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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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    Hint: what are the local rings of $\Bbb C^2$? – leoli1 Mar 26 '21 at 09:27
  • $\Bbb C\times 0$ and $0\times \Bbb C$? –  Mar 26 '21 at 09:29
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    Yes, the point is that these are fields. What can we say about modules over fields? – leoli1 Mar 26 '21 at 09:35
  • They are vector spaces –  Mar 26 '21 at 09:37
  • But what does that imply? –  Mar 26 '21 at 09:46
  • Vector spaces are always free. Do you know a characterization of projective modules using local properties? – leoli1 Mar 26 '21 at 09:59
  • I know that projective modules over a local ring are free –  Mar 26 '21 at 10:03
  • And any $\Bbb C^2$-module $A$ is isomorphic $A'=A_1\oplus A_2$, where the $A_i$ are $\mathbb{C}$-modules, hence they are free, so $A'$ is free –  Mar 26 '21 at 10:04
  • https://math.stackexchange.com/questions/2142672/modules-over-direct-sum-of-different-rings - I read it overhere –  Mar 26 '21 at 10:05
  • But this actually shows that $A$ is isomorphic to a free module, rather than a projective module. So that is a bit stronger statement. Is it correct nevertheless?? –  Mar 26 '21 at 10:11
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    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. – leoli1 Mar 26 '21 at 10:12
  • It is true that the sum of free modules is again free. Here we are in the situation where we consider different ground rings: $A_1=\Bbb C$ is free as a $\Bbb C$-module but not as a $\Bbb C^2$-module (where the action is given by $(x,y)\cdot z = xz$) since e.g. $(0,1)$ annihilates $A_1$ – leoli1 Mar 26 '21 at 10:17
  • Oh yes of course :P –  Mar 26 '21 at 10:18
  • Could we then say that $A_1\oplus A_2$ is projective, by saying that $(A_1\oplus A_2)\oplus (A_1\oplus A_2)$ is a free $\mathbb{C}^2$-module, where the action of $(z,w)\in\mathbb{C}^2$ is such that $z$ acts on the first copy of $A_1\oplus A_2$, and $w$ on the second copy of $A_1\oplus A_2$ –  Mar 26 '21 at 10:20
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    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 $(A_1\oplus A_2)\oplus (A_1\oplus A_2)$ – leoli1 Mar 26 '21 at 11:18
  • Side note: $\mathbb C^2$ is a semisimple ring, so all of its modules are projective. So the results sought here are really much weaker, and you can expect much more to be true. I realize the exercise may have been meant to find a more elementary approach. – rschwieb Mar 26 '21 at 11:49

1 Answers1

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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.

Bernard
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