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I am now reading a paper about Castelnuovo-Mumford regularity and in this paper, there is a notation as following:

Let $S=k[x_1,...,x_{n}]$, by Hilbert's syzygy theorem, if $N$ is a graded module over $S$ then $N$ has a free graded resolution over $S$ of the form :

$$o\longrightarrow F_{k}\longrightarrow ...\longrightarrow F_{1}\longrightarrow F_0\longrightarrow N\longrightarrow 0$$ where $F_{i}=\bigoplus_{j=1}^{t_{i}}S(-a_{ij})$.

My question is what is $S(-a_{ij})$ ? What does it mean ?

I am so sorry if the title is not gentle and soft enough. Moderator please do not delete my question, edit it please.


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I don't see anything ungentle or hard about the title, but it's also not very specific. Something like "Notation for free graded resolutions of graded modules" would be much more informative. If you imagine the main page full of titles of one or the other kind, you can probably see why such a title would be preferable. It's also likely to attract more people to your question who actually know stuff about free graded resolutions of graded modules. –  joriki Aug 12 '12 at 17:10
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1 Answer

If $M$ is a graded module and $a\in \mathbb Z$ an integer , the notation $M(a)$ denotes the same module with its gradation shifted as follows: $(M(a))_n=M_{a+n}$.
This is not profound at all but is a trick that allows one to say that some morphisms will be graded of degree zero while they wouldn't be without that shift.

A simple example
Take $S=k[x,y]$ and suppose you want to resolve the ideal $I=(x,y)\subset S$.
You write $$ 0\to S(-2)\xrightarrow {u} S(-1)\oplus S(-1)\xrightarrow {v} I\to 0 $$ where $u(R)=(yR,-xR), \; v(P,Q)=xP+yQ$
The point is that if $R$ is a homogeneous polynomial of degre $5$, say, it will live in $S(-2)_7$ and its image $(xR,-yR)$ consisting of two polynomials of degree $6$ will live in $(S(-1)\oplus S(-1))_7$: the degrees of the polynomials have changed but not the grading of $7$.
This is the advantage of the shift: it ensures that $u$ and $v$ are graded of degree zero.

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Dear Georges, maybe you should just use $x,y$ instead of $x_0,x_1$ but not both? –  Manos Dec 10 '13 at 22:46
Dear @Manos: of course! Thanks a lot for catching this typo, which I just corrected. –  Georges Elencwajg Dec 11 '13 at 11:29
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