MA554
Workshop 4
GCD (or HCF if you prefer)
Recall the definition: if R is a commutative ring and a, b ∈ R, then h ∈ R is a highest
common factor of a and b, abbreviated to HCF (or greatest common divisor, GCD), if
(1) h divides a and h divides b
(2) for any g ∈ R which divides both a and b, g also divides h.
Beware that highest common factors do not necessarily exist in most rings. But if R = Z,
then highest common factors always exists, and moreover, if h is a highest common factor
of a, b ∈ Z, then there are u, v ∈ Z such that ua + vb = h—this is called extended GCD.
This all follows from the Euclidean algorithm (or division with remainder). Since R[t]
also has a Euclidean algorithm, analogous statements hold in R[t], or indeed in any F [t]
where F is a field.
1. Compute h = HCF(48, 432) and find u, v ∈ Z such that 48u + 432v = h.
Repeat for 48 and 434. Repeat again for two numbers of your own choice.
2. Working in R[t], let f = (t + 1)(t2 + 2) and g = (t + 1)(t − 2)(t3 − 1) and compute
h = HCF(f, g) and find u, v ∈ R[t] such that uf + vg = h.
3. Let f, g, h ∈ R[t] and suppose f, g are coprime—i.e. HCF(f, g) = 1. Using the
extended GCD, show that if f |gh then f |h.
Field of fractions
4. Show that multiplication (as defined in lectures) is well defined in a field of fractions.
That is, if ab = ab11 and dc = dc11 , then
a c
a1
c1
× =
× .
b d
b1
d1
5. Consider the integral domain R = Q[t]. The field of fractions FR of R is often denoted
Q(t)—its elements are (equivalence classes) of quotients of polynomials.
Let f = t3 − 3t + 2, g = t − 1 and f 0 = 2t3 + 1, g 0 = 2t2 − 1 in R. Compute
f0
f
+ 0
g
g
in FR .
Show that fg can be regarded as an element of R inside FR —that is, it can be written as
h
for suitable h ∈ R.
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6. Consider R1 = Z[t] and R2 = Q[t], and note that R1 ⊂ R2 . Show that R1 and R2
have the same field of fractions.
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