Showing a degree formula $\dim_{\mathbb{C}} R^{2} / L$ 
If $a,b,c,d$ are in $R=\mathbb{C}[t]$ and  $ad-bc \ne 0$, $L= R(a,b)+R(c,d)$ in $R^{2}$. I want to show that $\dim_{\mathbb{C}}R^{2}/L = \deg(ad-bc)$.

In a previous theorem it was shown that : $\dim_{\mathbb{C}}R /tR = \deg(t)$. So this can be used, if it helps. 
Does anybody see how to do this? Please, do tell me the right path.
 A: By the theory of congruent matrices over a principal ideal domain, we can assume $b=c=0$, and then the statement is clear.
EDIT. Here are some more details. Put
$$
A:=\begin{pmatrix}a&b\\ c&d\end{pmatrix},
$$
and view this matrix as an $R$-linear transformation of $R^2$. Then we have 
$$
AR^2=R\begin{pmatrix}a\\ c\end{pmatrix}+R\begin{pmatrix}b\\ d\end{pmatrix}.
$$
What we want to check is $\dim(R^2/AR^2)=\deg(\det A)$. 
Clearly, we can replace $A$ by any matrix congruent to $A$, that is by any matrix $BAC$, where $B$ and $C$ are invertible elements of $M_2(R)$. 
It suffices to show that $A$ is congruent to a diagonal matrix.
Note that 
$(*)$ we can swap the rows of $A$, and add to a row of $A$ a multiple of the other row, and similarly for the columns. 
Among all the matrices congruent to $A$, consider one whose first entry is nonzero and has minimum degree among all the nonzero entries of all the matrices congruent to $A$. 
We may assume that this matrix is $A$ itself. 
It suffices to show that $A$ is diagonal.
If $a$ divides $b$ and $c$, we are done by $(*)$. 
Assume by contradiction that $a$ doesn't divide $b$, and write 
$$
b=aq+r,\quad 0\le\deg r<\deg a.
$$
In view of $(*)$ we get the sought-for contradiction be subtracting $q$ times the first column to the second column.
If $a$ doesn't divide $c$, the argument is similar.
As I said in a comment, for further information one can take a look at this related answer.
