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My question refers to following conclusion in Liu's "Algebraic Geometry" at page 290 (Thm 4.16):

enter image description here

Why does $e'_x= length_{\hat{B}}(W_{\hat{B}/\hat{A}}/\hat{B})$ hold?

Here $e'_x$ is introduced as $e'_x = length_{\mathcal{O}_{X,x}}(\omega_f/ \mathcal{O}_X)_x +1$.

As the excerps explains we get $(\omega_f/ \mathcal{O}_X)_x = Hom_{\hat{A}}(\hat{B},\hat{A})/\hat{B}$ and Exercise 6.4.8 provides the identification: $Hom_{\hat{A}}(\hat{B},\hat{A}) \cong W_{\hat{B}/\hat{A}}$, as see below:

enter image description here

Combine this both results we get:

$e'_x = length_{\mathcal{O}_{X,x}}(W_{\hat{B}/\hat{A}}/\hat{B}) +1$.

But how to concude the final step $length_{\mathcal{O}_{X,x}}(W_{\hat{B}/\hat{A}}/\hat{B}) +1= length_{\hat{B}}(W_{\hat{B}/\hat{A}}/\hat{B})$; especially how to cope the "change" of the "base modules" from $\mathcal{O}_{X,x}$ to $\hat{B}$ under contolling the $length$ of modules?

user267839
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  • It seems to me that $A$ is a DVR (normal of dimension 1), so both $A, \hat{A}$ are PID. Let $(t)$ be the maximal ideal of both $A, \hat{A}$. Every $\hat{A}$ module of finite length decomposes as $\osum_i M_i$ by structure theorem for fg PID modules, where $M_i = \hat{A}$ or $\hat{A}/(t^n)$. On one hand, $length_A(\hat{A}) = \infty$, so if equation holds there are no free factors. On the other hand, $length_A(\hat{A}/t^n) = length_A(A/t^n) = 1 \neq length_{\hat{A}}(\hat{A}/t^n) = 0$ and length is additive, so for torsion modules it holds $length_A(M) = length_{\hat{A}}(M)$. What am I missing? – frame95 Jan 17 '18 at 21:34
  • @frame95: Hi thank you for the answer. Two questions: Why do we have the equalities $length_A (\hat{A}/t^n) = length_A (A/t^n)$ and $length_{\hat{A}}(\hat{A}/t^n)=0$? – user267839 Jan 19 '18 at 11:12
  • $\hat{A}/t^n$ and $A/t^n$ are isomorph, so they have equal length. Think about $k[t]/t^n$ and $k[[t]]/t^n$: they always are polynomials with little degree! And $length_{\hat{A}}(\hat{A}/t^n)-1 =0$, I forgot a -1 which was in the equation you required. – frame95 Jan 25 '18 at 01:33

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