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2.3. GREATEST COMMON DIVISORS 45
Lemma 1 In these circumstances, the result of Euclid’s algorithm is a linear
combination of f and g, i.e. a i−1 = λ i−1 f + µ i−1 g: λ i−1 , µ i−1 ∈ K[x].
Proof. a 0 and a 1 are certainly such combinations: a 0 = 1·f +0·g or 1·g +0·f
and similarly for g. Then a 2 = a 0 − q 1 a 1 is also such a combination, and so on
until a i−1 , which is the result.
The above theory, and algorithm, are all very well, but we would like to
compute (assuming they exist!) greatest common divisors of polynomials with
integer coefficients, polynomials in several variables, etc. So now let R be any
g.c.d. domain.
Definition 29 If f = ∑ n
i=0 a ix i ∈ R[x], define the content of f, written
cont(f), or cont x (f) if we wish to make it clear that x is the variable, as
gcd(a 0 , . . . , a n ). Technically speaking, we should talk of a content, but in the
theory we tend to abuse language, and talk of the content. Similarly, the primitive
part, written pp(f) or pp x (f), is f/cont(f). f is said to be primitive if
cont(f) is a unit.
Proposition 14 If f divides g, then cont(f) divides cont(g) and pp(f) divides
pp(g). In particular, any divisor of a primitive polynomial is primitive.
The following result is in some sense a converse of the previous sentence.
Lemma 2 (Gauss) The product of two primitive polynomials is primitive.
Proof. Let f = ∑ n
i=0 a ix i and g = ∑ m
j=0 b jx j be two primitive polynomials,
and h = ∑ m+n
i=0 c ix i their product. Suppose, for contradiction, that h is not
primitive, and p is a prime 23 dividing cont(h). Suppose that p divides all the
coefficients of f up to, but not including, a k , and similarly for g up to but not
including b l . Now consider
k−1
∑
c k+l = a k b l + a i b k+l−i +
i=0
∑k+l
i=k+1
a i b k+l−i (2.9)
(where any indices out of range are deemed to correspond to zero coefficients).
Since p divides cont(h), p divides c k+l . By the definition of k, p divides
every a i in ∑ k−1
i=0 a ib k+l−i , and hence the whole sum. Similarly, by definition of
l, p divides every b k+l−i in ∑ k+l
i=k+1 a ib k+l−i , and hence the whole sum. Hence p
divides every term in equation (2.9) except a k b l , and hence has to divide a k b l .
But, by definition of k and l, it does not divide either a k or b l , and hence cannot
divide the product. Hence the hypothesis, that cont(h) could be divisible by a
prime, is false.
Corollary 2 cont(fg) = cont(f)cont(g).
23 The reader may complain that, in note 22, we said that the ability to compute g.c.d.s
was not equivalent to the ability to compute unique factors, and hence primes. But we are
not asking to factorise cont(f), merely supposing, for the sake of contradiction that it is
non-trivial, and therefore has a prime divisor.