Control roots and quotient coordinates

Try

Divide x³+2x²−x+1 by x²+1 over Q. Cancel the leading x³ term with x times the divisor, then cancel the remaining 2x² term with two times the divisor. The quotient is x+2 and the remainder is −2x−1. Multiply back to verify the result coefficient by coefficient.

Division is a statement about formal polynomial coefficients. It does not require choosing numerical values of x or drawing a graph.

Predict

Could two different quotients have remainders of degree below the divisor’s degree? Predict what happens after subtracting the two proposed division equations. One side would have degree at least the divisor’s degree, while the other would have smaller degree, unless both vanish.

Experiment

Record the intermediate remainder rather than only the final quotient.

Step	Subtract	Remaining polynomial
First cancellation	x(x²+1)	2x²−2x+1
Second cancellation	2(x²+1)	−2x−1

The degree decreases at each step. Over a field, the leading coefficient of a nonzero divisor is invertible, making every cancellation possible. Over Z, the same procedure can stop immediately: x divided by 2x would require coefficient one half, which is not an integer.

Understand

For f,g∈F[x] with g nonzero, division gives unique q,r with f=qg+r and either r=0 or deg r<deg g. Existence follows by repeated leading-term cancellation and termination of decreasing degrees. For uniqueness, subtract f=qg+r=q′g+r′. If q−q′ is nonzero, then (q−q′)g has degree at least deg g, while r′−r has smaller degree or is zero. This contradiction forces q=q′ and then r=r′.

Dividing by x−a leaves a constant remainder r. Evaluating at a gives r=f(a), the remainder theorem. Therefore f(a)=0 exactly when x−a divides f, the factor theorem.

A nonzero degree-n polynomial over a field has at most n distinct roots. Prove this by induction. If there is no root, the assertion holds. If a is a root, factor f=(x−a)q. For another root b≠a, the field equation (b−a)q(b)=0 forces q(b)=0. The degree-n−1 polynomial q has at most n−1 distinct roots by induction. Add a to obtain the bound. The no-zero-divisors hypothesis is essential to the step involving b−a.

This bound does not say a polynomial has all its roots in the base field. The polynomial x²+1 has no real roots but two complex roots. Nor does it equate polynomials with functions on a finite field. In F_p, xᵖ−x vanishes at every field element but is a nonzero formal polynomial. It has p roots and degree p, so the root bound is respected.

For a monic polynomial m of degree d, every coset in F[x]/(m) has a unique representative of degree below d. Division gives existence; uniqueness follows because a nonzero multiple of m cannot have smaller degree. Thus the quotient has an F-vector-space basis represented by 1,x,…,x^(d−1), whether or not it is a field.

Apply

Reduce a product modulo x²+1 and compare the answer with direct expansion followed by x²=−1. Explain why both methods produce the same unique low-degree representative.

When a root count appears to fail, check the coefficient structure. Over a ring with zero divisors, a nonzero polynomial can have more roots than its degree. Identify the exact cancellation step that the field proof used, rather than treating the counterexample as a contradiction.

Continue

Gauss’s lemma next controls coefficients when moving between a UFD and its fraction field. It explains why rational factorizations of primitive integer polynomials can be converted into integer factorizations.