Finite Morphisms of Affine Curves are Surjective The Next CEO of Stack OverflowStationary sequence of proper birational morphismsSimple questions about morphisms of finite type and proper morphismsCan a birational morphism surject from an affine to a projective variety?Can we recover étale, fpqc etc. morphisms of schemes from the affine versions?Reference request: Fibre functor for elliptic curves is pro-representableImage of finite locally free morphism is openConfusion about Vakil's proposition 17.4.5, on finite morphisms between curvesIntuition Behind, or Canonical Examples of Finite Type MorphismsFinite Morphisms ClosedFinite Dimensional $k$-Algebra has finitely many Maximal Ideals
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Finite Morphisms of Affine Curves are Surjective
The Next CEO of Stack OverflowStationary sequence of proper birational morphismsSimple questions about morphisms of finite type and proper morphismsCan a birational morphism surject from an affine to a projective variety?Can we recover étale, fpqc etc. morphisms of schemes from the affine versions?Reference request: Fibre functor for elliptic curves is pro-representableImage of finite locally free morphism is openConfusion about Vakil's proposition 17.4.5, on finite morphisms between curvesIntuition Behind, or Canonical Examples of Finite Type MorphismsFinite Morphisms ClosedFinite Dimensional $k$-Algebra has finitely many Maximal Ideals
$begingroup$
I have a question about a step used in Szamuely's "Galois Groups and Fundamental Groups" in the excerpt below (see page 109):
The statement is that a finite morphism of integral affine curves is surjective.
Indeed I'm familar with some proofs to show this statement but the point I'm really interested is concretely the argument given in this source by the author.
He claims that the Case 2 of the proof of 4.3.5 cannot occure in the case of a finite morphism.
My question is why?
Here the proof of 4.3.5:
algebraic-geometry
asked Mar 26 at 14:50
KarlPeterKarlPeter
5911316
$endgroup$
add a comment |
$begingroup$
I have a question about a step used in Szamuely's "Galois Groups and Fundamental Groups" in the excerpt below (see page 109):
The statement is that a finite morphism of integral affine curves is surjective.
Indeed I'm familar with some proofs to show this statement but the point I'm really interested is concretely the argument given in this source by the author.
He claims that the Case 2 of the proof of 4.3.5 cannot occure in the case of a finite morphism.
My question is why?
Here the proof of 4.3.5:
algebraic-geometry
asked Mar 26 at 14:50
KarlPeterKarlPeter
5911316
$endgroup$
1
$begingroup$
Over an algebraically closed field, use that finite morphisms have finite fibers.
$endgroup$
– Samir Canning
Mar 26 at 15:02
$begingroup$
@SamirCanning: if we asseme the case $2$ then we deduce following the proof of 4.3.5 that the map is a constant map and with your argument $Y$ (=preimage) is a finite set. This contradicts that $Y$ was assumed to be a curve. Do you mean this argument?
$endgroup$
– KarlPeter
Mar 26 at 15:09
$begingroup$
Yes that is what I mean.
$endgroup$
– Samir Canning
Mar 26 at 15:10
add a comment |
$begingroup$
I have a question about a step used in Szamuely's "Galois Groups and Fundamental Groups" in the excerpt below (see page 109):
The statement is that a finite morphism of integral affine curves is surjective.
Indeed I'm familar with some proofs to show this statement but the point I'm really interested is concretely the argument given in this source by the author.
He claims that the Case 2 of the proof of 4.3.5 cannot occure in the case of a finite morphism.
My question is why?
Here the proof of 4.3.5:
algebraic-geometry
asked Mar 26 at 14:50
KarlPeterKarlPeter
5911316
$endgroup$
I have a question about a step used in Szamuely's "Galois Groups and Fundamental Groups" in the excerpt below (see page 109):
The statement is that a finite morphism of integral affine curves is surjective.
Indeed I'm familar with some proofs to show this statement but the point I'm really interested is concretely the argument given in this source by the author.
