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Suppose that $R:=k[x_0,\dots,x_n]$ and $I$ is an ideal. Is there any relation between finding the minimal generators of $I$ and the graded betti numbers of the module $R/I$?

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The answer to your question ('yes') is basically part of the defintion of graded Betti numbers. You might want to review that. – J.C. Ottem Oct 22 '12 at 1:27
As far as I understand, betti number will give the number of generator of specific degree to each module in the free resolution, how that can contribute to the minimal number of generators of the whole R/I. Any extra hint or reference? – abd Oct 22 '12 at 2:42
up vote 1 down vote accepted

To support J.C. Ottem's answer, let me present one example.

Let $R = \mathbb{C}[x,y]$ and $I = (x,y^2)R$. What is the minimal graded free resolution of $R/I$, equivalently $I$? That is,

$0 \rightarrow R(-3) \stackrel{d_1}{\rightarrow} R(-1) \oplus R(-2) \stackrel{d_0}\rightarrow R \rightarrow R/I \rightarrow 0 $

where $d_1 = (-y^2 \;\; x)$ and $d_0 = (x \;\; y^2)$.

Now, ask what are the graded Betti numbers and minimal number of generators for $R/I$. I agree with J.C. Ottem's opinion on reviewing the definitions. I hope this helps.

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Thanks Youngsu, your example helped me to understand the definition more. please correct me if I understood that wrong, now from the free resolution we can find the betti numbers easily and now the first module in the sequence $R(-1)\oplus R(-2)$ will be generated by two polynomials of degree 1 and 2 respectively. Is that means the ideal $I$ will be generated by polynomials of degree 1 and 2? Also how the differential map relate to the generators of $I$? and what we can get from the betti number about $R/I=<1,y>$? Is there any thing we can extract about the quotient? – abd Oct 23 '12 at 17:28
Hi. You always need to put the condition "graded minimal free resolution". Check this condition with your definition of (graded) Betti numbers. The entries of $d_0$ are a generating set of $I$. Here $(x, y^2) = I$. This can be understood how you build a resolution. I am a bit confused by your example. $R/I=<1,y>$. Did you mean that $I = (1,y)$? If so then you wouldn't get anything since $I = R$. – Youngsu Oct 24 '12 at 14:42

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