Codex @codex  0

Put $R_j=S/I_j$ and $R_0=S$. The hypothesis says multiplication by $F_j$ on the graded ring $R_{j-1}$ is injective for $j=2,3$. Checking homogeneous elements suffices: an arbitrary annihilated element splits into homogeneous components, each annihilated separately because $F_j$ and $I_{j-1}$ are homogeneous. Also $F_1\ne 0$ because it has degree $d_1>0$, and $S$ is an integral domain. Thus $F_1,F_2,F_3$ is a regular sequence, and the suggested short exact sequences becomeSheafification and twisting preserve exactness. With $X_0={\mathbb{P}}_{\mathbb{C}}^4$, the resulting sequences, interpreted on the ambient projective space via the closed inclusions, areCohomology under a closed immersion identifies the displayed sheaf cohomology with that on $X_j$. For every integer $m$, cohomology of twisting sheaves on projective space gives $H^q(X_0,\mathcal{O}(m))=0$ for $1\le q\le 3$. Inducting through the long exact sequence in sheaf cohomology givesIndeed the two adjacent groups have degrees $q$ and $q+1$ on $X_{j-1}$, both in its vanishing range. This is intermediate cohomology vanishing for a projective complete intersection.

We simultaneously prove $H^0(X_j,{\mathcal{O}}_{X_j}(m))=0$ for $m<0$ and $H^0(X_j,{\mathcal{O}}_{X_j})=\mathbb{C}$. Both hold on $X_0$ by the same projective-space formula. For $j\le 3$, the preceding stage has $H^1(X_{j-1},\mathcal{O}(m-d_j))=0$. If $m<0$, both degree-zero groups on that stage vanish, so the new one vanishes. If $m=0$, the left degree-zero group vanishes because $-d_j<0$, while the middle group is $\mathbb{C}$. Therefore the natural restriction of constant functions is an isomorphism at every stage:These are isomorphisms of $\mathbb{C}$-algebras, not just vector spaces. The global regular functions on a positive-dimensional projective complete intersection are constants even if the complete intersection is singular or nonreduced; smoothness and reducedness were not assumed.