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In a practice exercise on algebra, I want to show that there is an equivalence between the category of all finitely generated $\mathbb{C}^2$-modules and the finitely generated projective $\mathbb{C}^2$-modules. But I can hardly believe it is true (there are so many modules....), and I actually have no clue how I would go about tackling this problem. A module $P$ is projective if every short exact sequence of modules $$ 0\to A\to B\to P\to 0 $$ is split, or, if $P$ is a direct summand of a free $\mathbb{C}^2$-module. This last description seems to be the one I should go for, so how can I prove that any $\mathbb{C}^2$-module is the direct summand of a free $\mathbb{C}^2$-module?

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    $\begingroup$ Hint: what are the local rings of $\Bbb C^2$? $\endgroup$
    – leoli1
    Commented Mar 26, 2021 at 9:27
  • $\begingroup$ $\Bbb C\times 0$ and $0\times \Bbb C$? $\endgroup$
    – user821819
    Commented Mar 26, 2021 at 9:29
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    $\begingroup$ Yes, the point is that these are fields. What can we say about modules over fields? $\endgroup$
    – leoli1
    Commented Mar 26, 2021 at 9:35
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    $\begingroup$ A' doesn't need to be free (over $\Bbb C^2$!), take for example $A_1=\Bbb C, A_2=0$. However in this case we can add $0\oplus \Bbb C$ to get $\Bbb C^2$, a free module!. Something similar should work more generally. I was originally thinking of the following property: A finitely presented module is projective iff it is locally free. $\endgroup$
    – leoli1
    Commented Mar 26, 2021 at 10:12
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    $\begingroup$ Yes, that should work. We then identify $A=A_1\oplus A_2$ with the direct summand $(A_1\oplus 0)\oplus (0\oplus A_2)$ of $(A_1\oplus A_2)\oplus (A_1\oplus A_2)$ $\endgroup$
    – leoli1
    Commented Mar 26, 2021 at 11:18

1 Answer 1

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Hint:

A product of fields is an absolutely flat and noetherian ring, and a finitely generated flat module over a noetherian ring is projective.

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