What is free family?

Example 18.16

The canonical basis of $K^n$: $e_i = (0, \dots, 1, \dots, 0)$ ($1$ in slot $i$). The monomials $(1, X, X^2, \dots, X^n)$: a basis of $K_n[X]$ (freeness: a null combination is the zero polynomial, so all coefficients vanish, Definition 8.1). In $\C$ over $\R$: the basis $(1, \iu)$.

Example 18.18 (Testing a candidate basis, start to finish)

Is $\mathcal{F} = (1 + X,\ 1 + X^2,\ X + X^2)$ a basis of $\R_2[X]$? Write $u_1, u_2, u_3$ for the three polynomials. Freeness: a null combination $a\,u_1 + b\,u_2 + c\,u_3 = 0$ gives, coefficient by coefficient,

subtracting the first two, $b = c$, then the third gives $2b = 0$: $a = b = c = 0$, free. Generating: instead of solving three systems, notice the symmetric combination

so $1 = \frac12(u_1 + u_2 - u_3)$; then

The monomials lie in the span, so everything does: $\mathcal{F}$ is a basis. As a bonus, assembling the three displays gives the coordinates of any $P = \alpha + \beta X + \gamma X^2$:

(Sanity check with $P = X$: coordinates $\bigl(\frac12, -\frac12, \frac12\bigr)$, as found above.) Two lessons: symmetry in the family usually hides a shortcut combination; and once dimension is available (Chapter 19), the whole generating half of this work will come free of charge — three free vectors of a $3$-dimensional space always form a basis.

Example 18.20 (The staircase principle)

Let $P_0, P_1, \dots, P_n \in K_n[X]$ with $\deg P_k = k$ for each $k$ (a “staircase” of degrees). Then $(P_0, \dots, P_n)$ is a basis of $K_n[X]$. Freeness is Proposition 18.19 (1). For the generating property, argue by finite descent on the degree: let $Q \in K_n[X]$, $Q \neq 0$, of degree $d$, with leading coefficient $a$, and let $b \neq 0$ be the leading coefficient of $P_d$. Then $Q - \frac ab P_d$ has degree $< d$ (the top terms cancel); replacing $Q$ by this difference and iterating, after at most $n + 1$ steps one reaches the zero polynomial, and unwinding the subtractions expresses $Q$ as a combination of the $P_k$. Two staircases already met: the shifted powers $\bigl((X-a)^k\bigr)_{0 \leq k \leq n}$ (Exercise 18.4), and the Newton products $\bigl((X - x_0)(X - x_1)\cdots(X - x_{k-1})\bigr)_{0 \leq k \leq n}$, put to work in the weekend problem.