# Unramified Extensions of Discrete Valuation Rings

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)

denote

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

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

and

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

$$k_A := \kappa(\sigma_A)$$

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:

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):

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.
• 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? – KarlPeter Dec 29 '18 at 4:18
• One way to show this is by observing $A/\mathfrak{m}_{\sigma_A}=A\otimes R/\mathfrak{m}_{\sigma_R}$ by using unramifiedness. – asdq Dec 29 '18 at 5:13
• 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. – KarlPeter Dec 29 '18 at 5:54
• Does the information about finiteness of integral closure of $R$ help here? – KarlPeter Dec 29 '18 at 5:54
• Well exercise 8 tells you that $A/R$ is finite since $A$ is precisely the integral closure of $R$ in $K_A$. – asdq Dec 29 '18 at 11:53