Uniform regularity for all quotients with a fixed Hilbert polynomial

Extend to an infinite field as in [F2]; it suffices to bound regularity there. Induct on $n$. In dimension zero every coherent sheaf is regular for every integer, so take $R=0$. For $n>0$, choose a hyperplane avoiding the associated points of both $K$ and ${\mathcal{O}}^p\mathrm{/}K$. Its equation is injective on both, so the Tor exact sequence gives $K_H\subseteq {\mathcal{O}}_H^p$, and $0\to K(t-1)\to K(t)\to K_H(t)\to 0$ is exact. The polynomial of $K_H$ is $\mathrm{\Delta }Q(t)=Q(t)-Q(t-1)$; hence by induction it is $a$-regular with $a=\max (0,R(n-1,p,\mathrm{\Delta }Q))$, depending only on the fixed data.

For $i\ge 2$ and $t\ge a-i$, both $H^{i-1}(K_H(t+1))$ and $H^i(K_H(t+1))$ vanish by [F1]. Thus $H^i(K(t))\to H^i(K(t+1))$ is an isomorphism. Iterating to a Serre-vanishing twist proves $H^i(K(t))=0$ in this entire range. For $t\ge a$, $H^1(K(t-1))\twoheadrightarrow H^1(K(t))$, so $h^1$ is nonincreasing. If two consecutive dimensions agree, the restriction $H^0(K(t))\to H^0(K_H(t))$ is surjective. The multiplication argument in [F1] propagates that surjectivity to every larger twist; the same exact sequence then makes $h^1$ constant thereafter, so Serre vanishing forces it to be zero. Therefore every positive value of $h^1(K(t))$ strictly decreases at the next twist for $t\ge a$.

At twist $a$, the higher groups with index at least two vanish, so $h^1(K(a))=h^0(K(a))-Q(a)\le p\left(\genfrac{}{}{0px}{}{n+a}{n}\right)-Q(a)$. Put $B=\max (0,p\left(\genfrac{}{}{0px}{}{n+a}{n}\right)-Q(a))$ and $R=a+B+1$. Step 2.1 gives $H^1(K(R-1))=0$ after at most $B$ decreases and gives $H^i(K(R-i))=0$ for $i\ge 2$. Hence $K$ is $R$-regular. This recursive integer is enough; no polynomial formula for the bound is claimed. For ${\mathcal{O}}^p\twoheadrightarrow F$, use $Q=p\left(\genfrac{}{}{0px}{}{n+t}{n}\right)-P(t)$, increase $R$ to at least one, and use the long exact sequence: regularity of $K$ and ${\mathcal{O}}^p$ makes $F$ $R$-regular.

For $E={⨁}_j\mathcal{O}(a_j)$ choose $b\ge {\max }_j{a}_j$. On each summand multiplication by $x_0^{b-a_j}$ embeds $\mathcal{O}(a_j)$ into $\mathcal{O}(b)$; multiplication is injective since projective space over a field is integral. Thus $K(-b)\subseteq {\mathcal{O}}^p$ has the fixed polynomial $\chi (E(t-b))-P(t-b)$, and step 3.1 bounds its regularity uniformly. Twisting back bounds $K$; increase the bound to make every summand of $E$ regular as well, and the exact sequence bounds $F$. This argument needs no information about the individual quotient beyond its polynomial.