fiber is 1-dimensional.

\[1fiber\] Assume that the $f_i$’s are linearly independent, the fiber over a closed point $\pp$ of coordinates $(p_0:p_1:p_2:p_3)$ is of dimension 1, and its unmixed component is defined by $h_\pp\in S$. Let $\ell_\pp$ be a linear form with $\ell_\pp (p_0,\ldots ,p_3 )=1$ and set $\ell_i (x):=x_i-p_i \ell_\pp (x)$. Then, $h_\pp =\gcd (\ell_0(f),\ldots ,\ell_3(f))$ and $$I=(\ell_\pp (f))+h_\pp (g_0,\ldots ,g_3)$$ with $\ell_i (f)=h_\pp g_i$ and $\ell_\pp (g_0,\ldots ,g_3 )=0$. In particular $$(\ell_\pp (f))+h_\pp (g_0,\ldots ,g_3 )^{sat}\subseteq I^{sat} \subseteq (\ell_\pp (f))+(h_\pp ).$$

A syzygy $L=\sum_{i} a_{i}x_i =\sum_{i} a_{i}(p_i \ell_\pp (x)+\ell_i (x))$ provides an equation for the fiber : $\overline{L}=\sum_{i} a_i p_{i}$. Recall that $h_\pp =\gcd (\overline{L_1},\ldots ,\overline{L_t})$ where the $L_j$ are generators of the syzygies of $I$.

The particular syzygy $L_{(i)}:=\ell_i (f)\ell_\pp (x)-\ell_\pp (f)\ell_i (x)$ satisfies $\overline{L_{(i)}}=\ell_i (f)$. It follows that $h_\pp$ divides $h:=\gcd (\ell_0(f),\ldots ,\ell_3(f))$. Set $\ell_i (f)=hg_i$ for some $g_i\in S$. The $\ell_i$’s span the linear forms vanishing at $p$, and are related by the equation $\ell_\pp (\ell_0,\ldots ,\ell_3 )=0$. It follows that $$I=(\ell_\pp (f),\ell_0(f),\ldots ,\ell_3 (f))=(\ell_\pp (f))+h(g_0,\ldots ,g_3),$$ and $\ell_\pp (g_0,\ldots ,g_3)=0$. Now if one has a relation $a\ell_\pp (f)+\sum_i b_i hg_i=0$, then $h$ divides $a$, as the $f_i$’s have no common factor. This in turn shows 