Understanding the blow up of $mathbb{A}^2$ in $left$.
1
$begingroup$
The following is an example from Gathmann's notes on algebraic geometry:
I am having problems with showing, rigorously, that $tilde X$ is given by the prescribed equations. First, I do not get why the RHS is even a closed subset. Yes, it is given by the zero set of a polynomial equation- but the topology on $mathbb{A}^2timesmathbb{P}^1$ is given by the product topology, and I do not see how to show that a zero set of a polynomial equation is closed in this product topology. Second, why can't we have a smaller closed subset containing $Gamma$?
Thanks in advance.
algebraic-geometry blowup
edited Dec 21 '18 at 11:42
Yanko
7,0081629
asked Dec 21 '18 at 11:40
MadarbMadarb
351211
$endgroup$
add a comment |
1
$begingroup$
The following is an example from Gathmann's notes on algebraic geometry:
I am having problems with showing, rigorously, that $tilde X$ is given by the prescribed equations. First, I do not get why the RHS is even a closed subset. Yes, it is given by the zero set of a polynomial equation- but the topology on $mathbb{A}^2timesmathbb{P}^1$ is given by the product topology, and I do not see how to show that a zero set of a polynomial equation is closed in this product topology. Second, why can't we have a smaller closed subset containing $Gamma$?
Thanks in advance.
algebraic-geometry blowup
edited Dec 21 '18 at 11:42
Yanko
7,0081629
asked Dec 21 '18 at 11:40
MadarbMadarb
351211
$endgroup$
4
$begingroup$
The topology on a product of varieties is not the topological product topology. We want $Bbb A_k^1times Bbb A^1_k$ to be homeomorphic to $Bbb A^2_k$, after all, and for that the product topology just won't do (the diagonal ${(x, x)mid xin k}$ wouldn't be closed, for instance). Presumably your book talks about this the first time it mentions products.
$endgroup$
– Arthur
Dec 21 '18 at 11:48
$begingroup$
That's a great comment, thank you! I didn't think about it. It does talk about it when it talk about product of affine variety - I checked now. But I don't find something about general varieties. I'll look it up on the internet, anyway.
$endgroup$
– Madarb
Dec 21 '18 at 11:52
1
$begingroup$
Topology is a local thing, and projective varieties are locally affine, so in some sense it's enough to discuss the topology on affine varieties. In this specific example, you can cover $Bbb P^1$ with two affine subvarieties $V_1$ and $V_2$, and then you can look at $Bbb A^2times V_1$ and $Bbb A_2times V_2$ separately to get their topologies, then take their union to get all of $Bbb A^2times Bbb P^1$.
$endgroup$
– Arthur
Dec 21 '18 at 12:05
$begingroup$
First of all thank you - your comments are very useful. I think I came up with a solution. Denote the affine patches of $mathbb{A}^2timesmathbb{P}^1$ by $U_1$,$U_2$. Then $tilde{X}$ is given on each patch by poly equations (on $y_0neq 0$ for example it is $x_0y_1=x_1$) hence is closed in each patch, and hence is closed in $tilde X$. It suffices therefore to check that $bar{Gamma}$ contains $tilde{X}$, which is true as $tilde{X}$ is the union of the closures on each patch, and the closure contains the union of closures on each patch by simple topology. Is it true? (...)
$endgroup$
– Madarb
Dec 24 '18 at 14:11
$begingroup$
(...) Also, it seem so complicated- is there a more simple way to see it?
$endgroup$
– Madarb
Dec 24 '18 at 14:13
add a comment |
1
1
1
$begingroup$
The following is an example from Gathmann's notes on algebraic geometry:
I am having problems with showing, rigorously, that $tilde X$ is given by the prescribed equations. First, I do not get why the RHS is even a closed subset. Yes, it is given by the zero set of a polynomial equation- but the topology on $mathbb{A}^2timesmathbb{P}^1$ is given by the product topology, and I do not see how to show that a zero set of a polynomial equation is closed in this product topology. Second, why can't we have a smaller closed subset containing $Gamma$?
