# relation between f.g. projective $k$-modules and f.g.projective $A$-modules

Let $$A$$ be a $$k$$-algebra such that $$A$$ is finitely generated projective as a $$k$$-module. Since $$A$$, hence $$A^∗$$, is finitely generated projective as a $$k$$-module, it follows that $$A^∗$$ is finitely generated projective as a left $$A$$-module.

I get that $$A^∗$$, is finitely generated projective as a $$k$$-module. How is it finitely generated projective as a $$A$$-module? Is there any relation between f.g. projective $$k$$-modules and f.g.projective $$A$$-modules? Thank you!

The $$k$$ action on an $$A$$ module by definition factors through the map $$\phi: k\to A$$, so any finite set of generators as a $$k$$ module is also a generating set as an $$A$$ module, just by passing from $$k$$ linear combinations to $$A$$ linear combinations via $$\phi$$

• Hello Ben thank you for the answer. Quick question: {f.g.$k$-modules} and {f.g. $A$-modules}. Theses two set are equal? Commented Sep 13, 2020 at 8:26
• Theyre proper classes and not equal. you ned to take into account the map between them coming from the algebra structure to relate the two types of object.
– Ben
Commented Sep 13, 2020 at 9:30
• The standard thing to say would be: the algebra structure map induces a functor from$k$ modules to $A$modules , and the image of a fg module is fg under that functor
– Ben
Commented Sep 13, 2020 at 9:33
• Hello Ben. So if we have a f.g. A-module, it can not be viewed as a f.g k-module? Commented Sep 13, 2020 at 9:50
• Youre correct, what I said before was inaccurate, the functor should go from A modules to k modules, not the other way around. (An A module can be viewed as a k module but not the other way around)
– Ben
Commented Sep 13, 2020 at 14:34