Trying to understand morphisms between varieties












2












$begingroup$


Let $k_1, k_2, k_0 in mathbb{N}$.
If $k_1 = k_2 = k_0$ then the map
which sends $[x_0:x_1:x_2] in mathbb{P}^2$ to $[x_0^{k_0}:x_1^{k_1}:x_2^{k_2}] in mathbb{P}^2$ is well defined, but
it's not well defined if all of the $k_i$'s are not the same.



(My understnading of morphisms between projective varieties:
$varphi: V to W$ is a map between projective varieties $V subset mathbb{P}^n$ and $W subset mathbb{P}^m$ given by $varphi([x_0 : ldots : x_n]) = [varphi_0([x_0 : ldots : x_n]): ldots : varphi_m([x_0 : ldots : x_n])]$, where the $varphi_i$ are homogeneous polynomials of the same degree that don't vanish simultaneously at any point of $V$.)



I was just wondering when not all of the $k_i$'s are equal, does the map 'make sense' if I change the domain to affine space?
i.e. if I define a map from $mathbb{A}^3 backslash { mathbf{0} }$ to $mathbb{P}^2$ by $(x_0, x_1, x_2) rightarrow [x_0^{k_0}:x_1^{k_1}:x_2^{k_2} ]$, is this just a weird map that sends of $mathbb{A}^3 backslash { mathbf{0} }$ to $mathbb{P}^2$? or does this become a morphism in an appropriate category (something that generalizes affine varieties and projective varities maybe?) Any comments would be appreciated. Thank you.



ps for simplicity I'm only thinking the affine space and the projective space over $mathbb{C}$






algebraic-geometry







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    The map as you define from affine space outside zero is indeed a morphism. The map you define can also be interpreted as a morphism of weighted projective spaces.
    $endgroup$
    – Mohan
    Dec 22 '18 at 20:28
















2












$begingroup$


Let $k_1, k_2, k_0 in mathbb{N}$.
If $k_1 = k_2 = k_0$ then the map
which sends $[x_0:x_1:x_2] in mathbb{P}^2$ to $[x_0^{k_0}:x_1^{k_1}:x_2^{k_2}] in mathbb{P}^2$ is well defined, but
it's not well defined if all of the $k_i$'s are not the same.



(My understnading of morphisms between projective varieties:
$varphi: V to W$ is a map between projective varieties $V subset mathbb{P}^n$ and $W subset mathbb{P}^m$ given by $varphi([x_0 : ldots : x_n]) = [varphi_0([x_0 : ldots : x_n]): ldots : varphi_m([x_0 : ldots : x_n])]$, where the $varphi_i$ are homogeneous polynomials of the same degree that don't vanish simultaneously at any point of $V$.)



I was just wondering when not all of the $k_i$'s are equal, does the map 'make sense' if I change the domain to affine space?
i.e. if I define a map from $mathbb{A}^3 backslash { mathbf{0} }$ to $mathbb{P}^2$ by $(x_0, x_1, x_2) rightarrow [x_0^{k_0}:x_1^{k_1}:x_2^{k_2} ]$, is this just a weird map that sends of $mathbb{A}^3 backslash { mathbf{0} }$ to $mathbb{P}^2$? or does this become a morphism in an appropriate category (something that generalizes affine varieties and projective varities maybe?) Any comments would be appreciated. Thank you.



ps for simplicity I'm only thinking the affine space and the projective space over $mathbb{C}$






algebraic-geometry







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    The map as you define from affine space outside zero is indeed a morphism. The map you define can also be interpreted as a morphism of weighted projective spaces.
    $endgroup$
    – Mohan
    Dec 22 '18 at 20:28














2












2








2





$begingroup$


Let $k_1, k_2, k_0 in mathbb{N}$.
If $k_1 = k_2 = k_0$ then the map
which sends $[x_0:x_1:x_2] in mathbb{P}^2$ to $[x_0^{k_0}:x_1^{k_1}:x_2^{k_2}] in mathbb{P}^2$ is well defined, but
it's not well defined if all of the $k_i$'s are not the same.



(My understnading of morphisms between projective varieties:
$varphi: V to W$ is a map between projective varieties $V subset mathbb{P}^n$ and $W subset mathbb{P}^m$ given by $varphi([x_0 : ldots : x_n]) = [varphi_0([x_0 : ldots : x_n]): ldots : varphi_m([x_0 : ldots : x_n])]$, where the $varphi_i$ are homogeneous polynomials of the same degree that don't vanish simultaneously at any point of $V$.)