Thanks in advance.
algebraic-geometry blowup
edited Dec 21 '18 at 11:42
Yanko
7,0081629
asked Dec 21 '18 at 11:40
MadarbMadarb
351211
$endgroup$
The following is an example from Gathmann's notes on algebraic geometry:
I am having problems with showing, rigorously, that $tilde X$ is given by the prescribed equations. First, I do not get why the RHS is even a closed subset. Yes, it is given by the zero set of a polynomial equation- but the topology on $mathbb{A}^2timesmathbb{P}^1$ is given by the product topology, and I do not see how to show that a zero set of a polynomial equation is closed in this product topology. Second, why can't we have a smaller closed subset containing $Gamma$?
Thanks in advance.
algebraic-geometry blowup
algebraic-geometry blowup
edited Dec 21 '18 at 11:42
Yanko
7,0081629
asked Dec 21 '18 at 11:40
MadarbMadarb
351211
edited Dec 21 '18 at 11:42
Yanko
7,0081629
asked Dec 21 '18 at 11:40
MadarbMadarb
351211
edited Dec 21 '18 at 11:42
Yanko
7,0081629
edited Dec 21 '18 at 11:42
Yanko
7,0081629
edited Dec 21 '18 at 11:42
Yanko
7,0081629
7,0081629
asked Dec 21 '18 at 11:40
MadarbMadarb
351211
asked Dec 21 '18 at 11:40
MadarbMadarb
351211
asked Dec 21 '18 at 11:40
MadarbMadarb
351211
351211
4
$begingroup$
The topology on a product of varieties is not the topological product topology. We want $Bbb A_k^1times Bbb A^1_k$ to be homeomorphic to $Bbb A^2_k$, after all, and for that the product topology just won't do (the diagonal ${(x, x)mid xin k}$ wouldn't be closed, for instance). Presumably your book talks about this the first time it mentions products.
$endgroup$
– Arthur
Dec 21 '18 at 11:48
$begingroup$
That's a great comment, thank you! I didn't think about it. It does talk about it when it talk about product of affine variety - I checked now. But I don't find something about general varieties. I'll look it up on the internet, anyway.
$endgroup$
– Madarb
Dec 21 '18 at 11:52
1
$begingroup$
Topology is a local thing, and projective varieties are locally affine, so in some sense it's enough to discuss the topology on affine varieties. In this specific example, you can cover $Bbb P^1$ with two affine subvarieties $V_1$ and $V_2$, and then you can look at $Bbb A^2times V_1$ and $Bbb A_2times V_2$ separately to get their topologies, then take their union to get all of $Bbb A^2times Bbb P^1$.
$endgroup$
– Arthur
Dec 21 '18 at 12:05
$begingroup$
First of all thank you - your comments are very useful. I think I came up with a solution. Denote the affine patches of $mathbb{A}^2timesmathbb{P}^1$ by $U_1$,$U_2$. Then $tilde{X}$ is given on each patch by poly equations (on $y_0neq 0$ for example it is $x_0y_1=x_1$) hence is closed in each patch, and hence is closed in $tilde X$. It suffices therefore to check that $bar{Gamma}$ contains $tilde{X}$, which is true as $tilde{X}$ is the union of the closures on each patch, and the closure contains the union of closures on each patch by simple topology. Is it true? (...)
$endgroup$
– Madarb
Dec 24 '18 at 14:11
$begingroup$
(...) Also, it seem so complicated- is there a more simple way to see it?
$endgroup$
– Madarb
Dec 24 '18 at 14:13
add a comment |
4
$begingroup$
The topology on a product of varieties is not the topological product topology. We want $Bbb A_k^1times Bbb A^1_k$ to be homeomorphic to $Bbb A^2_k$, after all, and for that the product topology just won't do (the diagonal ${(x, x)mid xin k}$ wouldn't be closed, for instance). Presumably your book talks about this the first time it mentions products.
