Isomorphism between an algebraic set and a cartesian product [closed]
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Let W, X and Y be algebraic sets and $gamma_1$: W $rightarrow$ X, $gamma_2$: W $rightarrow$ Y, two morphisms which verify that, given Z an algebraic set and two morphisms $alpha$: Z $rightarrow$ X and $beta$: Z $rightarrow$ Y, exists only one morphism $phi$: Z $rightarrow$ W verifying $alpha = gamma_1phi$ and $beta = gamma_2phi$. Prove that W is isomorphic to X x Y.
algebraic-geometry commutative-algebra
asked Nov 28 '18 at 11:48
Julia Vazquez EscobarJulia Vazquez Escobar
31
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closed as off-topic by amWhy, Alexander Gruber♦ Nov 30 '18 at 3:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let W, X and Y be algebraic sets and $gamma_1$: W $rightarrow$ X, $gamma_2$: W $rightarrow$ Y, two morphisms which verify that, given Z an algebraic set and two morphisms $alpha$: Z $rightarrow$ X and $beta$: Z $rightarrow$ Y, exists only one morphism $phi$: Z $rightarrow$ W verifying $alpha = gamma_1phi$ and $beta = gamma_2phi$. Prove that W is isomorphic to X x Y.
algebraic-geometry commutative-algebra
asked Nov 28 '18 at 11:48
Julia Vazquez EscobarJulia Vazquez Escobar
31
$endgroup$
closed as off-topic by amWhy, Alexander Gruber♦ Nov 30 '18 at 3:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
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Hi Julia, please write what did you try and what do you think about this problem if you would like useful feedback from the MSE community.
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– Nicolas Hemelsoet
Nov 28 '18 at 11:59
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I've tried choosing X x Y as Z, so $alpha$ and $beta$ would be the prejction morphisms. I've stated that exists $psi$: W $rightarrow$ X x Y expressed as $psi(t) = (gamma_1(t), gamma_2(t))$ that clearly is a morphism, and $phi$: X x Y $rightarrow$ W would be the inverse, we know that exists from the problem statements. But I'm having a hard time proving $phi circ psi$ is the identity.
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– Julia Vazquez Escobar
Nov 28 '18 at 12:08
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Thank you for your responses
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– Julia Vazquez Escobar
Nov 28 '18 at 14:05
add a comment |
$begingroup$
Let W, X and Y be algebraic sets and $gamma_1$: W $rightarrow$ X, $gamma_2$: W $rightarrow$ Y, two morphisms which verify that, given Z an algebraic set and two morphisms $alpha$: Z $rightarrow$ X and $beta$: Z $rightarrow$ Y, exists only one morphism $phi$: Z $rightarrow$ W verifying $alpha = gamma_1phi$ and $beta = gamma_2phi$. Prove that W is isomorphic to X x Y.
algebraic-geometry commutative-algebra
asked Nov 28 '18 at 11:48
Julia Vazquez EscobarJulia Vazquez Escobar
31
$endgroup$
Let W, X and Y be algebraic sets and $gamma_1$: W $rightarrow$ X, $gamma_2$: W $rightarrow$ Y, two morphisms which verify that, given Z an algebraic set and two morphisms $alpha$: Z $rightarrow$ X and $beta$: Z $rightarrow$ Y, exists only one morphism $phi$: Z $rightarrow$ W verifying $alpha = gamma_1phi$ and $beta = gamma_2phi$. Prove that W is isomorphic to X x Y.
algebraic-geometry commutative-algebra
algebraic-geometry commutative-algebra
asked Nov 28 '18 at 11:48
Julia Vazquez EscobarJulia Vazquez Escobar
31
asked Nov 28 '18 at 11:48
Julia Vazquez EscobarJulia Vazquez Escobar
31
asked Nov 28 '18 at 11:48
Julia Vazquez EscobarJulia Vazquez Escobar
31
asked Nov 28 '18 at 11:48
Julia Vazquez EscobarJulia Vazquez Escobar
31
asked Nov 28 '18 at 11:48
Julia Vazquez EscobarJulia Vazquez Escobar
31
31
closed as off-topic by amWhy, Alexander Gruber♦ Nov 30 '18 at 3:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Alexander Gruber♦ Nov 30 '18 at 3:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Hi Julia, please write what did you try and what do you think about this problem if you would like useful feedback from the MSE community.
