$\rank\gamma = m$, we obtain $\ker \gamma =\ker \theta_{1}= P$. Thus, $\theta_{1}\not\in B$, i.e. $\theta_{1}\in \OP(X,Y)$ and $\ker \theta_{1}\in \mathcal{Q}_{m}\cup \mathcal{P}_{m}$, contradicts $\ker \theta_{1} = P \in \mathcal{R}_{m}$.

  \
Lemma \[le9\] shows that $\rank(\T(X,Y):\OP(X,Y))\geq \left\vert \mathcal{R}_{m}\right\vert $. We will verify the equality.

\[le10\] $\left\vert \mathcal{R}_{m}\right\vert = S(m,n)-\left(
\begin{array}{c}
n \\
m\end{array}\right)$.

The cardinality of the set $\mathcal{D}_{m}:=\mathcal{R}_{m}\cup \mathcal{P}_{m}$ was determined in [@TK]. The authors show that $\left\vert \mathcal{D}_{m}\right\vert
=S(m,n)-\left(
\begin{array}{c}
n-1 \\
m-1\end{array}\right)$. Because of $\mathcal{R}_{m}\cap \mathcal{P}_{m}=\emptyset$, we obtain $\mathcal{R}_{m}=\mathcal{D}_{m}\setminus \mathcal{P}_{m}$. Since $|\mathcal{P}_{m}| =\left(
\begin{array}{c}
n-1 \\
m
\end{array}
\right)$ (see Remark \[re1\]) it follows $\left\vert \mathcal{R}_{m}\right\vert =\left\vert \mathcal{D}_{m}\right\vert
-\left\vert \mathcal{P}_{m}\right\vert =S(m,n)-\left(
\begin{array}{c}
n-1 \\
m-1\end{array}\right) -\left(
\begin{array}{c}
n-1 \\
m\end{array}\right) =S(m,n)-\left(
\begin{array}{c}
n \\
m\end{array}\right)$.

  \
Finally, we can state the relative rank of $\T(X,Y)$ modulo $\OP(X,Y)$.

\[th11\] $\rank(\T(X,Y):\OP(X,Y))=S(m,n)-\left(
\begin{array}{c}
n \\
m\end{array}\right)$.

If $m=1$ then $\T(X,Y)=\OP(X,Y)$, i.e. $\rank(\T(X,Y):\OP(X,Y))=0$. On the other hand, we have $S(1,n)=n=\left(
\begin{array}{c}
n \\
1\end{array}\right) $. Suppose now that $n\geq 2$. By Lemmas \[le9\] and \[le10\], we obtain $\rank(\T(X,Y):\OP(X,Y))\geq \left\vert
\mathcal{R}_{m}\right\vert =S(m,n)-\left(
\begin{array}{c}
n \\
m\end{array}\right) $. In order to prove the equality, we have to find a relative generating set $B$ of $\T(X,Y)$ modulo $\OP(X,Y)$ with $\left\vert
B \right\vert = |\mathcal{R}_{m}|$. We observe 