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I wanted to leave this as a comment, but for some reason I can't.

I gave a short answer to this question here: http://math.stackexchange.com/questions/209218/homogeneous-ideal-and-degree-of-generators.

A more general answer says the following: if $K$ is a field, $R$ is an $\mathbb{N}$-graded $K$-algebra finitely generated over $K$, and $M$ a $\mathbb{Z}$-graded finitely generated $R$-module, then

$$\beta_{ij}(M)=\dim_K\operatorname{Tor}_i^R(K,M)_j,$$

where $\beta_{ij}(M)$ are the graded Betti numbers of $M$.

8 added 58 characters in body

I wanted to leave this as a comment, but for some reason I can't.

I gave a short answer to this question here: http://math.stackexchange.com/questions/209218/homogeneous-ideal-and-degree-of-generators.

A more general answer says the following: if $R$ is a graded $K$-algebra, $K$ is a field, $R$ is an $\mathbb{N}$-graded $K$-algebra finitely generated over $K$, and $M$ a graded $\mathbb{Z}$-graded finitely generated $R$-module, then

$$\beta_{ij}(M)=\dim_K\operatorname{Tor}_i^R(K,M)_j,$$

where $\beta_{ij}(M)$ are the graded Betti numbers of $M$.

7 added 10 characters in body; deleted 103 characters in body; deleted 2 characters in body; added 2 characters in body

I wanted to leave this as a comment, but for some reason I can't.

I gave a short answer to this question here: http://math.stackexchange.com/questions/209218/homogeneous-ideal-and-degree-of-generators.

A more general answer says the following: if $R$ is a graded $K$-algebra, $K$ a field, and $M$ a graded finitely generated $R$-module, then

$$\beta_{ij}(M)=\dim_K\operatorname{Tor}_i^R(K,M)j,$$ $\beta_{ij}(M)=\dim_K\operatorname{Tor}_i^R(K,M)_j,$$where$\beta{ij}(M)$\beta_{ij}(M)$ are the graded Betti numbers of $M$.

PS I gave up with this LaTeX. Can't see any reason for looking like this. Maybe a moderator can help.

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