$endgroup$
– Nicolas Hemelsoet
Nov 28 '18 at 11:59
$begingroup$
I've tried choosing X x Y as Z, so $alpha$ and $beta$ would be the prejction morphisms. I've stated that exists $psi$: W $rightarrow$ X x Y expressed as $psi(t) = (gamma_1(t), gamma_2(t))$ that clearly is a morphism, and $phi$: X x Y $rightarrow$ W would be the inverse, we know that exists from the problem statements. But I'm having a hard time proving $phi circ psi$ is the identity.
$endgroup$
– Julia Vazquez Escobar
Nov 28 '18 at 12:08
$begingroup$
Thank you for your responses
$endgroup$
– Julia Vazquez Escobar
Nov 28 '18 at 14:05
add a comment |
$begingroup$
Hi Julia, please write what did you try and what do you think about this problem if you would like useful feedback from the MSE community.
$endgroup$
– Nicolas Hemelsoet
Nov 28 '18 at 11:59
$begingroup$
I've tried choosing X x Y as Z, so $alpha$ and $beta$ would be the prejction morphisms. I've stated that exists $psi$: W $rightarrow$ X x Y expressed as $psi(t) = (gamma_1(t), gamma_2(t))$ that clearly is a morphism, and $phi$: X x Y $rightarrow$ W would be the inverse, we know that exists from the problem statements. But I'm having a hard time proving $phi circ psi$ is the identity.
$endgroup$
– Julia Vazquez Escobar
Nov 28 '18 at 12:08
$begingroup$
Thank you for your responses
$endgroup$
– Julia Vazquez Escobar
Nov 28 '18 at 14:05
$begingroup$
Hi Julia, please write what did you try and what do you think about this problem if you would like useful feedback from the MSE community.
$endgroup$
– Nicolas Hemelsoet
Nov 28 '18 at 11:59
$begingroup$
Hi Julia, please write what did you try and what do you think about this problem if you would like useful feedback from the MSE community.
$endgroup$
– Nicolas Hemelsoet
Nov 28 '18 at 11:59
$begingroup$
I've tried choosing X x Y as Z, so $alpha$ and $beta$ would be the prejction morphisms. I've stated that exists $psi$: W $rightarrow$ X x Y expressed as $psi(t) = (gamma_1(t), gamma_2(t))$ that clearly is a morphism, and $phi$: X x Y $rightarrow$ W would be the inverse, we know that exists from the problem statements. But I'm having a hard time proving $phi circ psi$ is the identity.
$endgroup$
– Julia Vazquez Escobar
Nov 28 '18 at 12:08
$begingroup$
I've tried choosing X x Y as Z, so $alpha$ and $beta$ would be the prejction morphisms. I've stated that exists $psi$: W $rightarrow$ X x Y expressed as $psi(t) = (gamma_1(t), gamma_2(t))$ that clearly is a morphism, and $phi$: X x Y $rightarrow$ W would be the inverse, we know that exists from the problem statements. But I'm having a hard time proving $phi circ psi$ is the identity.
$endgroup$
– Julia Vazquez Escobar
Nov 28 '18 at 12:08
$begingroup$
Thank you for your responses
$endgroup$
– Julia Vazquez Escobar
Nov 28 '18 at 14:05
$begingroup$
Thank you for your responses
$endgroup$
– Julia Vazquez Escobar
Nov 28 '18 at 14:05
add a comment |
1 Answer
1
active
oldest
votes
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Well, you have to show that $Xtimes Y$ satisfies the same condition as $Z$. Doing this, you will have that there is a unique morphism $Xtimes Yto Z$ and a unique morphism $Zto Xtimes Y$ (such that the respective diagrams commute), since there is a unique morphism such that makes the diagram commutes and such that $Zto Z$ is the identity (same argument for $Xtimes Yto Xtimes Y$). You will have that $Z$ is isomorphic (with unique isomorphism) to $Xtimes Y$.
