every prime ideal ${\mathfrak m}\supsetneq \pp \supset I$, then ${\mathcal Z}_\bullet$ is acyclic, $a^*({\mathcal S}_I)+1\leq\nu_0:=2d-2-\operatorname{indeg}(I^{sat})$, and $$\operatorname{reg}({\mathcal S}_I)\leq \nu_0,$$ unless $I$ is a complete intersection of two forms of degree $d$[^1], in which case $a^*({\mathcal S}_I)= d-3$ and $\operatorname{reg}({\mathcal S}_I)=d-1=\nu_0+1$.

The acyclicity of ${\mathcal Z}_\bullet$ is proved in [@BC05 Lemma 4.2]. Furthermore, $H^i_{\mathfrak m}({\mathcal S}_I)=0$ for $i\notin \{0,1,2,3\}$, so we only need to examine $H^i_{\mathfrak m}({\mathcal S}_I)$ for $i\in \{0,1,2,3\}$. The case $i=0$ is treated in [@BC05 Theorem 4.1]. The proof uses the spectral sequence $$\label{eq:spectseq}
    H^j_{\mathfrak m}(Z_{i+j})_{\mu +(i+j)d}\otimes_k R\Rightarrow H^i_{\mathfrak m}( {\mathcal S}_I)_\mu$$ from which we will also deduce the vanishing for $i>0$.

For $i=1$, let $\mu \geq \nu_0$. We will show that $H^2_{\mathfrak m}(Z_1)_{\mu +d}=H^3_{\mathfrak m}(Z_2)_{\mu +2d}=0$ as by it implies that $H^1_{\mathfrak m}({\mathcal S}_I)_\mu=0$ as claimed (recall that $H^1_{\mathfrak m}(Z_0)=H^1_{\mathfrak m}(S)=0$).

By [@BC05 Lemma 4.1], $H^3_{\mathfrak m}(Z_2) \simeq (Z_1)^*[3-4d]$ and we deduce that $$H^3_{\mathfrak m}(Z_2)_{\mu +2d}\simeq \left((Z_1)^*[3-4d]\right)_{\mu+2d}\simeq (Z_1)^*_{\mu-2d+3}.$$ But $2d-\mu -3< \operatorname{indeg}( I^{sat})\leq d$ as $\mu\geq \nu_0$. Hence $(Z_1)_{2d-\mu -3}=0$ since $Z_1\subset S(-d)^4$, proving that $H^3_{\mathfrak m}(Z_2)_{\mu +2d}=0$.

The exact sequence $0\to Z_1\to S(-d)^4\to S\to S/I\to 0$ provides a graded isomorphism $H^2_{\mathfrak m}(Z_1) \simeq  H^0_{\mathfrak m}(S/I)$. By Lemma \[regdim0\], $\operatorname{reg}(S/I)\leq 3d-3-\operatorname{indeg}(I^{sat})=d+\nu_0 -1$, hence $H^2_{\mathfrak m}(Z_1)_{\mu +d}=0$.

For 