Finitely generated modules over $\mathbb{C}^2$ are projective.

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?

• Hint: what are the local rings of $\Bbb C^2$? Commented Mar 26, 2021 at 9:27
• $\Bbb C\times 0$ and $0\times \Bbb C$?
– user821819
Commented Mar 26, 2021 at 9:29
• Yes, the point is that these are fields. What can we say about modules over fields? Commented Mar 26, 2021 at 9:35
• 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. Commented Mar 26, 2021 at 10:12
• 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 $(A_1\oplus A_2)\oplus (A_1\oplus A_2)$ Commented Mar 26, 2021 at 11:18