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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.

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up vote 2 down vote accepted

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.

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Merci, I don't know the theory of congruent matrices over a principal ideal doamin. Can you tell me what that Theorem says? – VVV Dec 22 '11 at 20:17
Dear @VVV: Tout le plaisir est pour moi! Two $m$ by $n$ matrices with entries in a PID $A$ are congruent iff they represent the same $A$-linear map from $A^n$ to $A^m$ with respect to appropriate basis. You may take a look at this answer. – Pierre-Yves Gaillard Dec 22 '11 at 20:30
Merci monsieur Gaillard! I will look and try to understand! – VVV Dec 22 '11 at 20:35

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