$endgroup$
– Arthur
Dec 21 '18 at 11:48
$begingroup$
That's a great comment, thank you! I didn't think about it. It does talk about it when it talk about product of affine variety - I checked now. But I don't find something about general varieties. I'll look it up on the internet, anyway.
$endgroup$
– Madarb
Dec 21 '18 at 11:52
1
$begingroup$
Topology is a local thing, and projective varieties are locally affine, so in some sense it's enough to discuss the topology on affine varieties. In this specific example, you can cover $Bbb P^1$ with two affine subvarieties $V_1$ and $V_2$, and then you can look at $Bbb A^2times V_1$ and $Bbb A_2times V_2$ separately to get their topologies, then take their union to get all of $Bbb A^2times Bbb P^1$.
$endgroup$
– Arthur
Dec 21 '18 at 12:05
$begingroup$
First of all thank you - your comments are very useful. I think I came up with a solution. Denote the affine patches of $mathbb{A}^2timesmathbb{P}^1$ by $U_1$,$U_2$. Then $tilde{X}$ is given on each patch by poly equations (on $y_0neq 0$ for example it is $x_0y_1=x_1$) hence is closed in each patch, and hence is closed in $tilde X$. It suffices therefore to check that $bar{Gamma}$ contains $tilde{X}$, which is true as $tilde{X}$ is the union of the closures on each patch, and the closure contains the union of closures on each patch by simple topology. Is it true? (...)
$endgroup$
– Madarb
Dec 24 '18 at 14:11
$begingroup$
(...) Also, it seem so complicated- is there a more simple way to see it?
$endgroup$
– Madarb
Dec 24 '18 at 14:13
4
4
$begingroup$
The topology on a product of varieties is not the topological product topology. We want $Bbb A_k^1times Bbb A^1_k$ to be homeomorphic to $Bbb A^2_k$, after all, and for that the product topology just won't do (the diagonal ${(x, x)mid xin k}$ wouldn't be closed, for instance). Presumably your book talks about this the first time it mentions products.
$endgroup$
– Arthur
Dec 21 '18 at 11:48
$begingroup$
The topology on a product of varieties is not the topological product topology. We want $Bbb A_k^1times Bbb A^1_k$ to be homeomorphic to $Bbb A^2_k$, after all, and for that the product topology just won't do (the diagonal ${(x, x)mid xin k}$ wouldn't be closed, for instance). Presumably your book talks about this the first time it mentions products.
$endgroup$
– Arthur
Dec 21 '18 at 11:48
$begingroup$
That's a great comment, thank you! I didn't think about it. It does talk about it when it talk about product of affine variety - I checked now. But I don't find something about general varieties. I'll look it up on the internet, anyway.
$endgroup$
– Madarb
Dec 21 '18 at 11:52
$begingroup$
That's a great comment, thank you! I didn't think about it. It does talk about it when it talk about product of affine variety - I checked now. But I don't find something about general varieties. I'll look it up on the internet, anyway.
$endgroup$
– Madarb
Dec 21 '18 at 11:52
1
1
$begingroup$
Topology is a local thing, and projective varieties are locally affine, so in some sense it's enough to discuss the topology on affine varieties. In this specific example, you can cover $Bbb P^1$ with two affine subvarieties $V_1$ and $V_2$, and then you can look at $Bbb A^2times V_1$ and $Bbb A_2times V_2$ separately to get their topologies, then take their union to get all of $Bbb A^2times Bbb P^1$.
$endgroup$
– Arthur
Dec 21 '18 at 12:05
$begingroup$
Topology is a local thing, and projective varieties are locally affine, so in some sense it's enough to discuss the topology on affine varieties. In this specific example, you can cover $Bbb P^1$ with two affine subvarieties $V_1$ and $V_2$, and then you can look at $Bbb A^2times V_1$ and $Bbb A_2times V_2$ separately to get their topologies, then take their union to get all of $Bbb A^2times Bbb P^1$.