Now, forget the algebraic structure for a second. $Z$ satisfies the product universal property for sets, so that $Z$ is isomorphic to $Xtimes Y$ (see the above argument) with a unique isomorphism. Now, if $phi:Xtimes Yto Z$ is the biyection (iso of sets) you have to prove that it is an isomorphism of algebraic varieties. This is, if $psi:Zto Xtimes Y$ if the (set) inverse of $phi$: Probe that $phi$ and $psi$ are morphisms. But this will follow from the diagrams and using the definition of algebraic set morphism.
answered Nov 28 '18 at 14:16
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
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Thank you very much :)
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– Julia Vazquez Escobar
Nov 28 '18 at 20:39
$begingroup$
You are welcome. And good luck with algebraic geometry, it's a really beatiful (and hard) subject
$endgroup$
– José Alejandro Aburto Araneda
Nov 29 '18 at 0:01
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Well, you have to show that $Xtimes Y$ satisfies the same condition as $Z$. Doing this, you will have that there is a unique morphism $Xtimes Yto Z$ and a unique morphism $Zto Xtimes Y$ (such that the respective diagrams commute), since there is a unique morphism such that makes the diagram commutes and such that $Zto Z$ is the identity (same argument for $Xtimes Yto Xtimes Y$). You will have that $Z$ is isomorphic (with unique isomorphism) to $Xtimes Y$.
Now, forget the algebraic structure for a second. $Z$ satisfies the product universal property for sets, so that $Z$ is isomorphic to $Xtimes Y$ (see the above argument) with a unique isomorphism. Now, if $phi:Xtimes Yto Z$ is the biyection (iso of sets) you have to prove that it is an isomorphism of algebraic varieties. This is, if $psi:Zto Xtimes Y$ if the (set) inverse of $phi$: Probe that $phi$ and $psi$ are morphisms. But this will follow from the diagrams and using the definition of algebraic set morphism.
answered Nov 28 '18 at 14:16
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
$begingroup$
Thank you very much :)
$endgroup$
– Julia Vazquez Escobar
Nov 28 '18 at 20:39
$begingroup$
You are welcome. And good luck with algebraic geometry, it's a really beatiful (and hard) subject
$endgroup$
– José Alejandro Aburto Araneda
Nov 29 '18 at 0:01
add a comment |
$begingroup$
Well, you have to show that $Xtimes Y$ satisfies the same condition as $Z$. Doing this, you will have that there is a unique morphism $Xtimes Yto Z$ and a unique morphism $Zto Xtimes Y$ (such that the respective diagrams commute), since there is a unique morphism such that makes the diagram commutes and such that $Zto Z$ is the identity (same argument for $Xtimes Yto Xtimes Y$). You will have that $Z$ is isomorphic (with unique isomorphism) to $Xtimes Y$.
Now, forget the algebraic structure for a second. $Z$ satisfies the product universal property for sets, so that $Z$ is isomorphic to $Xtimes Y$ (see the above argument) with a unique isomorphism. Now, if $phi:Xtimes Yto Z$ is the biyection (iso of sets) you have to prove that it is an isomorphism of algebraic varieties. This is, if $psi:Zto Xtimes Y$ if the (set) inverse of $phi$: Probe that $phi$ and $psi$ are morphisms. But this will follow from the diagrams and using the definition of algebraic set morphism.
answered Nov 28 '18 at 14:16
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
$begingroup$
Thank you very much :)
$endgroup$
– Julia Vazquez Escobar
Nov 28 '18 at 20:39
$begingroup$
You are welcome. And good luck with algebraic geometry, it's a really beatiful (and hard) subject
$endgroup$
– José Alejandro Aburto Araneda
Nov 29 '18 at 0:01
add a comment |
$begingroup$
Well, you have to show that $Xtimes Y$ satisfies the same condition as $Z$. Doing this, you will have that there is a unique morphism $Xtimes Yto Z$ and a unique morphism $Zto Xtimes Y$ (such that the respective diagrams commute), since there is a unique morphism such that makes the diagram commutes and such that $Zto Z$ is the identity (same argument for $Xtimes Yto Xtimes Y$). You will have that $Z$ is isomorphic (with unique isomorphism) to $Xtimes Y$.
