Prove or disprove: there exists a basis p0,p1,p2,p3p_0,p_1,p_2,p_3 of P3(F)\mathcal P_3(\mathbf F) such that none of the polynomials p0,p1,p2,p3p_0,p_1,p_2,p_3 has degree 22.

Let

(p0,p1,p2,p3)=(1,x,x2+x3,x2−x3)(p_0,p_1,p_2,p_3) = (1,x,x^2+x^3,x^2-x^3)

It's clearly spanning as (p2+p3)=2x2(p_2+p_3) = 2x^2 and (p2−p3)=2x3(p_2-p_3) = 2x^3. to see independance consider

a0+a1x+a2(x2+x3)+a3(x2−x3)=0a_0 + a_1x + a_2(x^2 + x^3) + a_3(x^2-x^3) = 0

Distribute

a0+a1x+(a2+a3)x2+(a2−a3)x3=0a_0 + a_1x + (a_2+a_3)x^2 + (a_2-a_3)x^3 = 0

Since (1,x,x2,x3)(1,x,x^2,x^3) is independent this implies

a0=a1=(a2+a3)=(a2−a3)=0a_0 = a_1 = (a_2 + a_3) = (a_2 - a_3) = 0

Which implies a2=a3=0a_2=a_3=0 (add and subtract to isolate a2,a3a_2,a_3 like we did before).

Thus we have found a basis as requested, and are done.

I really hate this computational proof of independance which should have been a reference to 2.42. It's exactly the same style as 2a/6