### Appendix: Polynomial

##### 5. Coprime polynomials

Two integers are coprime if their gcd is 1. Similarly, two polynomials p(t) and q(t) are coprime if gcd(p(t), q(t)) = 1 (or any nonzero number). Specifically this means that

c(t) divides p(t) and q(t) ⇒ c(t) is a nonzero constant.

From this example, we find that the polynomials t4 + t3 + 3t2 + t + 1 and t2 - t + 1 are coprime.

For two complex polynomials p(t) and q(t), the following are equivalent.

• p(t) and q(t) are coprime.
• There are polynomials u(t) and v(t), such that u(t)p(t) + v(t)q(t) = 1.
• p(t) and q(t) do not have common roots.

Proof Applying this result to the special case that d(t) = 1, we find that the first property implies the second.

Now assume u(t)p(t) + v(t)q(t) = 1 for some polynomials u(t) and v(t). If r is a of p(t), then we have p(r) = 0, so that v(r)q(r) = u(r)p(r) + v(r)q(r) = 1. In particular, this implies q(r) ≠ 0. This proves that any root of p(t) cannot be a root of q(t). This completes the proof that the second property implies the third.

Finally, if p(t) and q(t) are not coprime, then gcd(p(t), q(t)) = d(t) is a polynomial of positive degree. By the fundamental theorem of algebra, d(t) must have a complex root r. Since d(t) divides p(t) and q(t), by this result, r is also a common root of p(t) and q(t). This completes the proof that the third property implies the first.

How can we determine whether two given polynomials have common roots? Let

p(t) = a0 + a1t + a2t2 + ... + am-1tm-1 + tm = (t - r1)(t - r2)...(t - rm),
q(t) = b0 + b1t + b2t2 + ... + bn-1tn-1 + tn = (t - s1)(t - s2)...(t - sn).

Then for p(t) and q(t) to have a common root means that the following expression, called the resultant, is zero.

Res(p, q) = ∏ij(ri - sj) = q(r1)q(r2) ... q(rm) = (-1)mnp(s1)p(s2) ... p(sn)
= (r1 - s1)(r1 - s2)...(r1 - sn) (r2 - s1)(r2 - s2)...(r2 - sn) ... ... (rm - s1)(rm - s2)...(rm - sn).

Note that the expression is symmetric in the roots r1, r2, ... rm of p(t), and is also symmetric in the roots s1, s2, ..., sn of q(t). On the other hand, we have seen in Viete theorem that the coefficients of a polynomial are also the symmetric polynomials of the roots. This suggests that Res(p, q) may be expressed in terms of the coefficients in p(t) and q(t).

Example Let p(t) = a0 + a1t + t2 and q(t) = b0 + b1t + t2 be of second order. Then by r1 + r2 = - a1 and r1r2 = a0, we have

Res(p, q) = (b0 + b1r1 + r12)(b0 + b1r2 + r22)
= b02 + b0b1(r1 + r2) + b0(r12 + r22) + b12r1r2 + b1r1r2(r1 + r2) + r12r22
= b02 + b0b1(- a1) + b0((- a1)2 - 2a0) + b12a0 + b1a0(- a1) + a02
= a02 + b02 - 2a0b0 + a12b0 + a0b12 - a1b0b1 - a0a1b1.