Now, forget the algebraic structure for a second. $Z$ satisfies the product universal property for sets, so that $Z$ is isomorphic to $Xtimes Y$ (see the above argument) with a unique isomorphism. Now, if $phi:Xtimes Yto Z$ is the biyection (iso of sets) you have to prove that it is an isomorphism of algebraic varieties. This is, if $psi:Zto Xtimes Y$ if the (set) inverse of $phi$: Probe that $phi$ and $psi$ are morphisms. But this will follow from the diagrams and using the definition of algebraic set morphism.
answered Nov 28 '18 at 14:16
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
$endgroup$
Well, you have to show that $Xtimes Y$ satisfies the same condition as $Z$. Doing this, you will have that there is a unique morphism $Xtimes Yto Z$ and a unique morphism $Zto Xtimes Y$ (such that the respective diagrams commute), since there is a unique morphism such that makes the diagram commutes and such that $Zto Z$ is the identity (same argument for $Xtimes Yto Xtimes Y$). You will have that $Z$ is isomorphic (with unique isomorphism) to $Xtimes Y$.
Now, forget the algebraic structure for a second. $Z$ satisfies the product universal property for sets, so that $Z$ is isomorphic to $Xtimes Y$ (see the above argument) with a unique isomorphism. Now, if $phi:Xtimes Yto Z$ is the biyection (iso of sets) you have to prove that it is an isomorphism of algebraic varieties. This is, if $psi:Zto Xtimes Y$ if the (set) inverse of $phi$: Probe that $phi$ and $psi$ are morphisms. But this will follow from the diagrams and using the definition of algebraic set morphism.
answered Nov 28 '18 at 14:16
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
edited Nov 28 '18 at 23:58
answered Nov 28 '18 at 14:16
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
answered Nov 28 '18 at 14:16
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
answered Nov 28 '18 at 14:16
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
802110
802110
$begingroup$
Thank you very much :)
$endgroup$
– Julia Vazquez Escobar
Nov 28 '18 at 20:39
$begingroup$
You are welcome. And good luck with algebraic geometry, it's a really beatiful (and hard) subject
$endgroup$
– José Alejandro Aburto Araneda
Nov 29 '18 at 0:01
add a comment |
$begingroup$
Thank you very much :)
$endgroup$
– Julia Vazquez Escobar
Nov 28 '18 at 20:39
$begingroup$
You are welcome. And good luck with algebraic geometry, it's a really beatiful (and hard) subject
$endgroup$
– José Alejandro Aburto Araneda
Nov 29 '18 at 0:01
$begingroup$
Thank you very much :)
$endgroup$
– Julia Vazquez Escobar
Nov 28 '18 at 20:39
$begingroup$
Thank you very much :)
$endgroup$
– Julia Vazquez Escobar
Nov 28 '18 at 20:39
$begingroup$
You are welcome. And good luck with algebraic geometry, it's a really beatiful (and hard) subject
$endgroup$
– José Alejandro Aburto Araneda
Nov 29 '18 at 0:01
$begingroup$
You are welcome. And good luck with algebraic geometry, it's a really beatiful (and hard) subject
$endgroup$
– José Alejandro Aburto Araneda
Nov 29 '18 at 0:01
add a comment |
$begingroup$
Hi Julia, please write what did you try and what do you think about this problem if you would like useful feedback from the MSE community.
$endgroup$
– Nicolas Hemelsoet
Nov 28 '18 at 11:59
$begingroup$
I've tried choosing X x Y as Z, so $alpha$ and $beta$ would be the prejction morphisms. I've stated that exists $psi$: W $rightarrow$ X x Y expressed as $psi(t) = (gamma_1(t), gamma_2(t))$ that clearly is a morphism, and $phi$: X x Y $rightarrow$ W would be the inverse, we know that exists from the problem statements. But I'm having a hard time proving $phi circ psi$ is the identity.
$endgroup$
– Julia Vazquez Escobar
Nov 28 '18 at 12:08
$begingroup$
Thank you for your responses
$endgroup$
– Julia Vazquez Escobar
Nov 28 '18 at 14:05