He claims that the Case 2 of the proof of 4.3.5 cannot occure in the case of a finite morphism.
My question is why?
Here the proof of 4.3.5:
algebraic-geometry
algebraic-geometry
asked Mar 26 at 14:50
KarlPeterKarlPeter
5911316
asked Mar 26 at 14:50
KarlPeterKarlPeter
5911316
edited Mar 26 at 15:02
KarlPeter
asked Mar 26 at 14:50
KarlPeterKarlPeter
5911316
asked Mar 26 at 14:50
KarlPeterKarlPeter
5911316
asked Mar 26 at 14:50
KarlPeterKarlPeter
5911316
5911316
1
$begingroup$
Over an algebraically closed field, use that finite morphisms have finite fibers.
$endgroup$
– Samir Canning
Mar 26 at 15:02
$begingroup$
@SamirCanning: if we asseme the case $2$ then we deduce following the proof of 4.3.5 that the map is a constant map and with your argument $Y$ (=preimage) is a finite set. This contradicts that $Y$ was assumed to be a curve. Do you mean this argument?
$endgroup$
– KarlPeter
Mar 26 at 15:09
$begingroup$
Yes that is what I mean.
$endgroup$
– Samir Canning
Mar 26 at 15:10
add a comment |
1
$begingroup$
Over an algebraically closed field, use that finite morphisms have finite fibers.
$endgroup$
– Samir Canning
Mar 26 at 15:02
$begingroup$
@SamirCanning: if we asseme the case $2$ then we deduce following the proof of 4.3.5 that the map is a constant map and with your argument $Y$ (=preimage) is a finite set. This contradicts that $Y$ was assumed to be a curve. Do you mean this argument?
$endgroup$
– KarlPeter
Mar 26 at 15:09
$begingroup$
Yes that is what I mean.
$endgroup$
– Samir Canning
Mar 26 at 15:10
1
1
$begingroup$
Over an algebraically closed field, use that finite morphisms have finite fibers.
$endgroup$
– Samir Canning
Mar 26 at 15:02
$begingroup$
Over an algebraically closed field, use that finite morphisms have finite fibers.
$endgroup$
– Samir Canning
Mar 26 at 15:02
$begingroup$
@SamirCanning: if we asseme the case $2$ then we deduce following the proof of 4.3.5 that the map is a constant map and with your argument $Y$ (=preimage) is a finite set. This contradicts that $Y$ was assumed to be a curve. Do you mean this argument?
$endgroup$
– KarlPeter
Mar 26 at 15:09
$begingroup$
@SamirCanning: if we asseme the case $2$ then we deduce following the proof of 4.3.5 that the map is a constant map and with your argument $Y$ (=preimage) is a finite set. This contradicts that $Y$ was assumed to be a curve. Do you mean this argument?
$endgroup$
– KarlPeter
Mar 26 at 15:09
$begingroup$
Yes that is what I mean.
$endgroup$
– Samir Canning
Mar 26 at 15:10
$begingroup$
Yes that is what I mean.
$endgroup$
– Samir Canning
Mar 26 at 15:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The argument in the book is too complicated:
A finite morphism of schemes is a closed map, hence the image of Y is closed and irreducible in $X$ and thus that image is equal to $X$.
By the way, my argument does not use that $X$ or $Y$ is affine, nor that $Y$ has dimension $1$!
answered Mar 28 at 9:41
Georges ElencwajgGeorges Elencwajg
120k7181334
$endgroup$
1
$begingroup$
To prevent misunderstandings: I have only used that $Y$ is not a point (in order that its irreducible image not be a point in $X$) but in fact $Y$ must a posteriori have dimension one, because dimension is preserved under finite surjective morphisms.
