# My question

Let $A$ and $B$ be polynomials of degrees $d$ and $e$ respectively. If there exists a pair of polynomials $(P, Q) \neq (0, 0)$ such that $AP+BQ = 0$, with $\deg P < e$ and $\deg Q < d$, do $A$ and $B$ necessarily have a common root? Why?

(I think they do, and the sources I have access to insinuate this, but don’t prove it.)

# Background

(Let $\mathcal P_i$ denote the $i$-dimensional vector space of polynomials of degree less than $i$.)

I’m reading the Wikipedia article about the resultant, and it defines the resultant $R(A, B)$ rather elegantly, as the determinant of a linear map $\varphi : \mathcal P_e \times \mathcal P_d \to \mathcal P_{d+e}$ sending $(P, Q)$ to $AP+BQ$, written as a matrix in the basis of powers of $x$:

If $A=a_0x^d +a_1x^{d-1} + \cdots + a_d$ and $B=b_0x^e +b_1x^{e-1} + \cdots + b_e$ then

$$R(A, B) = \begin{vmatrix} a_0 & & & & b_0 & & & \\ a_1 & a_0 & & & b_1 & b_0 & & \\ a_2 & a_1 & \ddots & & b_2 & b_1 & \ddots & \\ \vdots &\vdots & \ddots & a_0 & \vdots &\vdots & \ddots & b_0 \\ a_d & a_{d-1} & \ddots & \vdots & b_e & b_{e-1} & \ddots & \vdots\\ & a_d & \ddots & \vdots & & b_e & \ddots & \vdots \\ & & \ddots & a_{d-1} & & & \ddots & b_{e-1} \\ & & & a_d & & & & b_e \end{vmatrix}.$$

From what I know about the resultant, $A$ and $B$ share a root precisely when this determinant is zero.

From what I know about linear algebra, this means that $A$ and $B$ share a root precisely when $\ker \varphi \neq 0$: that is, there exist some $(P, Q) \neq (0, 0)$ in $\mathcal P_e \times \mathcal P_d$ such that $AP+BQ=0$. (In fact, assuming this is true, I would expect it to be the observation that leads to the above definition of the resultant, and the choice of $\varphi$.)

I can prove this in one direction:

If $A = a_0 (x-\alpha) A'$ and $B = b_0 (x-\alpha)B'$, then pick $P = b_0 B'$ and $Q = -a_0A'$: it follows that $$AP+BQ = a_0b_0(x-\alpha)A'B' - a_0b_0(x-\alpha)A'B' = 0.$$

Since $\deg P = e-1$ and $\deg Q = d-1$, this choice is valid.

But in the other direction, I’m stuck, leading to the question above.

• what if $P=B$ and $Q=-A$?
– daw
Dec 20, 2017 at 16:05
• @daw Good point: I had missed the requirement that $(P, Q) \in \mathcal P_e \times \mathcal P_d$ implies $\deg P < e = \deg B$ and $\deg Q < d = \deg A$. I’ve updated the question.
– Lynn
Dec 20, 2017 at 16:14

If $$AP=-BQ$$ then one of the factors of $A$ must also be a factor of $B$, by consideration of the degees.
• Ah, I see! Considering the degrees, in $(A, P; B, Q)$ there are respectively $(d, e-1; e, d-1)$ factors. So if both sides have the same factors, then when redistributing the factors from the LHS to the RHS, $Q$ can hold at most $d-1$ factors of $A$, but one of them must end up in $B$. Is that correct?