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I have some problems solving the following task:

Let $R = \mathbb{Q}[x,y,z]$ and: $$A = \begin{pmatrix} x & y \\ 0 & z \end{pmatrix} \in M_{2,2}(R) \qquad B = \begin{pmatrix} x & 0 \\ y & z \end{pmatrix} \in M_{2,2}(R).$$ Show that $A$ is not equivalent to $B$, that is, there are no invertible matrices $C,D \in M_{2,2}(R)$ such that $B=CAD$.

I've already tried substituting $C = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and the same for $D$ and then drawing conclusions from the products of the matrices. But I wasn't able to find something helpful.

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  • $\begingroup$ Welcome to MSE! You will probably have better luck getting a helpful response if you explain why you can't solve this problem yourself. Also, what does "equivalent" mean here, similar/conjugate? $\endgroup$ – pjs36 May 11 '15 at 15:58
  • $\begingroup$ @MarkusScheuer Matrices $A$ and $B$ are already fixed and them differ from $I$. $\endgroup$ – Alex Ravsky May 17 '15 at 11:47
  • $\begingroup$ @AlexRavsky: Thanks for your hint. Together with EwanDelanoy's answer it's now clear to me. $\endgroup$ – Markus Scheuer May 17 '15 at 21:51
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Let $R'$ be the ideal of $R$ generated by $x,y,z$ (in other words, $R'$'s elements are the polynomials in $R$ with no constant coefficient). It will suffice to show the following :

Lemma. Suppose that $C,D$ are two matrices in $M_{2,2}(R)$ such that $CB=AD$. Then $\det(C)$ and $\det(D)$ are in $R'$ ; in particular those determinants are not invertible in $R$.

Proof of lemma Write $D=(d_{ij})_{1\leq i,j \leq 2}$ in the usual notation. Comparing the $(1,2)$-coefficients in $CB$ and $AD$, we see that $(1) : xd_{12}+yd_{22}=c_{12}z$. So $xd_{12}\in (y,z)$, and since $(y,z)$ is a prime ideal of $R$ with $x\not\in (y,z)$, we see that $d_{12}\in (y,z)$ (thanks to user26857 for pointing this out) and hence $d_{12}\in R'$. Similarly, (1) forces $d_{22}\in R'$. So $\det(D) \in R'$. The proof for $\det(C)$ is similar.

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