$endgroup$
– Arthur
Dec 21 '18 at 12:05
$begingroup$
First of all thank you - your comments are very useful. I think I came up with a solution. Denote the affine patches of $mathbb{A}^2timesmathbb{P}^1$ by $U_1$,$U_2$. Then $tilde{X}$ is given on each patch by poly equations (on $y_0neq 0$ for example it is $x_0y_1=x_1$) hence is closed in each patch, and hence is closed in $tilde X$. It suffices therefore to check that $bar{Gamma}$ contains $tilde{X}$, which is true as $tilde{X}$ is the union of the closures on each patch, and the closure contains the union of closures on each patch by simple topology. Is it true? (...)
$endgroup$
– Madarb
Dec 24 '18 at 14:11
$begingroup$
First of all thank you - your comments are very useful. I think I came up with a solution. Denote the affine patches of $mathbb{A}^2timesmathbb{P}^1$ by $U_1$,$U_2$. Then $tilde{X}$ is given on each patch by poly equations (on $y_0neq 0$ for example it is $x_0y_1=x_1$) hence is closed in each patch, and hence is closed in $tilde X$. It suffices therefore to check that $bar{Gamma}$ contains $tilde{X}$, which is true as $tilde{X}$ is the union of the closures on each patch, and the closure contains the union of closures on each patch by simple topology. Is it true? (...)
$endgroup$
– Madarb
Dec 24 '18 at 14:11
$begingroup$
(...) Also, it seem so complicated- is there a more simple way to see it?
$endgroup$
– Madarb
Dec 24 '18 at 14:13
$begingroup$
(...) Also, it seem so complicated- is there a more simple way to see it?
$endgroup$
– Madarb
Dec 24 '18 at 14:13
add a comment |
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4
$begingroup$
The topology on a product of varieties is not the topological product topology. We want $Bbb A_k^1times Bbb A^1_k$ to be homeomorphic to $Bbb A^2_k$, after all, and for that the product topology just won't do (the diagonal ${(x, x)mid xin k}$ wouldn't be closed, for instance). Presumably your book talks about this the first time it mentions products.
$endgroup$
– Arthur
Dec 21 '18 at 11:48
$begingroup$
That's a great comment, thank you! I didn't think about it. It does talk about it when it talk about product of affine variety - I checked now. But I don't find something about general varieties. I'll look it up on the internet, anyway.
$endgroup$
– Madarb
Dec 21 '18 at 11:52
1
$begingroup$
Topology is a local thing, and projective varieties are locally affine, so in some sense it's enough to discuss the topology on affine varieties. In this specific example, you can cover $Bbb P^1$ with two affine subvarieties $V_1$ and $V_2$, and then you can look at $Bbb A^2times V_1$ and $Bbb A_2times V_2$ separately to get their topologies, then take their union to get all of $Bbb A^2times Bbb P^1$.
$endgroup$
– Arthur
Dec 21 '18 at 12:05
$begingroup$
First of all thank you - your comments are very useful. I think I came up with a solution. Denote the affine patches of $mathbb{A}^2timesmathbb{P}^1$ by $U_1$,$U_2$. Then $tilde{X}$ is given on each patch by poly equations (on $y_0neq 0$ for example it is $x_0y_1=x_1$) hence is closed in each patch, and hence is closed in $tilde X$. It suffices therefore to check that $bar{Gamma}$ contains $tilde{X}$, which is true as $tilde{X}$ is the union of the closures on each patch, and the closure contains the union of closures on each patch by simple topology. Is it true? (...)
$endgroup$
– Madarb
Dec 24 '18 at 14:11
$begingroup$
(...) Also, it seem so complicated- is there a more simple way to see it?
$endgroup$
– Madarb
Dec 24 '18 at 14:13