About Betti Numbers I'm studying the book 'The Geometry of Syzygies' of David Eisenbud, but I'm having problem with the following step, in page 7 he says the we have a free resolution to the set of ten points in $\mathbb{P}^2$ of the form:
$0 \to S(-6) \oplus S(-5) \to S(-4) \oplus S(-4) \oplus S(-3) \to S$
Where $S = \mathbb{K}[x_0,x_1,x_2]$. So we have $F_1 = S(-4) \oplus S(-4) \oplus S(-3)$ and $F_2 = S(-6) \oplus S(-5)$ and $F_0 = S$. Then he defines the betti number $\beta_{i,j}$ of the free module $F_i$ as the minimal generator of degree j that $F_i$ requires.
Then he says that $\beta_{0,0} = 1$ (this part is ok to me because $F_0 = S$) and for example, $\beta_{1,2} = 1$ but I don't understand this, precisely, I don't know how to determinate the degree of the generators of the free modules $F_i$. Does anyone have any tip or reference?
Thank you so much in any advance!
 A: 
Then he defines the betti number $\beta_{i,j}$ of the free module $F_i$ as the minimal generator of degree j that $F_i$ requires.

No, he defines the betti number to be the number of minimal generators of degree $j$ required to generate $F_i$.
The module $F_1$ is just a direct sum of three copies of $S$, so it is generated by $(1,0,0), (0,1,0), (0,0,1)$. Usually we think of $1\in S$ as a degree zero element, i.e. $1\in S_0$. But $S(-d)$ shifts the grading such that $S(-d)_e = S_{e-d}$. For example, in $S(-4)$ we have to look in degree $4$ to find the constants, i.e. $1\in S(-4)_4 = S_{4-4} = S_0$, as mentioned. Thus, in $F_1$ the generators $(1,0,0), (0,1,0)$ have degree $4$, while $(0,0,1)$ has degree $3$.
The twists are cooked up so that the maps in the resolution have degree $0$. So $(1,0,0)$ and $(0,1,0)$ are mapped to degree $4$ elements of $S$, and $(0,0,1)$ goes to a degree $3$ element. Since the image of the last map is the ideal we're interested in, we can read the degrees of its minimal generators right away from the twists in $F_1$.
Note that $\beta_{1,2} = 1$ since $F_1$ has a single generator in degree $1+2 = 3$. Similarly, $\beta_{1,3} = 2$ since $F_1$ has two generators in degree $1+3 = 4$. All other $\beta_{1,d}$'s are zero since there are no more minimal generators for $F_1$. To get the $\beta_{2,d}$'s we play the same game for the kernel of the previous map, known as the first syzygies.
