6. A glimpse at metric spaces 21
This equivalence relation partitions the set R[t] into equivalence classes; the
situation is strikingly similar to modular arithmetic. Given a polynomial p(t), we
use the division algorithm to write p(t) = q(t)(1 +
+ r(t), where the remainder
r has degree at most one. Thus r(t) = a + bt for some a,b, and this r is the
unique ﬁrst-degree polynomial equivalent to p. In the case of modular arithmetic
we used the remainder upon division by the modulus; here we use the remainder
upon division by
Exercise 1.31. Verify the transitivity property of equivalence modulo I.
Standard notation in algebra writes R[t]/(1 +
for the set of equivalence
classes. We can add and multiply in R[t]/(1 +
As usual, the sum (or product)
of equivalence classes P and Q is deﬁned to be the equivalence class of the sum
p + q (or the product pq) of members; the result is independent of the choice. An
equivalence class then can be identiﬁed with a polynomial a + bt, and the sum and
product of equivalence classes satisﬁes (15) and (16). In this setting we deﬁne C
as the collection of equivalence classes with this natural sum and product:
(25) C = R[t]/(1 +
Deﬁnition (25) allows us to set t2 = -1 whenever we encounter a term of
degree at least two. The irreducibility of t2 + 1 matters. If we form R[t]/(p(t)) for
a reducible polynomial p, then the resulting object will not be a ﬁeld. The reason
is precisely parallel to the situation with modular arithmetic. If we consider Z/(n),
then we get a ﬁeld (written Fn) if and only if n is prime.
Exercise 1.32. Show that R[t]/(t3 + 1) is not a ﬁeld.
Exercise 1.33. A polynomial
cktk in R[t] is equivalent to precisely one
polynomial of the form A + Bt in the quotient space. What is A + Bt in terms of
the coeﬃcients ck?
Exercise 1.34. Prove the division algorithm in R[t]. In other words, given
polynomials p and g, with g not the zero polynomial, show that one can write
p = qg + r where either r = 0 or the degree of r is less than the degree of g. Show
that q and r are uniquely determined by p and g.
Exercise 1.35. For any polynomial p and any x0, show that there is a polyno-
mial q such that p(x) = (x - x0)q(x) + p(x0).
6. A glimpse at metric spaces
Both the real number system and the complex number system provide intuition for
the general notion of a metric space. This section can be omitted without impacting
the logical development, but it should appeal to some readers.
Deﬁnition 6.1. Let X be a set. A distance function on X is a function δ : X×X →
R such that the following hold:
1) δ(x,y) ≥ 0 for all x,y ∈ X (distances are nonnegative).
2) δ(x,y) = 0 if and only if x = y (distinct points have positive distance between
them; a point has 0 distance to itself).