How Does Condition (2) Imply Condition (3) in Bland's Proposition 3.2.7?

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SUMMARY

The forum discussion centers on understanding Proposition 3.2.7 from Paul E. Bland's "Rings and Their Modules," specifically the proof that condition (2) implies condition (3). Participants clarify that if $$M$$ is expressed as a direct sum of $$\text{Ker } g$$ and a submodule $$N$$, then by the First Isomorphism Theorem, $$M/\text{Ker } g$$ is isomorphic to $$g(M)$$. The discussion also references Rotman's counterexample from "An Introduction to Homological Algebra," highlighting the nuances between exact sequences and split sequences in module theory.

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  • #31
steenis said:
The proof of (1)=>(3) may be looking correct, but I want to know if it is correct. In contrary, the proof confuses me.

Ok, sorry, here I was wrong which might be clear, because I was quoting you.

So we agree that (3)=>(1) is wrong. What must we do to finish this thread? I think a proof of the non-equivalence of S1 and S0 in post #27. And, if possible, a direct and easy counterexample of (3)=>(1).
 
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  • #32
I already gave a direct counterexample to $(3)\implies (1)$, which is actually the same as the the one in the MSE link you provided.
 
  • #33
Where is your counterexample, I cannot find it?
 
  • #34
It's at the end of post #28, but it's the same as the k.stm's example in your link.
 
  • #35
That specific example is a direct counterexample of "(3) => the ses splits" and therefore an indirect counterexample of (3) => (1). A direct counterexample of (3) => (1) is example in which (3) is true and (1) is not true, that is how I have learned it in my university years. Of course that all in the context of Bland's proposition and if possible very easy. It is not necessary, not mandatory, because the job is already done. It is just meant to enlighten that (3)=>(1) is faulty. I do not know if it is possible or it is too difficult.

I want to stop this discussion now, the only thing that is open is the non-equivalence of S1 and S0 in post #27.
 
  • #36
That does not make sense, since any counterexample to $(3)\implies (1)$ is a a counterexample to the implication that $(3)$ implies the ses splits; for a split exact sequence necessarily satisfies $(1), (2)$, and $(3)$. Seeing that you want to stop the discussion, I'll leave it to other users address your concerns.
 
  • #37
Euge said:
That does not make sense, since any counterexample to $(3)\implies (1)$ is a a counterexample to the implication that $(3)$ implies the ses splits; for a split exact sequence necessarily satisfies $(1), (2)$, and $(3)$. Seeing that you want to stop the discussion, I'll leave it to other users address your concerns.
Thanks to Euge and Steenis for clarifying matters ...

There are some further helpful and informative posts on the Physics Forums in the sub-forum Linear and Abstract Algebra ... here

https://www.physicsforums.com/threa...-bland-proposition-3-2-7.881174/#post-5538854

Hope that helps ...

Peter
 

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