I'm searching for nice isomorphism theorems for non-finitely generated $R$-modules.

I guess that if $A$ is an $R$-module which is not finitely generated, then there is an isomorphism between $A\oplus A$ and $A$.

I really hope it works, I'm able to prove this for countable generated modules, but I don't know how to prove it in general.

Another question is,

if I have an automorphism $\varphi:A\to A$ is then $\varphi\oplus \varphi: A\oplus A\to A\oplus A$ an isomorphism which is conjugated to $\varphi \oplus \text{id}_A$? [Like above $A$ is not finitely generated.]

This question is motivated by the definition of $K_1$ as universal determined and so on..., because in the defintion it is restricted to finitely generated projective $R$-modules, but I think it is no problem if we allow $A$ to be a not finitely generated projective $R$-module, because the above statement/conjecture is true.

Thanks for your help.

  • $\begingroup$ I think you are talking about $K_0$ rather than $K_1$. And I think Eilenberg–Mazur swindle (here) is helpful for your understanding of the definition. $\endgroup$
    – asatzhh
    Aug 4, 2013 at 13:02
  • $\begingroup$ No its $K_1(R)$: It's the free abelian group generated by the conjugacy classes of automorphisms of finitely generated projective $R-$modules, such that we have a composition formula $[\varphi_2\circ\varphi_1]=[\varphi_2]+[\varphi_1]$ and if we have a short exact sequence and automorphism, such that it commutes, then we want that the sum of the first and the last is the conjugacy class of the automorphism in the middle. $\endgroup$
    – Mebat
    Aug 4, 2013 at 14:20
  • $\begingroup$ Well, thanks. It's the first time I meet such a definition of $K_1$, what I have learned about $K_1$ before is the maximal abelian quotient of $GL(R)$. $\endgroup$
    – asatzhh
    Aug 4, 2013 at 14:33
  • $\begingroup$ yeah, both definitions are equivalent, but this is nicer. $\endgroup$
    – Mebat
    Aug 4, 2013 at 17:17
  • $\begingroup$ Would you please give me some ideas about the reason of being nicer? $\endgroup$
    – asatzhh
    Aug 5, 2013 at 8:47

1 Answer 1


I am not sure about your first statement. Let $p$ be a prime, and consider the $\mathbb{Z}$-module $$ A = \mathbb{Z} / p^{2} \mathbb{Z} \oplus \mathbb{Z} / p \mathbb{Z} \oplus \mathbb{Z} / p \mathbb{Z} \oplus \mathbb{Z} / p \mathbb{Z} \oplus \dots. $$ It does not look like $A \cong A \oplus A$.

Then, if you mean that two automorphisms $\alpha, \beta$ are conjugated if there is another automorphism $\gamma$ such that $\beta = \gamma \alpha \gamma^{-1}$, then I don't think the second statement is valid either.

Note first that if $F$ is the set of fixed point of $\alpha$, then $\gamma(F)$ is the set of fixed points of $\beta$.

Consider $A$ to be the vector space over $\mathbb{Z} / p \mathbb{Z}$ with countable basis $e_{1}, e_{2}, \dots$. Take $\varphi : e_{i} \to e_{i} + e_{i+1}$. This has no non-trivial fixed points, and the same holds true for $\varphi \oplus \varphi$ on $A \oplus A$. So $\varphi \oplus \varphi$ cannot be conjugated to $\varphi \oplus \operatorname{id}_{A}$, which has plenty of fixed points.

  • $\begingroup$ thanks for your help, I#ll check it. $\endgroup$
    – Mebat
    Aug 4, 2013 at 14:18
  • $\begingroup$ You're right. Do you know what happens, if you have projective modules? $\endgroup$
    – Mebat
    Aug 4, 2013 at 14:27
  • $\begingroup$ then it is torsion free, isn't it. then should it work. Do I need flatness? What about the second question? $\endgroup$
    – Mebat
    Aug 4, 2013 at 17:18

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