Skip to main content
added 147 characters in body
Source Link

Let $A$ be a normal ring with quotient field $K$. Let $L/K$ be a finite separable extension. Let $E_1/K$ and $E_2/K$ be extensions of $K$ contained in $L$. Let $B_1$ (resp. $B_2$) be the normalization of $A$ in $E_1$ (resp. $E_2$) and let $C$ be the normalization of $A$ in $L$. Then $$K\subset E_i\subset L\ \mbox{and}\ A\subset B_i\subset C\ \mbox{for}\ i=1, 2.$$ Let $\mathfrak P$ be a prime ideal of $C$. Define $\mathfrak p_1:=\mathfrak P\cap B_1$, $\mathfrak p_2:=\mathfrak P\cap B_2$ and $\mathfrak p:=\mathfrak P\cap A$. Then, on the level of residue fields, we have $$k(\mathfrak p)\subset k(\mathfrak p_i)\subset k(\mathfrak P)\ \mbox{for}\ i=1, 2.$$

Assume that $B_1$ and $B_2$ are finite etale $A$-algebras.

Questions:

a) Does $L=E_1E_2$ (composite field) imply $k(\mathfrak P)=k(\mathfrak p_1) k(\mathfrak p_2)$?

b) Does $K=E_1\cap E_2$ imply $k(\mathfrak p)=k(\mathfrak p_1)\cap k(\mathfrak p_2)$?

CommentComments:

a) For my application it would be enough to know the answers in the special case where $E_1/K$ and $E_2/K$ are Galois extensions.

b) Somebody told me that the answer to a) would be ``no'', if we omitted the hypothesis that the algebras $B_i/A$ are etale, even in the case where $K$ is a number field.

Let $A$ be a normal ring with quotient field $K$. Let $L/K$ be a finite extension. Let $E_1/K$ and $E_2/K$ be extensions of $K$ contained in $L$. Let $B_1$ (resp. $B_2$) be the normalization of $A$ in $E_1$ (resp. $E_2$) and let $C$ be the normalization of $A$ in $L$. Then $$K\subset E_i\subset L\ \mbox{and}\ A\subset B_i\subset C\ \mbox{for}\ i=1, 2.$$ Let $\mathfrak P$ be a prime ideal of $C$. Define $\mathfrak p_1:=\mathfrak P\cap B_1$, $\mathfrak p_2:=\mathfrak P\cap B_2$ and $\mathfrak p:=\mathfrak P\cap A$. Then, on the level of residue fields, we have $$k(\mathfrak p)\subset k(\mathfrak p_i)\subset k(\mathfrak P)\ \mbox{for}\ i=1, 2.$$

Assume that $B_1$ and $B_2$ are finite etale $A$-algebras.

Questions:

a) Does $L=E_1E_2$ (composite field) imply $k(\mathfrak P)=k(\mathfrak p_1) k(\mathfrak p_2)$?

b) Does $K=E_1\cap E_2$ imply $k(\mathfrak p)=k(\mathfrak p_1)\cap k(\mathfrak p_2)$?

Comment:

Somebody told me that the answer to a) would be ``no'', if we omitted the hypothesis that the algebras $B_i/A$ are etale, even in the case where $K$ is a number field.

Let $A$ be a normal ring with quotient field $K$. Let $L/K$ be a finite separable extension. Let $E_1/K$ and $E_2/K$ be extensions of $K$ contained in $L$. Let $B_1$ (resp. $B_2$) be the normalization of $A$ in $E_1$ (resp. $E_2$) and let $C$ be the normalization of $A$ in $L$. Then $$K\subset E_i\subset L\ \mbox{and}\ A\subset B_i\subset C\ \mbox{for}\ i=1, 2.$$ Let $\mathfrak P$ be a prime ideal of $C$. Define $\mathfrak p_1:=\mathfrak P\cap B_1$, $\mathfrak p_2:=\mathfrak P\cap B_2$ and $\mathfrak p:=\mathfrak P\cap A$. Then, on the level of residue fields, we have $$k(\mathfrak p)\subset k(\mathfrak p_i)\subset k(\mathfrak P)\ \mbox{for}\ i=1, 2.$$

Assume that $B_1$ and $B_2$ are finite etale $A$-algebras.

Questions:

a) Does $L=E_1E_2$ (composite field) imply $k(\mathfrak P)=k(\mathfrak p_1) k(\mathfrak p_2)$?

b) Does $K=E_1\cap E_2$ imply $k(\mathfrak p)=k(\mathfrak p_1)\cap k(\mathfrak p_2)$?

Comments:

a) For my application it would be enough to know the answers in the special case where $E_1/K$ and $E_2/K$ are Galois extensions.

b) Somebody told me that the answer to a) would be ``no'', if we omitted the hypothesis that the algebras $B_i/A$ are etale, even in the case where $K$ is a number field.

Source Link

Composition and intersection of residue fields

Let $A$ be a normal ring with quotient field $K$. Let $L/K$ be a finite extension. Let $E_1/K$ and $E_2/K$ be extensions of $K$ contained in $L$. Let $B_1$ (resp. $B_2$) be the normalization of $A$ in $E_1$ (resp. $E_2$) and let $C$ be the normalization of $A$ in $L$. Then $$K\subset E_i\subset L\ \mbox{and}\ A\subset B_i\subset C\ \mbox{for}\ i=1, 2.$$ Let $\mathfrak P$ be a prime ideal of $C$. Define $\mathfrak p_1:=\mathfrak P\cap B_1$, $\mathfrak p_2:=\mathfrak P\cap B_2$ and $\mathfrak p:=\mathfrak P\cap A$. Then, on the level of residue fields, we have $$k(\mathfrak p)\subset k(\mathfrak p_i)\subset k(\mathfrak P)\ \mbox{for}\ i=1, 2.$$

Assume that $B_1$ and $B_2$ are finite etale $A$-algebras.

Questions:

a) Does $L=E_1E_2$ (composite field) imply $k(\mathfrak P)=k(\mathfrak p_1) k(\mathfrak p_2)$?

b) Does $K=E_1\cap E_2$ imply $k(\mathfrak p)=k(\mathfrak p_1)\cap k(\mathfrak p_2)$?

Comment:

Somebody told me that the answer to a) would be ``no'', if we omitted the hypothesis that the algebras $B_i/A$ are etale, even in the case where $K$ is a number field.