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Abstract Algebra: Theory and Applications

Thomas W. Judson

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Section 17.2 The Division Algorithm

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Recall that the division algorithm for integers (Theorem 2.9) says that if a and b are integers with b>0, then there exist unique integers q and r such that a=bq+r, where 0≤r<b. The algorithm by which q and r are found is just long division. A similar theorem exists for polynomials. The division algorithm for polynomials has several important consequences. Since its proof is very similar to the corresponding proof for integers, it is worthwhile to review Theorem 2.9 at this point.

Proof.

We will first consider the existence of q(x) and r(x). If f(x) is the zero polynomial, then
0=0⋅g(x)+0;
hence, both q and r must also be the zero polynomial. Now suppose that f(x) is not the zero polynomial and that deg⁡f(x)=n and deg⁡g(x)=m. If m>n, then we can let q(x)=0 and r(x)=f(x). Hence, we may assume that m≤n and proceed by induction on n. If
f(x)=anxn+an−1xn−1+⋯+a1x+a0g(x)=bmxm+bm−1xm−1+⋯+b1x+b0
the polynomial
f′(x)=f(x)−anbmxn−mg(x)
has degree less than n or is the zero polynomial. By induction, there exist polynomials q′(x) and r(x) such that
f′(x)=q′(x)g(x)+r(x),
where r(x)=0 or the degree of r(x) is less than the degree of g(x). Now let
q(x)=q′(x)+anbmxn−m.
Then
f(x)=g(x)q(x)+r(x),
with r(x) the zero polynomial or deg⁡r(x)<deg⁡g(x).
To show that q(x) and r(x) are unique, suppose that there exist two other polynomials q1(x) and r1(x) such that f(x)=g(x)q1(x)+r1(x) with deg⁡r1(x)<deg⁡g(x) or r1(x)=0, so that
f(x)=g(x)q(x)+r(x)=g(x)q1(x)+r1(x),
and
g(x)[q(x)−q1(x)]=r1(x)−r(x).
If q(x)−q1(x) is not the zero polynomial, then
deg⁡(g(x)[q(x)−q1(x)])=deg⁡(r1(x)−r(x))≥deg⁡g(x).
However, the degrees of both r(x) and r1(x) are strictly less than the degree of g(x); therefore, r(x)=r1(x) and q(x)=q1(x).
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Example 17.7.

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The division algorithm merely formalizes long division of polynomials, a task we have been familiar with since high school. For example, suppose that we divide x3−x2+2x−3 by x−2.
x2 + x + 4
x − 2 x3 − x2 + 2x − 3
x3 − 2x2
x2 + 2x − 3
x2 − 2x
4x − 3
4x − 8
5
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Hence, x3−x2+2x−3=(x−2)(x2+x+4)+5.
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Let p(x) be a polynomial in F[x] and α∈F. We say that α is a zero or root of p(x) if p(x) is in the kernel of the evaluation homomorphism ϕα. All we are really saying here is that α is a zero of p(x) if p(α)=0.

Proof.

Suppose that α∈F and p(α)=0. By the division algorithm, there exist polynomials q(x) and r(x) such that
p(x)=(x−α)q(x)+r(x)
and the degree of r(x) must be less than the degree of x−α. Since the degree of r(x) is less than 1, r(x)=a for a∈F; therefore,
p(x)=(x−α)q(x)+a.
But
0=p(α)=0⋅q(α)+a=a;
consequently, p(x)=(x−α)q(x), and x−α is a factor of p(x).
Conversely, suppose that x−α is a factor of p(x); say p(x)=(x−α)q(x). Then p(α)=0⋅q(α)=0.

Proof.

We will use induction on the degree of p(x). If deg⁡p(x)=0, then p(x) is a constant polynomial and has no zeros. Let deg⁡p(x)=1. Then p(x)=ax+b for some a and b in F. If α1 and α2 are zeros of p(x), then aα1+b=aα2+b or α1=α2.
Now assume that deg⁡p(x)>1. If p(x) does not have a zero in F, then we are done. On the other hand, if α is a zero of p(x), then p(x)=(x−α)q(x) for some q(x)∈F[x] by Corollary 17.8. The degree of q(x) is n−1 by Proposition 17.4. Let β be some other zero of p(x) that is distinct from α. Then p(β)=(β−α)q(β)=0. Since α≠β and F is a field, q(β)=0. By our induction hypothesis, q(x) can have at most n−1 zeros in F that are distinct from α. Therefore, p(x) has at most n distinct zeros in F.
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Let F be a field. A monic polynomial d(x) is a greatest common divisor of polynomials p(x),q(x)∈F[x] if d(x) evenly divides both p(x) and q(x); and, if for any other polynomial d′(x) dividing both p(x) and q(x), d′(x)∣d(x). We write d(x)=gcd(p(x),q(x)). Two polynomials p(x) and q(x) are relatively prime if gcd(p(x),q(x))=1.

Proof.

Let d(x) be the monic polynomial of smallest degree in the set
S={f(x)p(x)+g(x)q(x):f(x),g(x)∈F[x]}.
We can write d(x)=r(x)p(x)+s(x)q(x) for two polynomials r(x) and s(x) in F[x]. We need to show that d(x) divides both p(x) and q(x). We shall first show that d(x) divides p(x). By the division algorithm, there exist polynomials a(x) and b(x) such that p(x)=a(x)d(x)+b(x), where b(x) is either the zero polynomial or deg⁡b(x)<deg⁡d(x). Therefore,
b(x)=p(x)−a(x)d(x)=p(x)−a(x)(r(x)p(x)+s(x)q(x))=p(x)−a(x)r(x)p(x)−a(x)s(x)q(x)=p(x)(1−a(x)r(x))+q(x)(−a(x)s(x))
is a linear combination of p(x) and q(x) and therefore must be in S. However, b(x) must be the zero polynomial since d(x) was chosen to be of smallest degree; consequently, d(x) divides p(x). A symmetric argument shows that d(x) must also divide q(x); hence, d(x) is a common divisor of p(x) and q(x).
To show that d(x) is a greatest common divisor of p(x) and q(x), suppose that d′(x) is another common divisor of p(x) and q(x). We will show that d′(x)∣d(x). Since d′(x) is a common divisor of p(x) and q(x), there exist polynomials u(x) and v(x) such that p(x)=u(x)d′(x) and q(x)=v(x)d′(x). Therefore,
d(x)=r(x)p(x)+s(x)q(x)=r(x)u(x)d′(x)+s(x)v(x)d′(x)=d′(x)[r(x)u(x)+s(x)v(x)].
Since d′(x)∣d(x), d(x) is a greatest common divisor of p(x) and q(x).
Finally, we must show that the greatest common divisor of p(x) and q(x) is unique. Suppose that d′(x) is another greatest common divisor of p(x) and q(x). We have just shown that there exist polynomials u(x) and v(x) in F[x] such that d(x)=d′(x)[r(x)u(x)+s(x)v(x)]. Since
deg⁡d(x)=deg⁡d′(x)+deg⁡[r(x)u(x)+s(x)v(x)]
and d(x) and d′(x) are both greatest common divisors, deg⁡d(x)=deg⁡d′(x). Since d(x) and d′(x) are both monic polynomials of the same degree, it must be the case that d(x)=d′(x).
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Notice the similarity between the proof of Proposition 17.10 and the proof of Theorem 2.10.
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