16. The Division Theorem
Throughout this worksheet, we are going to focus our attention on polynomials with coefficients from a field F . Examples of rings that are fields are Q, R, C, and Zp where p is prime.
Recall that one of the most important properties of the integers is the existence of a Division
Theorem: given two integers a, b with b > 0 then there exists unique integers q and r, with
0 ≤ r < b such that a = bq + r. We now show the same holds for polynomials in F [x], with a
slight change in the condition on the remainder:
Theorem 1. (The Division Theorem for F [x]) Let F be a field and f (x), g(x) ∈ F [x] with
g(x) 6= 0. Then there exists unique elements q(x), r(x) ∈ F [x] such that f (x) = g(x)q(x) + r(x)
and deg r(x) < deg g(x).
Before we give the proof, let’s try some examples:
Exercises: For each of f and g below, find q and r such that f = gq + r and deg r < deg g.
1. f = 3x4 − 6x2 + x + 3 and g = 2x2 + 5 in Q[x].
2. f = 4x3 + x2 − 3x + 1 and g = 3x + 2 in Z5 [x].
3. f = 3x5 + x3 + 2x and g = 4x2 − 1 in Z7 [x].
Proof of the Division Theorem: There are two components to the proof: the existence of q
and r, and the uniqueness of q and r.
Existence: First, note that if f = 0 then we can set q = 0 and r = 0, so there’s nothing really to
do. So we can assume f 6= 0. Write f = an xn + · · · + a0 and g = bm xm + · · · + b0 where an 6= 0
and bm 6= 0. Then n = deg f and m = deg g. We will use induction on n to show there exists a
q and r.
Our base case (actually, bases cases) will be all n with n < m. In this case, we simply let
q = 0 and r = f . Then f = g(0) + r and deg r = deg f = m < n = deg g. For the induction
step, we assume n ≥ m and that the existence part of the theorem is true for all polynomials f
−1 n−m
with degree less than n. Since F is a field and bm 6= 0, b−1
and
m exists in F . Let q1 = an bm x
consider the polynomial q1 g:
n−m
q1 g = an b−1
g
m x
n−m
= an b−1
(bm xm + · · · + b0 )
m x
n
n−m
)
= an b−1
m (bm x + · · · + b0 x
n
−1
n−m
= an x + · · · + an b m b 0 x
.
Notice that the leading term of q1 g and f are the same. Hence, deg(f − q1 g) < n. By our
induction hypothesis, there exists polynomials q2 and r with deg r < deg g such that f − q1 g =
gq2 + r. But then f = q1 g + gq2 + r = g(q1 + q2 ) + r = gq + r, where q = q1 + q2 , which is what
we wanted to show.
Uniqueness: Suppose f = gq + r and f = gq0 + r0 for q, q0 , r, r0 ∈ F [x] with deg r < deg g and
deg r0 < deg g. Then gq + r = gq0 + r0 , so g(q − q0 ) = r0 − r. Now, deg(r0 − r) < deg g and
deg g(q − q0 ) = deg g + deg(q − q0 ). This leads to deg g + deg(q − q0 ) < deg g. Clearly, this
means that deg(q − q0 ) is negative, which implies that q − q0 = 0. Thus r0 − r = g(q − q0 ) = 0
also. Hence, q = q0 and r = r0 .
The Division Theorem has many consequences: The first is known as The Remainder Theorem. Let f (x) = an xn + · · · + a1 x + a0 be a polynomial. We can consider f (x) as a function
from F to F by for b ∈ F setting f (b) = an bn + an−1 bn−1 + a1 b + a0 ; i.e., we just substitute b in
for x. If f (b) = 0, we say b is a root of f (x).
Corollary 2. Let F be a field, f (x) ∈ F [x], and a ∈ F . Then the remainder upon dividing f (x)
by x − a is f (a).
Proof. By the Division Theorem, f (x) = (x − a)q(x) + r(x) for some q(x), r(x) ∈ F [x] with
deg r(x) < deg(x − a) = 1. Thus, deg r(x) = 0 or −∞. In either case, r(x) = r is a constant
(that is, an element of F ). To find r, plug in a for x: f (a) = (a − a)q(a) + r = 0 + r = r.
Therefore, the remainder is f (a).
Exercises: Find the remainders upon dividing f by g in the given ring:
1. f = 5x6 − 6x3 + 3x+ 7x − 2, g = x − 1 in Q[x].
2. f = (3x2 − x + 6)(x + π) + (x + 2)(x3 − x − 1), g = x + 2 in R[x].
3. f = 2x4 − 5x2 + x − 3, g = x − 3 in Z7 [x].
As another consequence, we have what is known as the Root Theorem:
Corollary 3. Let F be a field, f (x) ∈ F [x], and a ∈ F . Then x − a is a factor of f (x) if and
only if f (a) = 0; i.e., a is a root of f (x).
Proof. Suppose x − a is a factor of f (x). Then f (x) = (x − a)q(x) for some q(x) ∈ F [x].
Substituting x = a, we have f (a) = (a − a)q(x) = 0 · q(x) = 0. Conversely, suppose f (a) = 0.
Using the Remainder Theorem, the remainder upon dividing f (x) by x−a is 0; i.e., x−a divides
f (x).
While we are on the topic of roots, let’s prove the Rational Root Theorem:
Theorem 4. Let f (x) = an xn + · · · + a1 x + a0 where ai ∈ Z and an 6= 0. Suppose q ∈ Q is a
nonzero root of f (x). Then q = dc where c | a0 and d | an .
Proof. Write q = dc where c and d are nonzero integers. By cancelling out the gcd of c and d,
we can assume gcd(c, d) = 1. Then
c
c
c
c
0 = f ( ) = an ( )n + an−1 ( )n−1 + · · · + a1 + a0 .
d
d
d
d
Multiplying both sides by dn , we get an equation of integers
an cn + an−1 cn−1 d + · · · + a1 cdn−1 + a0 dn = 0.
Rewriting, we have
c(an cn−1 + · · · + a1 dn−1 ) = −a0 dn .
Thus, c | a0 dn . Since gcd(c, d) = 1, gcd(c, dn ) = 1. (Why?) Therefore, c | a0 . A similar
argument where one puts all the terms involving d on one side of the equation shows d | an .
This can be used to not only find rational roots, but to show certain polynomials have no
rational roots. Or to show certain algebraic numbers are not rational.
√
√ 2
√
Example: Let’s give another proof that 2 is irrational. We know 2 = 2, so 2 is a root of
the polynomial f (x) = x2 − 2. According to the Rational Root Theorem, if q is a rational root,
then q = dc where c | −2 and d | 1. Thus, 2 and −2 are the only possible rational roots. But
√
it is easily checked f (2) 6= 0 and f (−2) 6= 0. Thus, f (x) has no rational roots and hence 2 is
irrational.
Exercise: Find all rational roots of x5 + x + 2.
Homework:
1. Prove that
√
5
3 is irrational.
2. Suppose a and m are positive integers. Prove that either
irrational.
√
m
a is either an integer or
3. Let f = 5x4 + 3x3 − 2x + 3 and g = x − 2. Find a prime p such that g(x) divides f (x) in
Zp [x].