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I'am reading Eisenbud, Commutative Algebra, p.272, Proposition 12.2

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Why can we assume that $R$ is graded? Here, parameter ideal is defined in his book p.235. Or, in other version of his book, he call 'ideal of finite colength', instead of parameter ideal (If needed, I can upload more detailed definition and associated theorems).

Q. What does "first factor out $\operatorname{ann}(M)$ and then pass to $\operatorname{gr}_\mathfrak{q}R$ and $\operatorname{gr}_\mathfrak{q}M$" exactly means?

As first trial, I quess that first, consider $I := \mathfrak{q}+ \operatorname{ann}(M)$ ( then $R / I$ is artinian as above image (below the definition of Hilbert-Samuel function) ; if needed, I can provide a proof) and second, consider the associated graded ring $\bar{R} := \operatorname{gr}_IR$ and module $\bar{M}:=\operatorname{gr}_IM$ (his book p.146) ;

$$ \bar{R} := \operatorname{gr}_IR := R/I \oplus I/I^2 \oplus \cdots $$

$$ \bar{M} := \operatorname{gr}_IM := M/IM \oplus IM / I^{2}M \oplus \cdots $$

(Then maybe $\operatorname{gr}_IM$ is finitely generated $\operatorname{gr}_IR$- module by his book p.146, prop.5.2) But now I stuck at how to connect the graded case with local case.

For example, is there any relation between the Hilbert Samuel function $H_{\mathfrak{q},M}(n)$ (for local case) and $H_{\bar{\mathfrak{q}}, \operatorname{gr}_IM}(n)$, where

$$ \bar{\mathfrak{q}} := \bar{R}_{+}:= I/I^2 \oplus I^2/I^3 \oplus \cdots$$ is the irrelevant ideal of $\bar{R}:= \operatorname{gr}_{I}R$ (Then maybe we can show that $\bar{R}/(\bar{\mathfrak{q}}+\operatorname{ann}_{\bar{R}}(\bar{M}))$ is artinian ring) ?

Note that $$H_{\mathfrak{q},M}(n) := l_{R/(\mathfrak{q}+\operatorname{ann}_R(M))}(\mathfrak{q}^nM/\mathfrak{q}^{n+1}M) = l_{R/(\mathfrak{q}+\operatorname{ann}_R(M))}(\mathfrak{q}^nM) -l_{R/(\mathfrak{q}+\operatorname{ann}_R(M))}(\mathfrak{q}^{n+1}M)$$ and $$ H_{\bar{\mathfrak{q}}, \operatorname{gr}_IM}(n) := l_{\bar{R}/(\bar{\mathfrak{q}}+\operatorname{ann}_{\bar{R}}(\bar{M}))} (\bar{\mathfrak{q}}^n\bar{M} /\bar{\mathfrak{q}}^{n+1}\bar{M}) = l_{\bar{R}/(\bar{\mathfrak{q}}+\operatorname{ann}_{\bar{R}}(\bar{M}))} (\bar{\mathfrak{q}}^n\bar{M}) - l_{\bar{R}/(\bar{\mathfrak{q}}+\operatorname{ann}_{\bar{R}}(\bar{M}))} (\bar{\mathfrak{q}}^{n+1}\bar{M}) $$

So, now, I'm trying to relate $ l_{R/I}(\mathfrak{q}^nM) = l_{R/(\mathfrak{q}+\operatorname{ann}_R(M))}(\mathfrak{q}^nM)$ and $ l_{\bar{R}/(\bar{\mathfrak{q}}+\operatorname{ann}_{\bar{R}}(\bar{M}))} (\bar{\mathfrak{q}}^n\bar{M})$.

$$ l_{\bar{R}/(\bar{\mathfrak{q}}+\operatorname{ann}_{\bar{R}}(\bar{M}))} (\bar{\mathfrak{q}}^n\bar{M}) = l_{\bar{R}/(\bar{\mathfrak{q}}+\operatorname{ann}_{\bar{R}}(\bar{M}))}(I/I^2 \oplus I/I^3 \oplus \cdots)^n \cdot ((M/IM) \oplus (IM/I^2M) \oplus \cdots) $$

How can we calculate this length to obtain some relation?

Is there some isomorphism from $(I/I^2 \oplus I/I^3 \oplus \cdots)^n \cdot ((M/IM) \oplus (IM/I^2M) \oplus \cdots)$ to some simpler structure?

Plantation
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  • Here I think what he means is that one first replaces $R $ by $ R/Ann(M)$ and it preserves the length formula and we can thus just let $Ann(M)=0$. Then pass to the case of $gr_qR$ and $gr_qM$. But at first you should check if $gr_qR $ is local. For your isomorphism part of the question, I think the answer should be yes. The complicated graded modules and ideals can be simplified. Maybe you should first try to do the easier case, like for $n=1$ and use the property of graded degrees to see what you can get. – Display name Feb 04 '22 at 06:25
  • Actually, on page $ 275$, he mentioned the special case that $q$ is the maximal ideal, the Hilbert function is the same for $R$ and $gr_qR$. – Display name Feb 04 '22 at 06:29
  • And $q$ in $gr_q R$ should generated by the image of $q$ in $R_1$, i.e. the homogeneous $1$ part. And thus should be $q/q^2\bigoplus q^2/q^3\bigoplus \dots$ – Display name Feb 04 '22 at 06:37

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