positive integers $\ell_1,\dots,\ell_p$ and an epimorphism $\bigoplus_{i=1}^ps^{-\ell_i}P\to M$ in ${\mathcal C}$,

    2.  for every epimorphism $f:M\to N$ in ${\mathcal C}$ there exists an integer $\ell_0$ such that for every $\ell\geq\ell_0 $ the map $H^0(s^\ell f):H^0(s^\ell M)\to H^0(s^\ell N)$ is surjective.

Using this setup, we can construct the following ${\mathbb{Z}}$-graded ring: $$B := \bigoplus_{\ell=0}^\infty H^0(s^\ell P)$$ and we have the following crucial result, which can be viewed as a generalization of Serre’s theorem:

[@AZ Theorem 4.5]\[Artin-Zhang\] $B$ is right noetherian $k$-algebra, and there is an equivalence of categories $${\mathcal C}\simeq \operatorname{\mathsf{qgr}}B$$ given by $M\mapsto \bigoplus_{\ell=0}^{\infty}H^0(s^\ell M)$.

Now we are ready to prove Theorem \[graded vs order\].

Using Theorem \[graded Morita\] and Propositions \[B and R\^Zc\], we have equivalences $$\begin{aligned}
    \operatorname{\mathsf{coh}}{\mathbb{X}}&=&\operatorname{\mathsf{qgr}}R\\
    &\simeq&\operatorname{\mathsf{qgr}}R^{[{\mathbb{Z}}{\vec c}]}\\
    &\simeq&\operatorname{\mathsf{qgr}}B(\Lambda,L).
$$ We specialise Artin-Zhang Theorem \[Artin-Zhang\] to our case by letting ${\mathcal C}:={\operatorname{\mathsf{mod}}}\Lambda$, $s:=-\otimes_\Lambda L$ and choosing $\Lambda\in{\operatorname{\mathsf{mod}}}\Lambda$ as the distinguished object. It follows easily from ampleness of ${{\mathcal O}}(1)$ on ${\mathbb{P}}^d$ that $s$ is ample. Therefore we have $${\operatorname{\mathsf{mod}}}\Lambda\simeq\operatorname{\mathsf{qgr}}B(\Lambda,L),$$ which completes the proof.

Examples
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To get a better feel for the tilting bundle $T$ from Section \[Tilting\] let us compute ${\operatorname{End}\nolimits}_\Lambda(T)$ in the case $d=1$. In this situation, $[0,{\vec c}]=\{0,{\vec c},a_i{\vec x}_i \;|\;1\leq a_i\leq p_i-1\}$. If 