If V a closed subvariety of $P^n$ and $phi_i:A^nto U_i subset P^n$, then $phi_i^-1(V)$ is a closed subvariety of $A^n$
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I 'm new to algebraic geometry and I am a little bit confused with a proposition on Fulton, Algebraic Curves, page 70 proposition 3:
I will quote the proposition:
Let V be a closed subvariety of $P^n$, $phi_i : A^n to U_i subset P^n$, (whereas $phi_n+1(x_1,...,x_n)=(x_1:...:x_n:1)$ ). Then $V_i=phi_i^-1(V)$ is a closed subvariety of $A^n$, and $phi_i$ restricts to an isomorphism of $V_i$ with $Vcap U_i$. A projective variety is a union of a finite number of open affine varieties.
It seems straightforward to show that $V_i=(Vcap U_i)_ast$ and thus prove that it is a closed subset of $A^n$ and with a little bit of extra effort prove that it 's also irreducible. Skipping the isomorphism part, I am confused with the last statement that a projective variety is a finite union of open affine varieties (the finite part is easy to show I guess from Hilbert's Basis Theorem, what I m confused is that they 're open). From the above, $V_i$ is closed, not open.
In general, I have seen the reverse statement going like this: Let $X=V(f)$ be an affine variety of $A^n$ and $D(f)=P in A^n: f(P) neq 0$. Obviously $D(f)$ is open. Then consider the map $phi: X to A^n+1$, where $phi(x_1,...,x_n)=(x_1,...,x_n,1/f(x_1,..,x_n) )$. Then you can indeed prove that $phi(X)=V(fX_n+1-1)$ and thus a closed subvariety of $A^n+1$. It seems to me that you can somehow reverse engineer this logic for the above proposition but I am not sure how.
If anyone had the patience to help me understand it, consider the following trivial example: Let $n=1, A^1, P^1$ and $V=V(X-Y)=(x,x), xin A^1$. If $f=X-Y$, then $f_ast = X-1$, and $V_2=V_ast =V(X-1)=1$. How can you show in this example that V is a union of open affine varieties?
I 'm sorry in advance if I have misunderstood anything, any help would be appreciated.
algebraic-geometry
edited Aug 15 at 11:10
José Carlos Santos
116k1699178
asked Aug 15 at 11:10
Foivos
35929
add a comment |Â
up vote
1
down vote
favorite
I 'm new to algebraic geometry and I am a little bit confused with a proposition on Fulton, Algebraic Curves, page 70 proposition 3:
I will quote the proposition:
Let V be a closed subvariety of $P^n$, $phi_i : A^n to U_i subset P^n$, (whereas $phi_n+1(x_1,...,x_n)=(x_1:...:x_n:1)$ ). Then $V_i=phi_i^-1(V)$ is a closed subvariety of $A^n$, and $phi_i$ restricts to an isomorphism of $V_i$ with $Vcap U_i$. A projective variety is a union of a finite number of open affine varieties.
It seems straightforward to show that $V_i=(Vcap U_i)_ast$ and thus prove that it is a closed subset of $A^n$ and with a little bit of extra effort prove that it 's also irreducible. Skipping the isomorphism part, I am confused with the last statement that a projective variety is a finite union of open affine varieties (the finite part is easy to show I guess from Hilbert's Basis Theorem, what I m confused is that they 're open). From the above, $V_i$ is closed, not open.
In general, I have seen the reverse statement going like this: Let $X=V(f)$ be an affine variety of $A^n$ and $D(f)=P in A^n: f(P) neq 0$. Obviously $D(f)$ is open. Then consider the map $phi: X to A^n+1$, where $phi(x_1,...,x_n)=(x_1,...,x_n,1/f(x_1,..,x_n) )$. Then you can indeed prove that $phi(X)=V(fX_n+1-1)$ and thus a closed subvariety of $A^n+1$. It seems to me that you can somehow reverse engineer this logic for the above proposition but I am not sure how.
If anyone had the patience to help me understand it, consider the following trivial example: Let $n=1, A^1, P^1$ and $V=V(X-Y)=(x,x), xin A^1$. If $f=X-Y$, then $f_ast = X-1$, and $V_2=V_ast =V(X-1)=1$. How can you show in this example that V is a union of open affine varieties?
I 'm sorry in advance if I have misunderstood anything, any help would be appreciated.
algebraic-geometry
edited Aug 15 at 11:10
José Carlos Santos
116k1699178
asked Aug 15 at 11:10
Foivos
35929
1
The proposition says that $Vcap U_i$ is an affine variety. But $Vcap U_i$ is also open in $V$ (with respect to the subspace topology). Therefore, $V = bigcup_i=1^n+1 (Vcap U_i)$ is a finite union of open subvarieties.
– Claudius
Aug 15 at 12:13
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I 'm new to algebraic geometry and I am a little bit confused with a proposition on Fulton, Algebraic Curves, page 70 proposition 3:
I will quote the proposition:
Let V be a closed subvariety of $P^n$, $phi_i : A^n to U_i subset P^n$, (whereas $phi_n+1(x_1,...,x_n)=(x_1:...:x_n:1)$ ). Then $V_i=phi_i^-1(V)$ is a closed subvariety of $A^n$, and $phi_i$ restricts to an isomorphism of $V_i$ with $Vcap U_i$. A projective variety is a union of a finite number of open affine varieties.
