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Suppose $A$ is a commutative ring. Then there does not exist a surjective $A$-module homomorphism from $A^m\rightarrow A^n$ for $n>m$

Proof: Since $A$ is commutative it as a maximal ideal $M$. Therefore, suppose there exists surjection $f:A^m\rightarrow A^n$ where $n>m$. Therefore, $f\otimes Id:A^m\otimes A/M\rightarrow A^n\otimes A/M$ is surjective. This is a surjective linear map from a vector space of dimension $m$ to a vector space of dimension $n$. However, this is not possible.

Is this a valid proof? Proof verification.

Jhon Doe
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  • https://math.stackexchange.com/questions/106786/am-hookrightarrow-an-implies-m-leq-n-for-a-ring-a-neq-0 – Marcos Feb 18 '22 at 15:24
  • The existence of maximal ideals, however, requires using the axiom of choice in the form of Zorn's Lemma. Could the statement be proven in $ZF\neg{C}$ without using Zorn's Lemma? – Geoffrey Trang Feb 18 '22 at 15:27
  • If you are unsure if this is valid, then you suspect one of the steps (all two of them) is invalid. Is it the first step you are worried about? Because the second step is certainly valid. – rschwieb Feb 18 '22 at 15:35
  • I'm sorry which is the first step? – Jhon Doe Feb 18 '22 at 15:37
  • @JhonDoe The one where you conclude $f\otimes Id$ is the way you think it is. (I agree, the argument is hardly even two steps. The second step is just noting the contradiction...) – rschwieb Feb 18 '22 at 15:39
  • Actually never mind. This exact same argument is at the duplicate. – rschwieb Feb 18 '22 at 15:42
  • Oh I saw that it was surjective because every pure tensor has a pre-image. I was wondering about the dimensions of $A^m\otimes A/M$ but I realized that $A\otimes A/M$ is isomorphic to $A/M$ which has dimension 1. – Jhon Doe Feb 18 '22 at 15:42

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