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This question already has an answer here:

Let $A$ be a commutative ring with $1,$ and $\phi : A^n \to A^n $ be a surjective $A$-linear map for some natural number $n.$ Then show that $\phi$ is injective as well.

Tensoring with $A/m$ for some maximal ideal $m$ in $A$ will give that tensor map is onto and being $A/m$ linear is injective. But from this map I cannot recover $\phi$ and claim that $\phi$ is also injective. Any help will be appreciated. Thanks.

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marked as duplicate by user26857 commutative-algebra Feb 5 '17 at 16:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Notice that it follows from a general result by Vasconcelos: let $R$ be a commutative ring, and let $M$ be a finitely generated $R$-module. Let $u : M \to M$ be a surjective $R$-module endomorphism. Show that $u$ is injective $\endgroup$ – Watson Feb 5 '17 at 9:41
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    $\begingroup$ I know that..converting $M$ as an $R[X]$ module using $u$, that proof goes. Here I am looking for some independent proof. $\endgroup$ – user371231 Feb 5 '17 at 9:44
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Correct me where i am wrong : We have an exact sequence : $ 0 \to \ker(T) \to A^n \xrightarrow{\phi} A^n \to 0$ of $A$-modules. Now as $A^n$ is free the sequence splits, thus $ A^n = A^n \oplus \ker(T)$. And now tensoring with $A/m$ now might help in adding the dimensions and then by Nakayama lemma we can show that $\ker(T) =0$.

Also thinking in the way you have done , we get that $ \phi(a_1,a_2,...,a_n)= 0 $ when on tensoring these $a_i$'s lie in m for every maximal ideal m. That implies all the $a_i$'s lie in the Jacobson radical $\mathfrak{R}$. Now consider $ \ker (T)$. We see that $\mathfrak{R}\ker(T) = \ker (T)$. Thus by Nakayama lemma then $\ker(T) = 0 $. (I might be a bit wrong here to so please correct me.)

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    $\begingroup$ In your first paragraph, to use Nakayama, $m$ must be the Jacobson radical rather than a maximal ideal. Does it still work well then? $\endgroup$ – JWL Feb 5 '17 at 11:31
  • $\begingroup$ Also, I am not even sure that $\ker(T)$ is finitely generated. $\endgroup$ – JWL Feb 5 '17 at 11:32
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    $\begingroup$ @JWL in Atiyah Mcdonald Ex 12 of modules states that if we have a finitely generated $A$ module M and a surjective morphism $ \phi : M \to A^n$ ,then $ \ker(\phi)$ is finitey generated. $\endgroup$ – Chirantan Chowdhury Feb 5 '17 at 11:38
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    $\begingroup$ @ Watson May be the following happens. After tensoring with $A/m$ for some maximal ideal $m$(in the first paragraph) we have $N=mN$ where $N=ker(\phi)$ i.e., by NAK $Ann_{A}(N)+m=A$ for all maximal ideal $m.$ Now if $Ann_{A}(N)$ is proper ideal, it must be contained in some other maximal ideal which gives a contradiction. Thus $1\in Ann_{A}(N)$ and $N=0.$ $\endgroup$ – user371231 Feb 5 '17 at 14:50
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    $\begingroup$ @JWL About your second comment: the answerer wrote $ A^n = A^n \oplus \ker(T)$, but this means $ A^n \simeq A^n \oplus \ker(T)$. Then $\ker T$ is isomorphic to a quotient module of $A^n$. $\endgroup$ – user26857 Feb 5 '17 at 16:47

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