It seems straightforward to show that $V_i=(Vcap U_i)_ast$ and thus prove that it is a closed subset of $A^n$ and with a little bit of extra effort prove that it 's also irreducible. Skipping the isomorphism part, I am confused with the last statement that a projective variety is a finite union of open affine varieties (the finite part is easy to show I guess from Hilbert's Basis Theorem, what I m confused is that they 're open). From the above, $V_i$ is closed, not open.
In general, I have seen the reverse statement going like this: Let $X=V(f)$ be an affine variety of $A^n$ and $D(f)=P in A^n: f(P) neq 0$. Obviously $D(f)$ is open. Then consider the map $phi: X to A^n+1$, where $phi(x_1,...,x_n)=(x_1,...,x_n,1/f(x_1,..,x_n) )$. Then you can indeed prove that $phi(X)=V(fX_n+1-1)$ and thus a closed subvariety of $A^n+1$. It seems to me that you can somehow reverse engineer this logic for the above proposition but I am not sure how.
If anyone had the patience to help me understand it, consider the following trivial example: Let $n=1, A^1, P^1$ and $V=V(X-Y)=(x,x), xin A^1$. If $f=X-Y$, then $f_ast = X-1$, and $V_2=V_ast =V(X-1)=1$. How can you show in this example that V is a union of open affine varieties?
I 'm sorry in advance if I have misunderstood anything, any help would be appreciated.
algebraic-geometry
edited Aug 15 at 11:10
José Carlos Santos
116k1699178
asked Aug 15 at 11:10
Foivos
35929
I 'm new to algebraic geometry and I am a little bit confused with a proposition on Fulton, Algebraic Curves, page 70 proposition 3:
I will quote the proposition:
Let V be a closed subvariety of $P^n$, $phi_i : A^n to U_i subset P^n$, (whereas $phi_n+1(x_1,...,x_n)=(x_1:...:x_n:1)$ ). Then $V_i=phi_i^-1(V)$ is a closed subvariety of $A^n$, and $phi_i$ restricts to an isomorphism of $V_i$ with $Vcap U_i$. A projective variety is a union of a finite number of open affine varieties.
It seems straightforward to show that $V_i=(Vcap U_i)_ast$ and thus prove that it is a closed subset of $A^n$ and with a little bit of extra effort prove that it 's also irreducible. Skipping the isomorphism part, I am confused with the last statement that a projective variety is a finite union of open affine varieties (the finite part is easy to show I guess from Hilbert's Basis Theorem, what I m confused is that they 're open). From the above, $V_i$ is closed, not open.
In general, I have seen the reverse statement going like this: Let $X=V(f)$ be an affine variety of $A^n$ and $D(f)=P in A^n: f(P) neq 0$. Obviously $D(f)$ is open. Then consider the map $phi: X to A^n+1$, where $phi(x_1,...,x_n)=(x_1,...,x_n,1/f(x_1,..,x_n) )$. Then you can indeed prove that $phi(X)=V(fX_n+1-1)$ and thus a closed subvariety of $A^n+1$. It seems to me that you can somehow reverse engineer this logic for the above proposition but I am not sure how.
If anyone had the patience to help me understand it, consider the following trivial example: Let $n=1, A^1, P^1$ and $V=V(X-Y)=(x,x), xin A^1$. If $f=X-Y$, then $f_ast = X-1$, and $V_2=V_ast =V(X-1)=1$. How can you show in this example that V is a union of open affine varieties?
I 'm sorry in advance if I have misunderstood anything, any help would be appreciated.
algebraic-geometry
edited Aug 15 at 11:10
José Carlos Santos
116k1699178
asked Aug 15 at 11:10
Foivos
35929
edited Aug 15 at 11:10
José Carlos Santos
116k1699178
edited Aug 15 at 11:10
José Carlos Santos
116k1699178
edited Aug 15 at 11:10
José Carlos Santos
116k1699178
116k1699178
asked Aug 15 at 11:10
Foivos
35929
asked Aug 15 at 11:10
Foivos
35929
asked Aug 15 at 11:10
Foivos
35929
35929
1
The proposition says that $Vcap U_i$ is an affine variety. But $Vcap U_i$ is also open in $V$ (with respect to the subspace topology). Therefore, $V = bigcup_i=1^n+1 (Vcap U_i)$ is a finite union of open subvarieties.
– Claudius
Aug 15 at 12:13
add a comment |Â
1
The proposition says that $Vcap U_i$ is an affine variety. But $Vcap U_i$ is also open in $V$ (with respect to the subspace topology). Therefore, $V = bigcup_i=1^n+1 (Vcap U_i)$ is a finite union of open subvarieties.
– Claudius
Aug 15 at 12:13
1
1
The proposition says that $Vcap U_i$ is an affine variety. But $Vcap U_i$ is also open in $V$ (with respect to the subspace topology). Therefore, $V = bigcup_i=1^n+1 (Vcap U_i)$ is a finite union of open subvarieties.
– Claudius
Aug 15 at 12:13
The proposition says that $Vcap U_i$ is an affine variety. But $Vcap U_i$ is also open in $V$ (with respect to the subspace topology). Therefore, $V = bigcup_i=1^n+1 (Vcap U_i)$ is a finite union of open subvarieties.
– Claudius
Aug 15 at 12:13
add a comment |Â
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The proposition says that $Vcap U_i$ is an affine variety. But $Vcap U_i$ is also open in $V$ (with respect to the subspace topology). Therefore, $V = bigcup_i=1^n+1 (Vcap U_i)$ is a finite union of open subvarieties.
– Claudius
Aug 15 at 12:13