# $\phi : A^n \to A^n$ be a surjective $A$-linear map then $\phi$ is injective as well. [duplicate]

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.

## marked as duplicate by user26857 commutative-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 5 '17 at 16:37

• 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 – Watson Feb 5 '17 at 9:41
• I know that..converting $M$ as an $R[X]$ module using $u$, that proof goes. Here I am looking for some independent proof. – user371231 Feb 5 '17 at 9:44

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.)
• In your first paragraph, to use Nakayama, $m$ must be the Jacobson radical rather than a maximal ideal. Does it still work well then? – JWL Feb 5 '17 at 11:31
• Also, I am not even sure that $\ker(T)$ is finitely generated. – JWL Feb 5 '17 at 11:32
• @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. – Chirantan Chowdhury Feb 5 '17 at 11:38
• @ 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 $N=0.$ – user371231 Feb 5 '17 at 14:50
• @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 $A^n$. – user26857 Feb 5 '17 at 16:47