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The definition of the simple components of a semisimple ring which I learned is as follow (see Rotman's Advanced Modern Algebra):

Let $R$ be a left semisimple ring, and let $$R=L_1\oplus \cdots \oplus L_n,$$ where the $L_p$ are minimal left ideals. Reindex the summands so that no two of the first $r$ ideals $L_1, ..., L_r$ are isomorphic, while every $L_p$ in the given decomposition is isomorphic to some $L_i$ for $1\leq i\leq r$. The left ideals $$B_i=\bigoplus_{L_p\cong L_i}L_p$$ are called the simple components of $R$ relative to the decomposition $R=\bigoplus_p L_p$.

Now, suppose that $R=M_2(\Bbb{C})\times M_2(\Bbb{C})$. Let $L_1=\left\{\begin{pmatrix}a&0\\b&0\end{pmatrix}\mid a, b\in \Bbb{C}\right\}$ and $L_2=\left\{\begin{pmatrix}0&c\\0&d\end{pmatrix}\mid c, d\in \Bbb{C}\right\}$. $L_1$ and $L_2$ are both minimal left left ideals in $R$. According to this definition, there is only one simple component of $R$, which is $B=L_1\oplus L_2\oplus L_1\oplus L_2$ because $L_1\cong L_2$ as an $R$-module. However, according Lam's definition (see Lectures on Modules and Rings), there are two simple components of $R$. Are there something I misunderstand?

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Let $L_1'$ and $L_2'$ denote the "version" of $L_1$ and $L_2$ in the second copy of $M_2(\mathbb C)$.

It is true that $L_1\cong L_2\cong L_1'\cong L_2'$ as $M_2(\mathbb C)$ modules, and it is also true that $L_1\cong L_2$ and $L_1'\cong L_2'$ as $R$ modules.

But $L_1\ncong L_1'$ as $R$ modules!

$L_1$ is annihilated by $\{0\}\times M_2(\mathbb C)$ but not by $M_2(\mathbb C)\times\{0\}$, and $L_1'$ is annihilated by $M_2(\mathbb C)\times\{0\}$ but not by $\{0\}\times M_2(\mathbb C)$.

So, you still get two components using the definition in question.

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  • $\begingroup$ Ah! I see. Thanks. $\endgroup$ – bfhaha Nov 27 '17 at 15:48

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