct a finite list of all those polynomials g

of degree d in Q[X ] which can possibly be divisors of f .be more precise if m = deg f for any d ≤ m − 1 we calculate, f (0), f (1),...,f (d + m).

Since f has at most m roots, at least d + 1, among those are = 0. Let’s say that for

{a0, a1,...,ad} ⊂ {0, 1,...,d + m} we have f (ai) = 0. Using lemmas 7.5.3, 7.5.4 we produce afinite list of polynomials of degree d which are possible divisors of f . If g is such a polynomialwe check whether in fact g ∈ Z[X ], and then, by long division, whether g|f . If g|f for some

g then we write f = gh and we repeat the procedure for g, h. This gives us an algorithm todecompose any polynomial f ∈ Q[X ] as a product of irreducibles. Of course if no g from ourfinite list divides f we conclude that f is irreducible.

7.6 Factorization in R[X ] and C[X ]

According to the Fundamental Theorem of Algebra any f ∈ C[X ] has a root in C. Inparticular a polynomial f ∈ C[X ] is irreducible if and only if deg f = 1.

Let f ∈ R[X ] be a polynomial with deg f > 2. If f has areal root clearly it is notirreducible. If a non-real a ∈ C is a root of f then a is also a root. So (X − a)(X − a) divides

f . But (X − a)(X − a) = X 2 − 2ℜaX + |a|2 ∈ R[X ]. So again f is not irreducible. Hencethe irreducible polynomials in R[X ] are the polynomials of degree 1 and the polynomials of

degree 2 with no real roots.

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Which of the following are true?

1. If the polynomial f (x) ∈ Z[X ] is irreducible in Z[X ] then f (X ) is also irreducible inZ3[X ].

2. If the polynomial f (x) ∈ Z7[X ] is irreducible in Z7[X ] then f (X ) is also irreducible inZ[X ].

3. If K is a field, a polynomial of degree 3, f (X ) ∈ K[X ] is irreducible if and only if it hasno roots in K.

4. There is a monic polynomial f (X ) ∈ Z[X ] such that f ( 20082009

) = 0.

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