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Consider the following two properties for an $R$-module $M$:

  1. For every endomorphism $f:M\rightarrow M$, there exist central endomorphisms $g, h\in \operatorname{End}_R(M)$ such that the sequence $M_{R}\xrightarrow g M_{R} \xrightarrow f M_{R} \xrightarrow h M_R$ is exact.

  2. For every endomorphism $f:M\rightarrow M$, there exist a central endomorphism $\alpha:M\rightarrow M$ such that the sequence $M_{R}\xrightarrow\alpha M_{R} \xrightarrow f M_{R} \xrightarrow\alpha M_R$ is exact.

For an $R$-module $M$, does the property 1 imply the property 2?

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  • $\begingroup$ What do you mean by a central endomorphism of an $R$-module? $\endgroup$
    – user473423
    Commented Oct 18, 2023 at 7:51
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    $\begingroup$ @Echo My guess the element of the centralizer subring $Z(End_R(M))$ $\endgroup$
    – Sampah
    Commented Oct 18, 2023 at 10:45
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    $\begingroup$ Yes. This means the center of the ring endomorphism of M. Means Cent(${\rm End}_{R}(M)$ $\endgroup$
    – Najmeh Dehghani
    Commented Oct 18, 2023 at 11:01

1 Answer 1

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If $M$ has finite dimension over some field then this is true. This is because if $g,f$ is exact then the nullity of $f$ is equal to the rank of $g$. It follows that the rank of $f$ is equal to the nullity of $g$. By centrality, $gf=0$ is equivalent to $fg=0$ so $f,g$ is exact. So we can just take $\alpha=g$. If $M$ is not finite dimensional, I'm betting there's some sort of Eilenberg swindle you can pull to construct a counterexample, but I didn't yet manage to write it down.

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