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By definition, $R$ is a semisimple ring iff it is a semisimple left module over itself. Let $K$ be a two-sided ideal. The left $R$ modules of the form $R/K$ have some extra structure on them given by the fact that scalar multiplication is intimately tied up with multiplication in the ring. This especially helps us when we want to multiply by $1_R$, for ...


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Are these isomorphism classes themselves finite, so that there are only finitely many simple left ideals of any given isomorphism type? No, not in general. Let $F$ be a field, and consider the left ideal $L_\lambda$ of matrices of the form $\left[\begin{smallmatrix}\lambda a& a\\ \lambda b & b\end{smallmatrix}\right]$ where $\lambda,a,b \in F$, ...


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Here's a hint on your third thought. How does one differentiate a determinant? One row/ column at a time, holding the others fixed. Let $V = [v_1,v_2,\cdots, v_n]$ where $v_i'$ are columns. Then, look at the matrix $e^{tA}V.$ Can you differentiate this function at $t=0$ in two ways that correspond to the two sides?


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Hint: consider the descending chain of left ideals $Ra \supseteq Ra^2 \supseteq Ra^3 \supseteq \dots $ and use the left artinian hypothesis.


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Answer summary: The generating function $r(u) = 1 + 3 u + 21 u^2 + 183 u^3 + 1773 u^4 \cdots$ is the unique solution to $$ 4 - 3 r(u)^2 - r(u)^3 + 27 r(u)^3 u=0 \qquad (\ast) $$ with $r(0)=1$. We can use this formula to generate many terms quickly, and to determine the asymptotic behavior of the sequence. To a first approximation, the sequence grows like ...


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Show that for any simple left ideal $S \subseteq R$ and index $j$ $$R_{S, j} = \left\{A \in M_n(R) \ \middle| \ A_{ij} \in S \ \text{and} \ A_{ik} = 0 \ \forall i, k \neq j\right\}$$ is a simple left ideal in $M_n(R)$ and as a module over itself $M_n(R)$ is the direct sum of these when $S$ varies over all simples and $j$ varies over all indices.


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You're on the right track: $M_n(D^{op})\cong (M_n(D))^{op}$ via the transposition map. The only real challenge is checking if the map is multiplicative. Let's use $\circ$ abusively to denote "opposite multiplication" in both rings, and let's use the convention that $T[a_{ij}]=[a_{ji}]=[\overline{a_{ij}}]$. Take two matrices $[a_{ij}],[b_{ij}]\in ...


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If $M$ has some torsion, then $V$ has some torsion. This is nothing surprising, since we have an embedding $M \subseteq V$. On the other hand, suppose that $M$ is torsion-free and, for contradiction, that $x \in V$ is a torsion element, i.e. nonzero such that it is killed by some nonzero $r \in R$. Then $M \cap Rx =0$ (every nonzero element of $Rx$ is ...


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Usually this comes up in the context of Cayley-Hamilton. See that the left side is exactly the $n-1$ order term in $\lambda$ when one starts expanding $\det ( A - \lambda I)$. But Cayley-Hamilton can be derived not in terms of determinants but traces of linear operators on the exterior algebra. In particular, let $A_{(k)}$ denote the natural analogue of ...



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