A coprime pair of polynomials of maximal degree 2 has non-zero resultant [20]² − [21][10] (characteristic zero)

Let $K$ be a field of characteristic zero and $p, q \in K[X]$ coprime (that is, $ap + bq = 1$ for some $a, b \in K[X]$) with $\max(\deg p, \deg q) = 2$. Write $p_i, q_i$ for the coefficients of $X^i$ and $[ij] = p_iq_j - p_jq_i$. Then $\mathrm{Res} = [20]^2 - [21][10] \neq 0$, that is, $(p_2q_0 - p_0q_2)^2 - (p_2q_1 - p_1q_2)(p_1q_0 - p_0q_1) \neq 0$.

Up to sign, $\mathrm{Res}$ is the determinant of the Bezout matrix of $p$ and $q$, their resultant as binary quadratic forms.

Proof. By DiazModulus.quadratic_pencil_wronskian_identities, $J = p'q - pq' = [21]X^2 + 2[20]X + [10]$, so the matrix whose rows are the coefficient vectors of $q$, $p$ and $J$ in the basis $(1, X, X^2)$ has determinant $-2\,\mathrm{Res}$. If $\mathrm{Res} = 0$, a non-zero vector $(v_0, v_1, v_2)$ kills this matrix from the left, that is, $v_0q + v_1p + v_2J = 0$. If $v_2 = 0$, one of $p, q$ is a constant multiple of the other; coprimality makes the other a unit, and both have degree $0$, against $\max(\deg p, \deg q) = 2$. If $v_2 \neq 0$, the relation reads $(v_2p' + v_0)q = p(v_2q' - v_1)$, so by coprimality $p$ divides $v_2p' + v_0$ and $q$ divides $v_2q' - v_1$. Whichever of $p, q$ has degree $2$ then divides a polynomial of degree at most $1$, which must be $0$; but its coefficient of $X$ is $2v_2$ times the leading coefficient, which is non-zero in characteristic zero.

Novelty. Classical (the Bezoutian and the Wronskian of a pencil of binary quadratics); routine.