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Let $R = M_2(\mathbb{C})\times M_2(\mathbb{C})$ and let \begin{equation*} e_1 = \left( \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \right),~~ e_2 = \left( \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \right),~~ e_3 = \left( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \right), \end{equation*} Prove that, as $R$-module, $Re_1$ is isomorphic to $Re_2$ but is not isomorphic to $Re_3.$

What I did is \begin{equation} Re_1 = \begin{bmatrix} a_1 & b_1 \\ c_1 & d_1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} a_2 & b_2 \\ c_2 & d_2 \end{bmatrix} \cdot \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = \left( \begin{bmatrix} a_1 & 0 \\ c_1 & 0 \end{bmatrix} , \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \right) \end{equation} Similarly, \begin{equation} Re_2 = \left( \begin{bmatrix} 0 & b_2 \\ 0 & d_2 \end{bmatrix} , \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \right) , \quad Re_3 =\left( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} , \begin{bmatrix} a_3 & 0 \\ c_3 & 0 \end{bmatrix} \right) \end{equation}

I dont see why $Re_1 \cong Re_2$ and why $Re_1, Re_2 \not\cong Re_3$ as $R$-module. What would be the isomorphism in first case? Thanks in advance for any kind of help.

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You can see the answer by thinking of each $R$-module as $\mathbb C^2$ with a particular $R$-action. If you write $(x,y)$ for an element, you can calculate that the $R$-action is \begin{equation} \left(\begin{bmatrix} a_1 & b_1 \\ c_1 & d_1 \end{bmatrix}, \begin{bmatrix} a_2 & b_2 \\ c_2 & d_2 \end{bmatrix}\right) \cdot (x,y) = (a_1x+b_1y,c_1x+d_1y) \end{equation} for $Re_1$ and $Re_2$. So an isomorphism between those modules would identify $\left( \begin{bmatrix} x & 0 \\ y & 0 \end{bmatrix} , \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \right)$ with $\left( \begin{bmatrix} 0 & x \\ 0 & y \end{bmatrix} , \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \right)$. But for $Re_3$, the action is \begin{equation} \left(\begin{bmatrix} a_1 & b_1 \\ c_1 & d_1 \end{bmatrix}, \begin{bmatrix} a_2 & b_2 \\ c_2 & d_2 \end{bmatrix}\right) \cdot (x,y) = (a_2x+b_2y,c_2x+d_2y). \end{equation} I'll leave it to you to think about how to turn this into a proof that an isomorphism does not exist between this module and the other two.

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