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9
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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 $M$.
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8
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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 $R$ 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 $M$.
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7
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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 $R$ 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,$$
where $\beta{ij}(M)$ \beta_{ij}(M)$ are the graded Betti numbers of $M$.
PS I gave up with this LaTeX. Can't see any reason for looking like this. Maybe a moderator can help.
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6
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I wanted to leave this as a comment, but for some reason I can't. I gave a short answer to this question here: http://math.stackexchange.com/questions/209218/homogeneous-ideal-and-degree-of-generators.
A more general answer says the following: if $R$ is a graded $K$-algebra, $K$ a field, and $M$ a graded finitely generated $R$-module, then
$\beta_{ij}(M)=\dim_K\mathrm{Tor}i^R(K,M)$ $\beta_{ij}(M)=\dim_K\operatorname{Tor}_i^R(K,M)j,$$ where $\beta{ij}(M)$ 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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5
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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 $R$ is a graded $K$-algebra, $K$ a field, and $M$ a graded finitely generated $R$-module, then
$$\beta_{ij}(M)=\dim_K\mathrm{Tor}_i^R(K,M)j,$$
\beta_{ij}(M)=\dim_K\mathrm{Tor}i^R(K,M)$ where $\beta{ij}(M)$ are the graded Betti numbers of $M$.
PS I gave up with this LaTeX. Can't see any reason for looking like this.
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4
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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 $R$ is a graded $K$-algebra, $K$ a field, and $M$ a graded finitely generated $R$-module, then
$$\beta_{ij}(M)=\dim_K\mathrm{Tor}_i^R(K,M)j,$$
where $\beta{ij}(M)$ are the graded Betti numbers of $M$.
PS I gave up with your this LaTeX. Can't see any reason for looking like this.
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3
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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 $R$ is a graded $K$-algebra, $K$ a field, and $M$ a graded finitely generated $R$-module, then
$\beta_{ij}(M)=\dim_K$ $\mathrm{Tor}i^R$ $\beta_{ij}(M)=\dim_K\mathrm{Tor}_i^R(K,M)j,$$
where $\beta{ij}(M)$ are the graded Betti numbers of $M$.
PS I gave up with your LaTeX. Can't see any reason for looking like this.
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2
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I wanted to leave this as a comment, but for some reason I can't. I gave a short answer to this question here: http://math.stackexchange.com/questions/209218/homogeneous-ideal-and-degree-of-generators.
A more general answer says the following: if $R$ is a graded $K$-algebra, $K$ a field, and $M$ a graded finitely generated $R$-module, then
$$\beta_{ij}(M)=\dim_K\operatorname{Tor}_i^R(K,M)j,$$
\beta_{ij}(M)=\dim_K$ $\mathrm{Tor}i^R$ where $\beta{ij}(M)$ are the graded Betti numbers of $M$.
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1
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I wanted to leave this as a comment, but for some reason I can't. I gave a short answer to this question here: http://math.stackexchange.com/questions/209218/homogeneous-ideal-and-degree-of-generators.
A more general answer says the following: if $R$ is a graded $K$-algebra, $K$ a field, and $M$ a graded finitely generated $R$-module, then
$$\beta_{ij}(M)=\dim_K\operatorname{Tor}_i^R(K,M)j,$$
where $\beta{ij}(M)$ are the graded Betti numbers of $M$.
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