I was just wondering when not all of the $k_i$'s are equal, does the map 'make sense' if I change the domain to affine space?
i.e. if I define a map from $mathbb{A}^3 backslash { mathbf{0} }$ to $mathbb{P}^2$ by $(x_0, x_1, x_2) rightarrow [x_0^{k_0}:x_1^{k_1}:x_2^{k_2} ]$, is this just a weird map that sends of $mathbb{A}^3 backslash { mathbf{0} }$ to $mathbb{P}^2$? or does this become a morphism in an appropriate category (something that generalizes affine varieties and projective varities maybe?) Any comments would be appreciated. Thank you.



ps for simplicity I'm only thinking the affine space and the projective space over $mathbb{C}$






algebraic-geometry







share|cite|improve this question









$endgroup$




Let $k_1, k_2, k_0 in mathbb{N}$.
If $k_1 = k_2 = k_0$ then the map
which sends $[x_0:x_1:x_2] in mathbb{P}^2$ to $[x_0^{k_0}:x_1^{k_1}:x_2^{k_2}] in mathbb{P}^2$ is well defined, but
it's not well defined if all of the $k_i$'s are not the same.



(My understnading of morphisms between projective varieties:
$varphi: V to W$ is a map between projective varieties $V subset mathbb{P}^n$ and $W subset mathbb{P}^m$ given by $varphi([x_0 : ldots : x_n]) = [varphi_0([x_0 : ldots : x_n]): ldots : varphi_m([x_0 : ldots : x_n])]$, where the $varphi_i$ are homogeneous polynomials of the same degree that don't vanish simultaneously at any point of $V$.)



I was just wondering when not all of the $k_i$'s are equal, does the map 'make sense' if I change the domain to affine space?
i.e. if I define a map from $mathbb{A}^3 backslash { mathbf{0} }$ to $mathbb{P}^2$ by $(x_0, x_1, x_2) rightarrow [x_0^{k_0}:x_1^{k_1}:x_2^{k_2} ]$, is this just a weird map that sends of $mathbb{A}^3 backslash { mathbf{0} }$ to $mathbb{P}^2$? or does this become a morphism in an appropriate category (something that generalizes affine varieties and projective varities maybe?) Any comments would be appreciated. Thank you.



ps for simplicity I'm only thinking the affine space and the projective space over $mathbb{C}$






algebraic-geometry




algebraic-geometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 22 '18 at 20:11









Johnny T.Johnny T.

5821415




5821415








  • 2




    $begingroup$
    The map as you define from affine space outside zero is indeed a morphism. The map you define can also be interpreted as a morphism of weighted projective spaces.
    $endgroup$
    – Mohan
    Dec 22 '18 at 20:28














  • 2




    $begingroup$
    The map as you define from affine space outside zero is indeed a morphism. The map you define can also be interpreted as a morphism of weighted projective spaces.
    $endgroup$
    – Mohan
    Dec 22 '18 at 20:28








2




2




$begingroup$
The map as you define from affine space outside zero is indeed a morphism. The map you define can also be interpreted as a morphism of weighted projective spaces.
$endgroup$
– Mohan
Dec 22 '18 at 20:28




$begingroup$
The map as you define from affine space outside zero is indeed a morphism. The map you define can also be interpreted as a morphism of weighted projective spaces.
$endgroup$
– Mohan
Dec 22 '18 at 20:28










1 Answer
1






active

oldest

votes


















1












$begingroup$

You can consider the affine map $(x,y,z)mapsto (x^a,y^b,z^c)$ from $mathbb{A}^3-{0}$ to itself. (Observe that this map does not contain $0$ in its image). Hence you can compose it with the quotient map $(x,y,z)mapsto [x:y:z]$ (which is well-defined on non-zero points) to obtain the map you give.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is there a terminology for this composed map? (as in it's not a morphism between affine varieties or morphism between projective varieties. So do we still say this is a morphism between something?)
    $endgroup$
    – Johnny T.
    Dec 22 '18 at 20:34






  • 1




    $begingroup$
    It is a morphism of varieties :)
    $endgroup$
    – Levent
    Dec 22 '18 at 20:38










  • $begingroup$
    Thank you very much. I was wondering if I could ask you just one more question: is this (this composed map) a finite morphism?
    $endgroup$
    – Johnny T.
    Dec 22 '18 at 20:46








  • 1




    $begingroup$
    No it is not. Note that finite morphisms necessarily have finite fibers, where as for example when $k_0=k_1=k_2=1$ all fibers are infinite.
    $endgroup$
    – Levent
    Dec 22 '18 at 20:51






  • 1




    $begingroup$
    The key concept here, as Mohan mentions above, is weighted projective space.
    $endgroup$
    – user347489
    Dec 23 '18 at 2:10












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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1












