Birational Morphism to a Normal Scheme Isomorphism
My question refers the answer of @user18119 in following thread: Are these two notions of Galois morphism the same
We have $f:Xto Y$ a finite morphism of integral schemes and $G$ letautomorphism group of $X$ over $Y$.
There was showed that following statements where equivalent:
The function field extension $K(Y)subset K(X)$ is Galois (in the field-theoretic sense)
The quotient $X/G$ exists and the natural morphism $X/Gto Y$ is an isomorphism.
In the (2) implies (1) direction $Xto Y$ factoriezes through the finite morphism $g: X/Gto Y$. By (1) $g$ is birational. Futhermore $Y$ is normal.
Why these two conditions already imply that $g$ is an isomorphism?
Clearly that suffice to check it on the level of stalks.
algebraic-geometry schemes birational-geometry
asked Dec 26 at 0:53
KarlPeter
5881314
add a comment |
My question refers the answer of @user18119 in following thread: Are these two notions of Galois morphism the same
We have $f:Xto Y$ a finite morphism of integral schemes and $G$ letautomorphism group of $X$ over $Y$.
There was showed that following statements where equivalent:
The function field extension $K(Y)subset K(X)$ is Galois (in the field-theoretic sense)
The quotient $X/G$ exists and the natural morphism $X/Gto Y$ is an isomorphism.
In the (2) implies (1) direction $Xto Y$ factoriezes through the finite morphism $g: X/Gto Y$. By (1) $g$ is birational. Futhermore $Y$ is normal.
Why these two conditions already imply that $g$ is an isomorphism?
Clearly that suffice to check it on the level of stalks.
algebraic-geometry schemes birational-geometry
asked Dec 26 at 0:53
KarlPeter
5881314
3
1 and 2 are equivalent only if you assume $Y$ is normal. For your question, any finite birational morphism $h:Zto Y$ is an isomorphism if $Y$ is normal. To see this, notice that you may replace $Y$ by any affine open cover and assume that $Y$ is affine. Finite morphisms are affine, so $Z$ is also affine. If $Y,Z$ are spectrum of $A,B$, we have an integral birational extension $Ato B$ which is finite. Since $A$ is integrally closed, this means $A=B$.
– Mohan
Dec 26 at 1:10
add a comment |
My question refers the answer of @user18119 in following thread: Are these two notions of Galois morphism the same
We have $f:Xto Y$ a finite morphism of integral schemes and $G$ letautomorphism group of $X$ over $Y$.
There was showed that following statements where equivalent:
The function field extension $K(Y)subset K(X)$ is Galois (in the field-theoretic sense)
The quotient $X/G$ exists and the natural morphism $X/Gto Y$ is an isomorphism.
In the (2) implies (1) direction $Xto Y$ factoriezes through the finite morphism $g: X/Gto Y$. By (1) $g$ is birational. Futhermore $Y$ is normal.
Why these two conditions already imply that $g$ is an isomorphism?
Clearly that suffice to check it on the level of stalks.
algebraic-geometry schemes birational-geometry
asked Dec 26 at 0:53
KarlPeter
5881314
My question refers the answer of @user18119 in following thread: Are these two notions of Galois morphism the same
We have $f:Xto Y$ a finite morphism of integral schemes and $G$ letautomorphism group of $X$ over $Y$.
There was showed that following statements where equivalent:
The function field extension $K(Y)subset K(X)$ is Galois (in the field-theoretic sense)
The quotient $X/G$ exists and the natural morphism $X/Gto Y$ is an isomorphism.
In the (2) implies (1) direction $Xto Y$ factoriezes through the finite morphism $g: X/Gto Y$. By (1) $g$ is birational. Futhermore $Y$ is normal.
Why these two conditions already imply that $g$ is an isomorphism?
Clearly that suffice to check it on the level of stalks.
algebraic-geometry schemes birational-geometry
algebraic-geometry schemes birational-geometry
asked Dec 26 at 0:53
KarlPeter
5881314
asked Dec 26 at 0:53
KarlPeter
5881314
asked Dec 26 at 0:53
KarlPeter
5881314
asked Dec 26 at 0:53
KarlPeter
5881314
asked Dec 26 at 0:53
KarlPeter
5881314
5881314
3
1 and 2 are equivalent only if you assume $Y$ is normal. For your question, any finite birational morphism $h:Zto Y$ is an isomorphism if $Y$ is normal. To see this, notice that you may replace $Y$ by any affine open cover and assume that $Y$ is affine. Finite morphisms are affine, so $Z$ is also affine. If $Y,Z$ are spectrum of $A,B$, we have an integral birational extension $Ato B$ which is finite. Since $A$ is integrally closed, this means $A=B$.
– Mohan
Dec 26 at 1:10
add a comment |
3
1 and 2 are equivalent only if you assume $Y$ is normal. For your question, any finite birational morphism $h:Zto Y$ is an isomorphism if $Y$ is normal. To see this, notice that you may replace $Y$ by any affine open cover and assume that $Y$ is affine. Finite morphisms are affine, so $Z$ is also affine. If $Y,Z$ are spectrum of $A,B$, we have an integral birational extension $Ato B$ which is finite. Since $A$ is integrally closed, this means $A=B$.
– Mohan
Dec 26 at 1:10
3
3
1 and 2 are equivalent only if you assume $Y$ is normal. For your question, any finite birational morphism $h:Zto Y$ is an isomorphism if $Y$ is normal. To see this, notice that you may replace $Y$ by any affine open cover and assume that $Y$ is affine. Finite morphisms are affine, so $Z$ is also affine. If $Y,Z$ are spectrum of $A,B$, we have an integral birational extension $Ato B$ which is finite. Since $A$ is integrally closed, this means $A=B$.
– Mohan
Dec 26 at 1:10
1 and 2 are equivalent only if you assume $Y$ is normal. For your question, any finite birational morphism $h:Zto Y$ is an isomorphism if $Y$ is normal. To see this, notice that you may replace $Y$ by any affine open cover and assume that $Y$ is affine. Finite morphisms are affine, so $Z$ is also affine. If $Y,Z$ are spectrum of $A,B$, we have an integral birational extension $Ato B$ which is finite. Since $A$ is integrally closed, this means $A=B$.
– Mohan
Dec 26 at 1:10
add a comment |
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3
1 and 2 are equivalent only if you assume $Y$ is normal. For your question, any finite birational morphism $h:Zto Y$ is an isomorphism if $Y$ is normal. To see this, notice that you may replace $Y$ by any affine open cover and assume that $Y$ is affine. Finite morphisms are affine, so $Z$ is also affine. If $Y,Z$ are spectrum of $A,B$, we have an integral birational extension $Ato B$ which is finite. Since $A$ is integrally closed, this means $A=B$.
– Mohan
Dec 26 at 1:10