m
\end{array}
\right)$.

Now, we are able to state the main result of the section. The relative rank of $\OP(X,Y)$ modulo $\O(X,Y)$ depends of the fact whether both $1$ and $n$ belong to $Y$ or not.

\[th5\] For each $1 < m < n \in \mathbb{N}$,

1.  $\rank(\OP(X,Y):\O(X,Y))=\left(
    \begin{array}{c}
    n-1 \\
    m\end{array}\right)$ if $1\notin Y$ or $n\notin Y$;

2.  $\rank(\OP(X,Y):\O(X,Y))=1+\left(
    \begin{array}{c}
    n-1 \\
    m\end{array}\right)$ if $\{1,n\}\subseteq Y$.

1\. Note that $\ker \eta \in \mathcal{P}_{m}$ and $\eta = \alpha_{P_0}$. Hence, the set $\mathcal{A}=\{\alpha_{P}:P\in \mathcal{P}_{m}\}$ is a generating set of $\OP(X,Y)$ modulo $\O(X,Y)$ by Proposition \[le2\], i.e. the relative rank of $\OP(X,Y)$ modulo $\O(X,Y)$ is bounded by the cardinality of $\mathcal{P}_{m}$, which is $\left(
\begin{array}{c}
n-1 \\
m\end{array}\right)$ by Remark \[re1\]. But this number cannot be reduced by Lemma \[le3\].\
\
2. Let $B\subseteq \OP(X,Y)$ be a relative generating set of $\OP(X,Y)$ modulo $\O(X,Y)$. By Lemma \[le3\], we know that $\mathcal{P}_{m}\subseteq \{\ker \alpha
:\alpha \in B\}$. Assume that the equality holds. Note that $\ker \eta \in \mathcal{Q}_{m}$ and $\eta$ is not order-preserving. Hence, there are $\theta _{1},\ldots ,\theta _{l}\in
\O(X,Y)\cup B$, for a suitable natural number $l$, such that $\eta
=\theta _{1}\cdots \theta _{l}$. From $\rank\eta =m$, we obtain $\ker \theta
_{1}=\ker 