Math 115. Slides from the Lecture of September 29

This handout contains the slides from the lecture of September 29 (with a more complete Division Algorithm).

Polynomials with Coefficients in ℂ

The main line of today’s class will mimic the following statements and proofs for polynomials with coefficients in ℂ (or ℝ or ℚ).

Definition. ℂ[x] is the set of polynomials with coefficients in ℂ. ℝ[x] and ℚ[x] are defined analogously.

We prove here that a nonzero polynomial in ℂ[x] of degree n has at most n roots (in ℂ). (In fact, it has exactly n roots, when counted with multiplicities, but this will not be proved for congruences modulo p.)

For the rest of today’s class, we will use the convention that the zero polynomial in ℂ[x] has degree −∞.

Theorem (Division Algorithm for Polynomials in ℂ[x] ). Let f, g ∈ ℂ[x] with g ≠ 0 . Then there are polynomials q, r ∈ ℂ[x] such that f(x) = q(x)g(x) + r(x) and deg r < deg g. Also, q and r are unique with these properties.

Proof. Existence. Long division of polynomials.

Uniqueness. Suppose that q_{2}, r_{2} ∈ ℂ[x] also satisfy f(x) = q_{2}(x)g(x) + r_{2}(x) and deg r_{2} < deg g. Then

r(x) − r_{2}(x) = (q_{2}(x) − q(x))g(x) .  (*)

But now the left-hand side has degree < deg g, and the only way this can happen is for the factor q_{2} − q to be zero. Therefore q_{2} = q. Therefore the left-hand side of (*) must be zero, so r_{2} = r also. This gives uniqueness. □

Corollary. Let f ∈ ℂ[x] and a ∈ ℂ. Then a is a root of f (i.e., f(a) = 0 ) if and only if (x − a) | f (i.e., f(x) = (x − a)g(x) for some g ∈ ℂ[x] ).

Proof. Write f(x) = (x − a)g(x) + r(x) with deg r < 1 . Then r is a constant c, and substituting x = a gives f(a) = (a − a)g(a) + c = c (because a − a = 0 ), so f(a) = c. Therefore

f(a) = 0 ⟺ c = 0 ⟺ (x − a) | f . □

Corollary. If f ∈ ℂ[x] and a_{1}, …, a_{r} ∈ ℂ are distinct roots of f (with r > 0 ), then writing f(x) = (x − a_{1})g(x) , we have that a_{2}, …, a_{r} are distinct roots of g.

Proof. Exercise. □