$endgroup$
– Georges Elencwajg
Mar 28 at 9:54
1
$begingroup$
how do you excluded that the given finite morphism isn't constant? I don't see how conclude that without using the fact that finite morphisms have finite fibers
$endgroup$
– KarlPeter
Mar 28 at 13:17
1
$begingroup$
Dear Karl, indeed you have to know that finite morphisms have finite fibers. By the way in algebraic geometry you have the completely general characterization for morphisms: finite = proper with finite fibers (The same result is also true for holomorphic morphisms between complex analytic varieties)
$endgroup$
– Georges Elencwajg
Mar 28 at 14:52
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The argument in the book is too complicated:
A finite morphism of schemes is a closed map, hence the image of Y is closed and irreducible in $X$ and thus that image is equal to $X$.
By the way, my argument does not use that $X$ or $Y$ is affine, nor that $Y$ has dimension $1$!
answered Mar 28 at 9:41
Georges ElencwajgGeorges Elencwajg
120k7181334
$endgroup$
1
$begingroup$
To prevent misunderstandings: I have only used that $Y$ is not a point (in order that its irreducible image not be a point in $X$) but in fact $Y$ must a posteriori have dimension one, because dimension is preserved under finite surjective morphisms.
$endgroup$
– Georges Elencwajg
Mar 28 at 9:54
1
$begingroup$
how do you excluded that the given finite morphism isn't constant? I don't see how conclude that without using the fact that finite morphisms have finite fibers
$endgroup$
– KarlPeter
Mar 28 at 13:17
1
$begingroup$
Dear Karl, indeed you have to know that finite morphisms have finite fibers. By the way in algebraic geometry you have the completely general characterization for morphisms: finite = proper with finite fibers (The same result is also true for holomorphic morphisms between complex analytic varieties)
$endgroup$
– Georges Elencwajg
Mar 28 at 14:52
add a comment |
$begingroup$
The argument in the book is too complicated:
A finite morphism of schemes is a closed map, hence the image of Y is closed and irreducible in $X$ and thus that image is equal to $X$.
By the way, my argument does not use that $X$ or $Y$ is affine, nor that $Y$ has dimension $1$!
answered Mar 28 at 9:41
Georges ElencwajgGeorges Elencwajg
120k7181334
$endgroup$
1
$begingroup$
To prevent misunderstandings: I have only used that $Y$ is not a point (in order that its irreducible image not be a point in $X$) but in fact $Y$ must a posteriori have dimension one, because dimension is preserved under finite surjective morphisms.
$endgroup$
– Georges Elencwajg
Mar 28 at 9:54
1
$begingroup$
how do you excluded that the given finite morphism isn't constant? I don't see how conclude that without using the fact that finite morphisms have finite fibers
$endgroup$
– KarlPeter
Mar 28 at 13:17
1
$begingroup$
Dear Karl, indeed you have to know that finite morphisms have finite fibers. By the way in algebraic geometry you have the completely general characterization for morphisms: finite = proper with finite fibers (The same result is also true for holomorphic morphisms between complex analytic varieties)
$endgroup$
– Georges Elencwajg
Mar 28 at 14:52
add a comment |
$begingroup$
The argument in the book is too complicated:
A finite morphism of schemes is a closed map, hence the image of Y is closed and irreducible in $X$ and thus that image is equal to $X$.
By the way, my argument does not use that $X$ or $Y$ is affine, nor that $Y$ has dimension $1$!
answered Mar 28 at 9:41
Georges ElencwajgGeorges Elencwajg
120k7181334
$endgroup$
The argument in the book is too complicated:
A finite morphism of schemes is a closed map, hence the image of Y is closed and irreducible in $X$ and thus that image is equal to $X$.
By the way, my argument does not use that $X$ or $Y$ is affine, nor that $Y$ has dimension $1$!
answered Mar 28 at 9:41
Georges ElencwajgGeorges Elencwajg
120k7181334
answered Mar 28 at 9:41
Georges ElencwajgGeorges Elencwajg
120k7181334
answered Mar 28 at 9:41
Georges ElencwajgGeorges Elencwajg
120k7181334
answered Mar 28 at 9:41
Georges ElencwajgGeorges Elencwajg
120k7181334
120k7181334
1
$begingroup$
To prevent misunderstandings: I have only used that $Y$ is not a point (in order that its irreducible image not be a point in $X$) but in fact $Y$ must a posteriori have dimension one, because dimension is preserved under finite surjective morphisms.
