Prove or disprove: there exists a basis $p_0, p_1, p_2, p_3 \in P_3(F)$ such that none of the polynomials $p_0, p_1, p_2, p_3$ has degree 2 Prove or disprove: there exists a basis  $p_0, p_1, p_2, p_3 \in  P_3(F)$ such that none of the polynomials $p_0, p_1, p_2, p_3$ has degree 2
This is a repeat of Does there exist a basis $(p_0,p_1,p_2,p_3)\in P_3(\Bbb F)$ such that none of the polynomials $p_0,p_1,p_2,p_3$ has degree $2$?
But I just have a question in regards to the supposed basis vectors. 
My conclusion was that it could not occur because in order to characterize all of the  polynomials of degree 3, you will need a polynomial of degree 2.
But the solution said otherwise, particularly how are $x^2 + x^3, x^2$ going to be basis vectors. Do these not have a polynomial of degreee 2? Which is what we are trying to show cannot occur?

$p_0:=x^3$
$p_1:=x^2+x^3$
$p_2:=x+x^2+x^3$
$p_3:=1+x+x^2+x^3$
$\deg P_j = 3\neq 2$ for any $j$.
$p_1-p_0=x^2$
$p_2-p_1=x$
$p_3-p_2=1$
So, $p_0,p_1,p_2,p_3$ is a basis of $\mathcal{P}_3(\mathbb{F})$.
