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Let $S$ be a finite-dimensional vector space. Then $S$ is an irreducible $\operatorname{End}(S)$-module. Furthermore I was told that $S$ is the only irreducible $\operatorname{End}(S)$-module. I guess that this means that if $E$ is an irreducible $\operatorname{End}(S)$-module, then there is a unique isomorphism of modules, i.e. $\operatorname{Hom_{\operatorname{End}(S)}}(S,E)$ contains a unique invertible element. Since I have no prior training in algebra, I am hoping that someone can either name a reference or outline the proof.

Filippo
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    Does this answer your question? Simple $M_n(D)$-module with $D$ a division ring $\text{End}(S)$ is just the matrix algebra. – Amateur_Algebraist Feb 10 '24 at 11:01
  • @Amateur_Algebraist I don't see why that answers my question, could you elaborate please? – Filippo Feb 10 '24 at 11:16
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    Simple = irreducible, and instead of $D$ you have just a field $k$. $S$ is simple since e.g. $GL_n(k) \subset M_n(k) = R$ acts transitively on $S \setminus {0}$. If $S = Rs$, then $\varphi: _RR \to S$, $r \mapsto rs$, is an onto left $R$-module homomorphism; by the first isomorphism theorem $S \cong _RR/\ker \varphi$, but $R$ is semisimple, so $S$ is actually isomorphic to a submodule of $_RR$ = (minimal) left ideal. Then use any of the answers, say the third if you like it. In the interpretation you gave the "uniqueness" does not hold, i.e. an isomorphism should merely exist. – Amateur_Algebraist Feb 10 '24 at 12:28
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    @Amateur_Algebraist Thank you for the comment! Concerning the uniqueness: I think that if $T:S\to E$ does the job, then $\lambda T$ with $\lambda\neq 0$ also does the job, so indeed no uniqueness - but almost: I think that if $T_1$ and $T_2$ are both Clifford isomorphisms from $S$ to $E$, then there is $\lambda\neq 0$ with $T_2=\lambda T_1$. – Filippo Feb 13 '24 at 17:46
  • What's a Clifford isomorphism in this context? What you wrote is indeed correct, it's Schur's lemma. – Amateur_Algebraist Feb 13 '24 at 18:50
  • @Amateur_Algebraist Sorry for the confusion, I am considering the Clifford algebra $\operatorname{Cl}(V)$ of an even-dimensional Euclidean vector space $V$ and the complexification of $\operatorname{Cl}(V)$ is isomorphic to $\mathrm{End}(S)$ for some complex vector space $S$. So I just meant that $T_1,T_2\in\mathrm{Hom}_{\mathrm{End}(S)}(S,E)$ are invertible. Unfortunately I don't see why what I just said follows from Schur's lemma, could you elaborate please? Note that $\lambda$ was meant to be a scalar/a complex number. – Filippo Feb 13 '24 at 20:12
  • For any pair of isomorphisms $T_1$ and $T_2$, $T_2^{-1} T_1$ is an automorphism of $S$. In the link they prove the endomorphism ring to be a skew field, and later in the section they comment that if the base field is algebraically closed, then the skew field in question is equal to the base field. (See e.g. this.) – Amateur_Algebraist Feb 14 '24 at 06:55
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    @Amateur_Algebraist Thank you, but I think that there is a more elementary way which doesn't require the field to be algebraically closed: I used this. – Filippo Feb 14 '24 at 08:37
  • Sorry, I don't follow: why do they a priori have to commute? – Amateur_Algebraist Feb 14 '24 at 13:46
  • @Amateur_Algebraist Suppose that $T_1,T_2\in\mathrm{Hom}_{\mathrm{End}(S)}(S,E)$ are invertible and let $c$ be the action of $S$ on $E$. Then$$\forall A\in\operatorname{End}(S):c(A)=T_i\circ A\circ T^{-1}_i$$for $i=1,2$ and hence$$\forall A:T^{-1}_1T_2A=AT^{-1}_1T_2$$ – Filippo Feb 14 '24 at 14:12
  • So $T_1^{-1}T_2=\lambda 1$ and hence $T_2= \lambda T_1$. – Filippo Feb 14 '24 at 15:37
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    @Amateur_Algebraist Sorry, the first comment should say "action of $\operatorname{End}(S)$ on $E$". – Filippo Feb 15 '24 at 14:38

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