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Is there an example of a module over a non-commutative ring in which two maximal independet sets have distinct cardinality?

Let $M$ be a module over a ring $R$. As usually a subset of $M=\{a_1,\dots,\, a_n \}$ will be said free (and its elements independent) if and only if everytime we have something like this $\alpha_1a_1+\dots+\alpha_na_n=0$ (with $\alpha_i\in R$) it follows that all the $\alpha_i$ equals $0$.

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Since a free basis of a free module is a maximal independent set, any free module which does not have the unique rank property will do. In particular, the ring of endomorphisms of infinite-dimensional vector space as a regular module over itself will satisfy this. –  PavelC May 12 at 7:52
    
Let $V$ be an infinite dimensional vector space over the field $F$. Let $R:=End_F(V)$. I think $R$ is isomorphic to $R+R$ as $R$-modules but I cannot show why. Any suggestion? –  W4cc0 May 12 at 9:21
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@W4cc0: Since $V$ is infinite dimensional, there is an isomorphism $V\to V\oplus V$ of vector spaces, which induces an isomorphism $\operatorname{Hom}(V\oplus V,V)\to\operatorname{Hom}(V,V)$ of left $\operatorname{End}(V)$-modules (with endomorphisms acting on the left). –  Jeremy Rickard May 12 at 12:21
    
I've just solved the thing directly (showing two distinct bases) but your method is fine too! Thanks! –  W4cc0 May 12 at 12:37
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@W4cc0 FFR: "non commutative ring" is far more standard and recognizable than "non abelian ring." I know that in some books they are used to mean the same thing, but in most books Abelian isn't used this way. Moreover, some places use "Abelian ring" to mean rings in which idempotents are central. Everybody will be on the same page if you use "commutative" though. –  rschwieb May 12 at 12:43
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