Let $i: R \subset A$ be an integral extension of discrete valuation rings and denote by $$f: \operatorname{Spec}(A) \to \operatorname{Spec}(R)$$ the corresponding scheme morphism. Since $\operatorname{Spec}(R) = \{\sigma_R, \eta_R\}$ (resp. $\operatorname{Spec}(A) = \{\sigma_A, \eta_A\}$ where $\sigma$ is the unique closed and $\eta$ the unique generic points)


$K_R := \kappa(\eta_R) = \operatorname{Frac}(R)$

$K_A := \kappa(\eta_A) = \operatorname{Frac}(A)$


$k_R := \kappa(\sigma_R)= \mathcal{O}_{R, \sigma_R}/m_{\sigma_R}\mathcal{O}_{R, \sigma_R}$

$ k_A := \kappa(\sigma_A)$

we obtain field extensions $K_R\subset K_A$ and $k_R\subset k_A$.

I want to show that:

$f$ is a unramified morphism (in sense of scheme morphism) $\Leftrightarrow$

$K_A/K_R, k_A/k_R$ are finite separable and $[K_A:K_R]=[k_A: k_R]$.

Source: S. Bosch, Commutative Algebra and Algebraic Geometry, p. 374.

My attempts:

Since there are only two points it seems ok to apply the chararacterization (v) from Thm 3:

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So the the finite separability condition for $K_A/K_R, k_A/k_R$ works in both directions.

The only problem lies in the condition $[K_A:K_R]=[k_A: k_R]$ and that $m_{\sigma_R}$ generates $m_{\sigma_A}$ ($m_{\eta_R}$and $m_{\eta_A}=0$ so it's ok)

The author's hint was to use exersise 3.1.8 (page 90):

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So in our case $K= K_R$ and $L= K_A$. Then the integral closure of $R$ is in $K_A$ a finite $R$-module. How does it help to get $[K_A:K_R]=[k_A: k_R]$?

Btw: In Ex 3.1.8: What would fail with the trace function $Tr_{L/K}$ if $L/K$ wouldn't separable?


Hint: You can show that $A/R$ is in fact finite free with $\mathrm{rk}_R A=[K_A:K_R]$. This will create the missing link between the degrees of the two field extensions.

As to your last remark: If the field extension is not separable then the trace will be zero.

  • $\begingroup$ The freeness of $A/R$ commes from a structure theorem for modules over Dedekid rings (I never heard about a structure theorem over DVRs). Then $\mathrm{rk}_R A=[K_A:K_R]$ boils down to the general property of free extensions. But the link between $ [K_A:K_R]$ and $ [k_A:k_R]$ is still unclear. Is $\mathrm{rk}_R$ conserved (why?) if we quotient out someting that is contained in Jacobsen ideal? So some kind of Nakayama argument? $\endgroup$ – KarlPeter Dec 29 '18 at 4:18
  • $\begingroup$ One way to show this is by observing $A/\mathfrak{m}_{\sigma_A}=A\otimes R/\mathfrak{m}_{\sigma_R}$ by using unramifiedness. $\endgroup$ – asdq Dec 29 '18 at 5:13
  • $\begingroup$ Ah ok, this settle $rk_R(A)=[k_A: k_R]$. One point stays unclear: How do you get the finiteness of $A/R$? SInce after settled this $A$ considered as $R$ module without torsion is indeed finite free so $A = \oplus_{rk_R A} R$. Regarding equality $\mathrm{rk}_R A=[K_A:K_R]$ there occure following problem. Passing to $K_R$ is the same as localizing by $S := R \backslash \{0\}$ and that respects products. So $AS^{-1} = \oplus R S^{-1}= \oplus K_R$. But $S$ is not enough to get $AS^{-1} = K_A$. I guess I have to combine it with result from Ex 8. $\endgroup$ – KarlPeter Dec 29 '18 at 5:54
  • $\begingroup$ Does the information about finiteness of integral closure of $R$ help here? $\endgroup$ – KarlPeter Dec 29 '18 at 5:54
  • $\begingroup$ Well exercise 8 tells you that $A/R$ is finite since $A$ is precisely the integral closure of $R$ in $K_A$. $\endgroup$ – asdq Dec 29 '18 at 11:53

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