given ideal $J$ stands for its minimal number of generators.

The following lemma will be useful in some places (see [@SZ Corollary A12] for a self contained elementary proof) :

\[regdim0\] Let $J$ be a graded ideal in $S$, generated in degree $d$ such that $\dim(S/J)\leq 1$ and $\mu(J_\pp)\leq 3$ for every prime ideal ${\mathfrak m}\supsetneq \pp \supset J$. Then, $$\operatorname{reg}(S/J^{sat})\leq 2d-3$$ and $$\operatorname{reg}(S/J)\leq 3d-3-\operatorname{indeg}(J^{sat})$$ unless $\mu (J)=2$, in which case $J=J^{sat}$ and $\operatorname{reg}(S/J)=2d-2$.

We may assume that $k$ is infinite. Let $f$ and $g$ be two general forms of degree $d$ in $J$. They form a complete intersection. Hence, by liaison (see e.g. [@CU02 4.1 (a)]) $$\operatorname{reg}(S/J^{sat})=\operatorname{reg}(S/(f,g))-\operatorname{indeg}(((f,g):J^{sat})/(f,g)).$$ Hence $\operatorname{reg}(S/J^{sat})\leq 2d-3$, unless $(f,g)\supseteq J^{sat}$ in which case $J=(f,g)$.

Let $J'\subseteq J$ be an ideal generated by $3$ general forms of degree $d$. As $\mu(J_\pp)\leq 3$ for every prime ideal ${\mathfrak m}\supsetneq \pp $, $J'$ and $J$ have the same saturation. Therefore, it suffices to show that $H^0_{\mathfrak m}(S/J')_{\mu +d}=0$. Observe that by Koszul duality (see for instance [@Ch04 Lemma 5.8]) $H^0_{\mathfrak m}(S/J')_{\mu +d}=0$ is equivalent to $H^0_{\mathfrak m}(S/J')_{2d-3-\mu}=0$. But since $2d-3-\mu <d$ we indeed have $$H^0_{\mathfrak m}(S/J')_{2d-3-\mu}=(J')^{sat}_{2d-3-\mu}=J^{sat}_{2d-3-\mu}=0$$ since $2d-\mu -3<\operatorname{indeg}( J^{sat})$.

\[prop:regSym\] Assume that $\dim(S/I)\leq 1$ and $\mu(I_\pp)\leq 3$ for 