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I found this decomposition theorem used in a paper I'm reading, but it isn't referenced and I can't seem to find it in any of the books I have:

Every graded module $M$ over a graded PID decomposes uniquely into the form $$({\bigoplus\limits_{i=1}^n \Sigma^{\alpha_i}D}) \oplus ({\bigoplus\limits_{j=1}^m \Sigma^{\gamma_j}D/d_jD})$$ where $d_j \in D$ are homogenous elements so that $d_j\mid d_{j+1}, \alpha_i, \gamma_j \in\mathbb Z$, and $\Sigma^\alpha$ denotes an $\alpha$-shift upward in grading.

Now, it looks like a regular decomposition theorem for modules over PIDs but I am having trouble understanding this "$\alpha$-shift upward". Could you point me in a direction of a text were this is proved? An example would be great, but I suppose this is too much to ask.

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As I was also looking for the answer I found that the shift operator $\sum^{\alpha}$ simply moves an element of grading $i$ into an element of grading $i\ + \alpha$ for every element in D (as D is a graded PID and each homogenous element of D has a grade).

That's what I understand so far from this book http://bit.ly/1srXIs0.

I hope it will help

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  • $\begingroup$ did you by any chance find a proof of this anywhere? $\endgroup$ – user135963 Aug 5 '14 at 20:39
  • $\begingroup$ Okay, the best I can find for the moment is this link bit.ly/1y7iXiw $\endgroup$ – user3174754 Aug 7 '14 at 0:27
  • $\begingroup$ This is the version for the non graded case. $\endgroup$ – user135963 Aug 7 '14 at 7:46
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Indeed but the "proof" can be transpose to our case. For example in the second outline of a proof, $M/t(M)$ is a finitely generated module with a generating set $(e_{1},...,e_{n})$ of homogenous element (we can always find such a set). Then in the proof $R^{n}$ can be replaced by $({\bigoplus\limits_{i=1}^n e_{i}D})$.

Thus you can make a correspondence between $e_{i}D$ and $\Sigma^{\alpha_i} D$ where $\alpha_i$ is the degree of $e_{i}$ and considering that the action of $D$ on $M$ is defined by a bilinear pairing (i.e., $D_{n} \otimes M_{m} \rightarrow M_{n+m}$).

It may work the same way for the second part of the outline but I haven't check yet.

Please tell me if I'm wrong somewhere.

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