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I'd like to prove that if $M$ is a finitely generated module over a ring R then any surjective endomorphism $f$ of $M$ is injective. I do not want to appeal to Nakayama's lemma.

I have a "solution" which I think is incorrect, since I have nowhere used the surjection...

Make M into an $R[x]$ module via $f$, $f(m) = Xm$. Let $\{m_i, i = 1,...,n\}$ generate $M$. Let $f(m_i) = \sum_j\mu_{ij}m_j$ so that $\sum \delta_{ij}Xm_j = \sum\mu_{ij}m_j$ and therefore $\sum_j (\delta_{ij}X - \mu_{ij})m_j = 0$

Let $c_{ij} = (\delta_{ij}X - \mu_{ij})$ and $C = (c_{ij})$.

Then $adjC C = detC I_n$.

So, since $C(m_i) = 0$, $detC(m_i) = 0$ for all $i$, and as detC is $1+X$(stuff), $f^{-1} = X^{-1} =$ -(stuff)

probablystuck
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1 Answers1

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The key part of the proof is that if $ f : M \to M $ is a surjective endomorphism, then turning $ M $ into a $ R[X] $-module in the standard way, you have the equality $ XM = M $ (this is where we use surjectivity), so that for a generating set $ m_1, m_2, \ldots, m_n $ of the finitely generated $ R $-module $ M $, we can write

$$ m_i = \sum_{k=1}^n X c_{ik} m_k $$

with $ c_{ik} \in R $ for each $ i $. (If you don't see how this differs from what you did, the positions of $ X m_k $ and $ m_i $ have been reversed). Now, writing this as a matrix equation and using the same idea yields things of the form $ 1 - rX $ on the diagonal and $ rX $ off the diagonal; hence the determinant will clearly have constant term $ 1 $.

Ege Erdil
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