The span of two coprime polynomials of degree at most two, not both constant, is span(1, X), span(1, X² + αX), span(X, X² + β) or span(X + β, X² + γX)

Let $K$ be a field and $a_0, a_1 \in K[X]$ coprime with $1 \leq \max(\deg a_0, \deg a_1) \leq 2$. Then $\mathrm{span}_K(a_0, a_1) = \mathrm{span}_K(p, q)$ for a pair $(p, q)$ of one of the forms

• $(1, X)$;
• $(1, X^2 + \alpha X)$ with $\alpha \in K$;
• $(X, X^2 + \beta)$ with $\beta \neq 0$;
• $(X + \beta, X^2 + \gamma X)$ with $\beta \neq 0$ and $\gamma \neq \beta$.

Proof. Coprimality makes $a_0, a_1$ linearly independent, so $P = \mathrm{span}(a_0, a_1)$ is a plane. If neither has an $X^2$ term, $P \subseteq \mathrm{span}(1, X)$ and the two are equal by dimension. Otherwise $P$ contains a monic $f = X^2 + dX + e$ and a non-zero $r$ of degree at most one (the combination of $a_0, a_1$ that kills the $X^2$ terms): if $r$ is constant, $P = \mathrm{span}(1, X^2 + dX)$; otherwise $X + \beta \in P$, and $P = \mathrm{span}(X, X^2 + e)$ with $e \neq 0$ when $\beta = 0$, or $P = \mathrm{span}(X + \beta, X^2 + \gamma X)$ with $\gamma = d - e/\beta \neq \beta$ when $\beta \neq 0$, the inequalities because $X$, resp. $X + \beta$, is not a common factor. Each equality holds because two polynomials of distinct degrees in $P$ span the plane $P$.

Novelty. An elementary fact about polynomials over a field; routine.