Finite Integral Extension of DVRs $rk_R(A)$
$begingroup$
Let $R subset A$ be a finite integral extension of discrete valuation rings such that $A = oplus_{i=1} ^n R$ therefore $rk_R(A)=n$. Denote by $K_R,K_A$ the quotient fields of $R$ and $A$. Obviously this gives a rise for field extension $K_A / K_R$ with $d := [K_A: K_R]$.
My goal is to verify that $n =d$.
My ideas: $K_R$ arises from $R$ as localisation on the multiplicative set $S:= R backslash {0}$. Therefore $K_R= RS^{-1}$. Since localisations comute with direct sums we obtain
$$AS^{-1} = oplus_{i=1} ^n K_R$$
If we could deduce that $AS^{-1}= K_A$ we are done but I'm not sure if it true.
Is there maybe a more effective way to show the claim?
abstract-algebra ring-theory commutative-algebra extension-field
asked Dec 30 '18 at 0:23
KarlPeterKarlPeter
7141416
$endgroup$
add a comment |
$begingroup$
Let $R subset A$ be a finite integral extension of discrete valuation rings such that $A = oplus_{i=1} ^n R$ therefore $rk_R(A)=n$. Denote by $K_R,K_A$ the quotient fields of $R$ and $A$. Obviously this gives a rise for field extension $K_A / K_R$ with $d := [K_A: K_R]$.
My goal is to verify that $n =d$.
My ideas: $K_R$ arises from $R$ as localisation on the multiplicative set $S:= R backslash {0}$. Therefore $K_R= RS^{-1}$. Since localisations comute with direct sums we obtain
$$AS^{-1} = oplus_{i=1} ^n K_R$$
If we could deduce that $AS^{-1}= K_A$ we are done but I'm not sure if it true.
Is there maybe a more effective way to show the claim?
abstract-algebra ring-theory commutative-algebra extension-field
asked Dec 30 '18 at 0:23
KarlPeterKarlPeter
7141416
$endgroup$
add a comment |
$begingroup$
Let $R subset A$ be a finite integral extension of discrete valuation rings such that $A = oplus_{i=1} ^n R$ therefore $rk_R(A)=n$. Denote by $K_R,K_A$ the quotient fields of $R$ and $A$. Obviously this gives a rise for field extension $K_A / K_R$ with $d := [K_A: K_R]$.
My goal is to verify that $n =d$.
My ideas: $K_R$ arises from $R$ as localisation on the multiplicative set $S:= R backslash {0}$. Therefore $K_R= RS^{-1}$. Since localisations comute with direct sums we obtain
$$AS^{-1} = oplus_{i=1} ^n K_R$$
If we could deduce that $AS^{-1}= K_A$ we are done but I'm not sure if it true.
Is there maybe a more effective way to show the claim?
abstract-algebra ring-theory commutative-algebra extension-field
asked Dec 30 '18 at 0:23
KarlPeterKarlPeter
7141416
$endgroup$
Let $R subset A$ be a finite integral extension of discrete valuation rings such that $A = oplus_{i=1} ^n R$ therefore $rk_R(A)=n$. Denote by $K_R,K_A$ the quotient fields of $R$ and $A$. Obviously this gives a rise for field extension $K_A / K_R$ with $d := [K_A: K_R]$.
My goal is to verify that $n =d$.
My ideas: $K_R$ arises from $R$ as localisation on the multiplicative set $S:= R backslash {0}$. Therefore $K_R= RS^{-1}$. Since localisations comute with direct sums we obtain
$$AS^{-1} = oplus_{i=1} ^n K_R$$
If we could deduce that $AS^{-1}= K_A$ we are done but I'm not sure if it true.
Is there maybe a more effective way to show the claim?
abstract-algebra ring-theory commutative-algebra extension-field
abstract-algebra ring-theory commutative-algebra extension-field
asked Dec 30 '18 at 0:23
KarlPeterKarlPeter
7141416
asked Dec 30 '18 at 0:23
KarlPeterKarlPeter
7141416
asked Dec 30 '18 at 0:23
KarlPeterKarlPeter
7141416
asked Dec 30 '18 at 0:23
KarlPeterKarlPeter
7141416
asked Dec 30 '18 at 0:23
KarlPeterKarlPeter
7141416
7141416
add a comment |
add a comment |
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