is reducible.

If $t=2$, then $F$ simplifies into $$\begin{aligned}
F_0&=& -4 (-u^2+S_2)v +S_2^2 - u^4  + 2S_2 u^2 -2 S_4 \\
&=&  4 (u^2-S_2)v -  (u^2-S_2)^2    + 2S_2^2 -2 S_4.\end{aligned}$$ Since $F_0$ is linear in $v$, it follows that every factor of $F_0$ is a factor of $(u^2-S_2)$ and of $-(u^2-S_2)^2+2S_2^2 -2 S_4$. But any factor of $u^2-S_2$ must contain $u$, while $2S_2^2 -2 S_4$ does not. Therefore $F_0$ is reducible if and only if $2S_2^2 -2 S_4 = 0$, i.e., if and only if $n=3$, in which case $S_4=x_1^4=(x_1^2)^2=S_2^2$. In this case $F_0$ is as given in (\[uuu\]).

If $t \ne 2$, then $F$ is a quadratic in $v$ with coefficients in $k[x_1,\cdots,x_{n-2},u]$. Also, its leading coefficient $(4-2t)$ is a unit. By (i) and (ii) of Lemma \[Q\], it is reducible if and only if its discriminant $\Delta$ in $v$ is a square in $k[x_1,\cdots,x_{n-2},u]$. Using Maple, we find that $$\begin{aligned}
 \label{D}
\Delta &=&  8t((t-1)u^4-2S_2u^2+(2-t)S_4+S_2^2).\end{aligned}$$ If $t=1$, then $\Delta = 8(-2S_2 u^2 + S_4 + S_2^2)$. By (iii) of Lemma \[Q\], this cannot be a square, since its discriminant in $u$ is $8^3S_2(S_4 + S_2^2) \ne 0$. Thus suppose that $$t \ne 1.$$ Then for $\Delta$ to be a 