I am trying to understand the particular argument in Lemma 15.102.2's proof. It writes that if $M$ is invertible $R$-module, then

we have an automorphism $M \rightarrow M$ which factors as $$M \rightarrow R^n \rightarrow M.$$

Then it says $M$ is a direct summand of $R^n$. How is this so?


In this case, there are homomorphisms $\phi:R^n\to M$ and $\psi:M\to R^n$ with $\psi\circ\phi=\text{id}_M$. Then $R^n$ is the direct sum of $\text{im}\,\psi$ and $\ker\phi$.

  • $\begingroup$ The problem is I don't see where $\psi$ comes from. $\endgroup$ – CL. May 6 at 11:41
  • 1
    $\begingroup$ @CL. If $u\circ v=w$ and $w$ is an automorphism, then compose with $w^{-1}$ and get... $\endgroup$ – user26857 May 6 at 18:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.