as a subring of $M_{p_1\cdots p_\ell}(R)$.

Let $$M:=\left[\begin{array}{c}
R/(a_1,\ldots,a_\ell)\\ 0\\ \vdots\\ 0\\ 0
\end{array}\right]\simeq
\bigotimes_{i=1}^\ell\left[\begin{array}{c}
R/(a_i)\\ 0\\ \vdots\\ 0\\ 0
\end{array}\right]\in{\operatorname{\mathsf{mod}}}B.$$ Since $R/{\mathfrak m}$ is an $R/(a_1,\ldots,a_\ell)$-module with projective dimension $d-\ell$, we have an exact sequence $$0\to M_{d-\ell}\to\cdots\to M_0\to S\to0$$ of $B$-modules with $M_i\in\operatorname{\mathsf{add}}M$. On the other hand, the $B$-module $M$ has projective dimension $\ell$ since it has a projective resolution $$\bigotimes_{i=1}^\ell\left(
\left[\begin{array}{c}
(a_i)\\ R\\ \vdots\\ R\\ R
\end{array}\right]\to
\left[\begin{array}{c}
R\\ R\\ \vdots\\ R\\ R
\end{array}\right]
\right).$$ Thus the $B$-module $S$ has projective dimension $d$.

\[global dimension\]

-   The order $\Lambda_x$ has global dimension $d$ for all closed points $x\in{\mathbb{P}}^d$.

-   The order $\Lambda$ has global dimension $d$.

\(a) Note that ${{\mathcal O}}_x:={{\mathcal O}}_{{\mathbb{P}}^d,x}$ is a regular local ring of dimension $d$. Let $I_x:=\{i\mid 1\le i\le n,\; x\in L_i\}$. Then we have $$(\Lambda_i)_x=\left\{\begin{array}{cc}
{\rm T}_{p_i}({{\mathcal O}}_x,(a_i))&\mbox{if $i\in I_x$,}\\
{\rm M}_{p_i}({{\mathcal O}}_x)&\mbox{if $i\notin I_x$.}
\end{array}\right.$$ for $a_i\in {{\mathcal O}}_x$ defining $L_i$ locally. Our Assumption \[general position\] implies that $(a_i)_{i\in I_x}$ is a regular sequence of $R$. Thus $\Lambda_x$ is Morita-equivalent to $$\bigotimes_{i\in I_x} {\rm T}_{p_i}({{\mathcal O}}_x,(a_i)),$$ and the statement now follows from Lemma \[gldimmat\].

\(b) For $X,Y\in{\operatorname{\mathsf{mod}}}\Lambda$, we need show that ${\operatorname{Ext}\nolimits}^i_\Lambda(X,Y)=0$ for all $i>d$. We use an argument similar to 