$endgroup$
– Georges Elencwajg
Mar 28 at 9:54
1
$begingroup$
how do you excluded that the given finite morphism isn't constant? I don't see how conclude that without using the fact that finite morphisms have finite fibers
$endgroup$
– KarlPeter
Mar 28 at 13:17
1
$begingroup$
Dear Karl, indeed you have to know that finite morphisms have finite fibers. By the way in algebraic geometry you have the completely general characterization for morphisms: finite = proper with finite fibers (The same result is also true for holomorphic morphisms between complex analytic varieties)
$endgroup$
– Georges Elencwajg
Mar 28 at 14:52
add a comment |
1
$begingroup$
To prevent misunderstandings: I have only used that $Y$ is not a point (in order that its irreducible image not be a point in $X$) but in fact $Y$ must a posteriori have dimension one, because dimension is preserved under finite surjective morphisms.
$endgroup$
– Georges Elencwajg
Mar 28 at 9:54
1
$begingroup$
how do you excluded that the given finite morphism isn't constant? I don't see how conclude that without using the fact that finite morphisms have finite fibers
$endgroup$
– KarlPeter
Mar 28 at 13:17
1
$begingroup$
Dear Karl, indeed you have to know that finite morphisms have finite fibers. By the way in algebraic geometry you have the completely general characterization for morphisms: finite = proper with finite fibers (The same result is also true for holomorphic morphisms between complex analytic varieties)
$endgroup$
– Georges Elencwajg
Mar 28 at 14:52
1
1
$begingroup$
To prevent misunderstandings: I have only used that $Y$ is not a point (in order that its irreducible image not be a point in $X$) but in fact $Y$ must a posteriori have dimension one, because dimension is preserved under finite surjective morphisms.
$endgroup$
– Georges Elencwajg
Mar 28 at 9:54
$begingroup$
To prevent misunderstandings: I have only used that $Y$ is not a point (in order that its irreducible image not be a point in $X$) but in fact $Y$ must a posteriori have dimension one, because dimension is preserved under finite surjective morphisms.
$endgroup$
– Georges Elencwajg
Mar 28 at 9:54
1
1
$begingroup$
how do you excluded that the given finite morphism isn't constant? I don't see how conclude that without using the fact that finite morphisms have finite fibers
$endgroup$
– KarlPeter
Mar 28 at 13:17
$begingroup$
how do you excluded that the given finite morphism isn't constant? I don't see how conclude that without using the fact that finite morphisms have finite fibers
$endgroup$
– KarlPeter
Mar 28 at 13:17
1
1
$begingroup$
Dear Karl, indeed you have to know that finite morphisms have finite fibers. By the way in algebraic geometry you have the completely general characterization for morphisms: finite = proper with finite fibers (The same result is also true for holomorphic morphisms between complex analytic varieties)
$endgroup$
– Georges Elencwajg
Mar 28 at 14:52
$begingroup$
Dear Karl, indeed you have to know that finite morphisms have finite fibers. By the way in algebraic geometry you have the completely general characterization for morphisms: finite = proper with finite fibers (The same result is also true for holomorphic morphisms between complex analytic varieties)
$endgroup$
– Georges Elencwajg
Mar 28 at 14:52
add a comment |
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$begingroup$
Over an algebraically closed field, use that finite morphisms have finite fibers.
$endgroup$
– Samir Canning
Mar 26 at 15:02
$begingroup$
@SamirCanning: if we asseme the case $2$ then we deduce following the proof of 4.3.5 that the map is a constant map and with your argument $Y$ (=preimage) is a finite set. This contradicts that $Y$ was assumed to be a curve. Do you mean this argument?
$endgroup$
– KarlPeter
Mar 26 at 15:09
$begingroup$
Yes that is what I mean.
$endgroup$
– Samir Canning
Mar 26 at 15:10