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This is similar to this but different. The original argument requires Nakayama's lemma, I wonder if the following still holds:

Let $\varphi:A \rightarrow A$ be a surjective $k$-algebra homomorphism between finitely generated $k$-algebra $A$. Then $\varphi$ is injective.

Is this true? This would be useful in study of coordinate rings.

Bryan Shih
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    @user: Finitely generated doesn't mean finite-dimensional as $k$-vector spaces. – darij grinberg Oct 13 '18 at 20:09
  • @darijgrinberg Yes. I have missed this part for sure as I have somehow had field extension in mind. Then this does not guarantee f.g. as module. – user45765 Oct 13 '18 at 20:26
  • Are your $k$-algebras commutative? I suspect (but am not sure) that the answer is "no" without commutativity. – darij grinberg Oct 13 '18 at 20:39
  • I was actually working with coordinate rings of affine varieties - which I believe are commutative? So I am also interested in commutative case. – Bryan Shih Oct 13 '18 at 20:47

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For commutative rings, this is certainly true. The crucial hypothesis you need is that $A$ is Noetherian, which is true for finitely generated algebras over a field. You have a chain of ideals, $\ker \phi\subset\ker \phi^2\subset\ker \phi^3\subset\cdots$ and thus there is an $n$ such that $\ker \phi^n=\ker\phi^{n+1}=\cdots$. This can not happen unless $\ker\phi=0$ since $\phi$ is surjective.

Mohan
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  • Sorry, may you elaborate on the last step ? I don't see how $\ker \phi =0$... – Bryan Shih Oct 13 '18 at 23:48
  • Ok . let $a \in A$, so $a = \varphi^n(b_a)$ for some $b_a \in A$. Thus $\varphi(a) = \varphi^{n+1}(b_a)$. If $\varphi(a) =0$, $b_a \in \ker \varphi^{n+1}=\ker \varphi^n$ so $a = 0$. Is this what you had in mind? Or you had simpler proof? – Bryan Shih Oct 14 '18 at 10:55
  • @CL. Your argument is fine. – Mohan Oct 14 '18 at 13:43
  • Thanks a lot Mohan, I think I will leave the question open until someone answer the noncommutative case. – Bryan Shih Oct 14 '18 at 16:50