Elements of correction

$(1)$

With ${\alpha}_2 = 2$ , we have $u_f = x_1^3 + x_1^4 + x_1^2$ and with ${\alpha}_2 = 3$ , we have $u_f = x_1^4 + x_1^6 + x_1^2$ .

$(2)$

${\alpha}_2 \geq {\deg}(f,x_1) + 1$ .

$(3)$

${\alpha}_2 \geq d_1 + c_1 + 1$ .

$(4)$

Using ${\alpha}_2 = d_1 + c_1 + 1$ compute $u_f$ and $u_g$ the respective images of $f$ and $g$ , then the product $u_f u_g$ in ${\mbox{${\mathbb{K}}$}}[x_1]$ , then its (unique) preimage $fg$ . It is easy to check that the transformations $f \rightarrow u_f$ , $g \rightarrow u_g$ , and $u_f u_g \rightarrow f g$ have a linear cost in the number of terms of their results. So the dominant cost is the computation of the product $u_f u_g$ which is clearly in $M( ({\deg}(fg,x_1) +1) {\deg}(fg,x_2) ) = M( {\alpha}_2 {\deg}(fg,x_2) )$ operations in ${\mathbb{K}}$ .