Thus, it is easy to verify that $\ker \nu \in \mathcal{P}_{i}$. Hence, $\nu
\in \left\langle \O(X,Y),\mathcal{A}\right\rangle $ by the previous case and it remains to show that $\beta =\varphi \nu \in \left\langle \O(X,Y),\mathcal{A}\right\rangle$. For this let $x\in X$. Then $x\in b_{j}\beta ^{-1}$ for some $j\in
\{1,\ldots ,i\}$, i.e. $x\varphi \nu =a_{j+1}\nu=b_{j}=x\beta $.

  \
The previous proposition shows that $\mathcal{A}$ is a relative generating set for $\OP(X,Y)$ modulo $\O(X,Y)$. It remains to show that $\mathcal{A}$ is of minimal size.

\[le3\] Let $B\subseteq \OP(X,Y)$ be a relative generating set of $\OP(X,Y)$ modulo $\O(X,Y)$. Then $\mathcal{P}_{m}\subseteq \{\ker \alpha :\alpha \in B\}$.

Let $P\in \mathcal{P}_{m}$. Since $\alpha _{P}\in \OP(X,Y)=\left\langle
\O(X,Y),B\right\rangle $, there are $\theta _{1}\in \O(X,Y)\cup B$ and $\theta
_{2}\in \OP(X,Y)$ with $\alpha _{P}=\theta _{1}\theta _{2}$. Because of $\rank\alpha _{P}=m$, we obtain $\ker \alpha _{P}=\ker \theta _{1}$. Since $1\alpha _{P}=n\alpha _{P}$, we conclude that $\theta _{1}\notin \O(X,Y)$, i.e. $\theta _{1}\in B$ with $\ker \theta _{1}=\ker \alpha _{P}=P$.

  \
In order to find a formula for the number of elements in $\mathcal{P}_{m}$, we have to compute the number of possible partitions of $X$ into $m+1$ convex sets. This number is $\left(
\begin{array}{c}
n-1 \\
m\end{array}\right)$. Thus, we have

\[re1\] $\left\vert \mathcal{P}_{m}\right\vert =\left(
\begin{array}{c}
n-1 \\
