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The reason I ask this question is I found two different statements about Hilbert's syzygy theorem from Jacobson's Basic Algebras 2nd and Wikipedia. Please have a look at the following pictures. The first one is from the former and the second one is from the latter. enter image description here


enter image description here
In this two statements I found the modules required are different. One is called graded and one is called finitely generated. But the results are the same. So I want to ask are these two kinds of modules equivalent? Or there is some connection between them. Someone has told me that there are two editions about this theorem. If this is the case, then which one is more general or widely used?
At the same time, I found Quillen-Suslin theorem from wikipedia, which states that every finitely generated projective module over a polynomial ring is free. And from Jacobson's book mentioned above there is a corollary enter image description here
Here I consider the corollary as Any projective graded R-module for $R=F[x_1,x_2,...,x_m]$ is free. I don't know whether I understand it in a correct way or not. Then are these two statements about projective modules to be free the same?
Thank you for your assistance.

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Presumabl, the map $d_m$ in the first statement is begin assumed to be injective, for otherwise either it is false or does not make sense. –  Mariano Suárez-Alvarez May 15 '13 at 6:07
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1 Answer

up vote 1 down vote accepted

The two theorems are not the same. One is about graded modules and the other is about modules.

The two results are related, though. As you say, every projective graded module (that is every projective object of the category of graded modules and homogeneous maps of degree zero) is free as a module, and the conection between the two theorems stems from that.

The second theorem is true even if you remove the hypothesis that the module $M$ be finitely generated.

Finally, that corollary of Quillen-Suslin should be understood in the sense that every projective module is graded free for some grading which is to be found.

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Graded modules are modules. Then isn't the second Edition of the theorem more powerful? Then why do we still have the first edition? One more question, a module is graded free means that the graded module is free? I can't find the definition of graded free. Thanks. –  Andylang May 15 '13 at 7:29
    
The second version does not tell you that you can choose the maps in the resolution to be homogeneous of degree zero. –  Mariano Suárez-Alvarez May 15 '13 at 7:39
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