In Rotman's book "Intro to homological algebra" Theorem 3.62 Let $0\rightarrow K\rightarrow\ F\rightarrow A\rightarrow 0$ be an exact sequence of right R-modules, where $F$ is free. The following are equivalent:

  1. $A$ is flat

  2. For every $v \in K$, there is an $R$-map $\theta:F\rightarrow K$ with $\theta(v)=v$.

My problem is that isn't the second condition implies that $A$ is a direct summand of free module $F$, hence projective. Then isn't this theorem is saying that flat implies projective? And we already know that projective modules are flat. So isn't this theorem saying that projective are equivalent to flat which is not true in general. For example $\mathbb Q$ as a $\mathbb Z$-module. So where am I wrong?

  • 1
    $\begingroup$ In TeX and in MathJax, there are special environments for enumerated lists, so don't abuse TeX's {align} environment (meant for math mode) to achieve the effect. $\endgroup$ – LSpice Nov 27 '18 at 18:53
  • $\begingroup$ Why do you think that the second condition implies that $A$ is a direct summand of $F$? It would be if there is a single $\theta$ that works for all $v$; but if every $v$ requires different $\theta$ and you cannot find one $\theta$ that would work for all of them, then you cannot conclude that $A$ is a direct summand. $\endgroup$ – მამუკა ჯიბლაძე Dec 5 '18 at 9:09

The point is that the map $\theta$ is allowed to depend on $v$. So this is not a splitting of the exact sequence.


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.