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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!

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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 $\phi$

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  • $\begingroup$ Hello Ben thank you for the answer. Quick question: {f.g.$k$-modules} and {f.g. $A$-modules}. Theses two set are equal? $\endgroup$
    – scsnm
    Commented Sep 13, 2020 at 8:26
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    $\begingroup$ 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. $\endgroup$
    – Ben
    Commented Sep 13, 2020 at 9:30
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    $\begingroup$ 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 $\endgroup$
    – Ben
    Commented Sep 13, 2020 at 9:33
  • $\begingroup$ Hello Ben. So if we have a f.g. A-module, it can not be viewed as a f.g k-module? $\endgroup$
    – scsnm
    Commented Sep 13, 2020 at 9:50
  • $\begingroup$ 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) $\endgroup$
    – Ben
    Commented Sep 13, 2020 at 14:34

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