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Let $A$ be a commutative ring and let $M$ be a finitely generated $A$-module, say(possibly) by $x_1,...,x_n$. Let $\phi$ be a surjective endomorphism of $M$. The generators gives rise to an induced endomorphism $\phi^*$ of $A^n$ defined by one of the representing $n\times n$ matrices $(a_{ij})$ such that $\phi(x_i)=a_{i1}x_1+...+a_{in}x_n$.(Note that the induced endomorphism depends not only on the generators but also the coefficients $a_{ij}$ chosen, so for a fix set of generators we may have different coefficient matrices.)

I would like to show that there is a surjection from $\mathrm{Ker}(\phi^*)$ to $\mathrm{Ker}(\phi)$. I have tried using snake lemma but I am not able to prove the result. Actually I doubt this is correct: could anyone please give a hint for this problem, or a counterexample? Thank you.

I came up with this question after reading the last answer in this post. See the comments to that answer.(Since only surjection from the induced kernel sufficies, I formulate the question in a weaker way than the comment.) Please do not use the result that the surjection $\phi$ is bijective since the goal is to give another proof to this result without Cayley-Hamilton.

William Sun
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  • You should be more clear in your question about the "obvious way". It hides the fact that $\phi^*$ depends on choices, and is not uniquely defined. – Captain Lama Nov 20 '19 at 02:23
  • @CaptainLama I have tried to use the word “possibly”(maybe that’s still too vague?) I will try to edit the question. – William Sun Nov 20 '19 at 02:24
  • It's not just the choice of generators. You also have to choose a decomposition of $f(x_i)$, which is not unique if $M$ is not free. – Captain Lama Nov 20 '19 at 02:26

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