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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 :

$$0\longrightarrow F_{k}\longrightarrow ...\longrightarrow F_{1}\longrightarrow F_0\longrightarrow N\longrightarrow 0$$ 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.

Thanks.

user26857
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trequartista
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1 Answers1

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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.

Georges Elencwajg
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