$begingroup$

You can consider the affine map $(x,y,z)mapsto (x^a,y^b,z^c)$ from $mathbb{A}^3-{0}$ to itself. (Observe that this map does not contain $0$ in its image). Hence you can compose it with the quotient map $(x,y,z)mapsto [x:y:z]$ (which is well-defined on non-zero points) to obtain the map you give.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is there a terminology for this composed map? (as in it's not a morphism between affine varieties or morphism between projective varieties. So do we still say this is a morphism between something?)
    $endgroup$
    – Johnny T.
    Dec 22 '18 at 20:34






  • 1




    $begingroup$
    It is a morphism of varieties :)
    $endgroup$
    – Levent
    Dec 22 '18 at 20:38










  • $begingroup$
    Thank you very much. I was wondering if I could ask you just one more question: is this (this composed map) a finite morphism?
    $endgroup$
    – Johnny T.
    Dec 22 '18 at 20:46








  • 1




    $begingroup$
    No it is not. Note that finite morphisms necessarily have finite fibers, where as for example when $k_0=k_1=k_2=1$ all fibers are infinite.
    $endgroup$
    – Levent
    Dec 22 '18 at 20:51






  • 1




    $begingroup$
    The key concept here, as Mohan mentions above, is weighted projective space.
    $endgroup$
    – user347489
    Dec 23 '18 at 2:10
















1












$begingroup$

You can consider the affine map $(x,y,z)mapsto (x^a,y^b,z^c)$ from $mathbb{A}^3-{0}$ to itself. (Observe that this map does not contain $0$ in its image). Hence you can compose it with the quotient map $(x,y,z)mapsto [x:y:z]$ (which is well-defined on non-zero points) to obtain the map you give.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Is there a terminology for this composed map? (as in it's not a morphism between affine varieties or morphism between projective varieties. So do we still say this is a morphism between something?)
    $endgroup$
    – Johnny T.
    Dec 22 '18 at 20:34






  • 1




    $begingroup$
    It is a morphism of varieties :)
    $endgroup$
    – Levent
    Dec 22 '18 at 20:38










  • $begingroup$
    Thank you very much. I was wondering if I could ask you just one more question: is this (this composed map) a finite morphism?
    $endgroup$
    – Johnny T.
    Dec 22 '18 at 20:46








  • 1




    $begingroup$
    No it is not. Note that finite morphisms necessarily have finite fibers, where as for example when $k_0=k_1=k_2=1$ all fibers are infinite.
    $endgroup$
    – Levent
    Dec 22 '18 at 20:51






  • 1




    $begingroup$
    The key concept here, as Mohan mentions above, is weighted projective space.
    $endgroup$
    – user347489
    Dec 23 '18 at 2:10














1












1








1





$begingroup$

You can consider the affine map $(x,y,z)mapsto (x^a,y^b,z^c)$ from $mathbb{A}^3-{0}$ to itself. (Observe that this map does not contain $0$ in its image). Hence you can compose it with the quotient map $(x,y,z)mapsto [x:y:z]$ (which is well-defined on non-zero points) to obtain the map you give.






share|cite|improve this answer











$endgroup$



You can consider the affine map $(x,y,z)mapsto (x^a,y^b,z^c)$ from $mathbb{A}^3-{0}$ to itself. (Observe that this map does not contain $0$ in its image). Hence you can compose it with the quotient map $(x,y,z)mapsto [x:y:z]$ (which is well-defined on non-zero points) to obtain the map you give.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 22 '18 at 20:31

























answered Dec 22 '18 at 20:24









LeventLevent

2,741925




2,741925












  • $begingroup$
    Is there a terminology for this composed map? (as in it's not a morphism between affine varieties or morphism between projective varieties. So do we still say this is a morphism between something?)
    $endgroup$
    – Johnny T.
    Dec 22 '18 at 20:34






  • 1




    $begingroup$
    It is a morphism of varieties :)
    $endgroup$
    – Levent
    Dec 22 '18 at 20:38










  • $begingroup$
    Thank you very much. I was wondering if I could ask you just one more question: is this (this composed map) a finite morphism?
    $endgroup$
    – Johnny T.
    Dec 22 '18 at 20:46








  • 1




    $begingroup$
    No it is not. Note that finite morphisms necessarily have finite fibers, where as for example when $k_0=k_1=k_2=1$ all fibers are infinite.
    $endgroup$
    – Levent
    Dec 22 '18 at 20:51






  • 1




    $begingroup$
    The key concept here, as Mohan mentions above, is weighted projective space.
    $endgroup$
    – user347489
    Dec 23 '18 at 2:10


















  • $begingroup$
    Is there a terminology for this composed map? (as in it's not a morphism between affine varieties or morphism between projective varieties. So do we still say this is a morphism between something?)
    $endgroup$
    – Johnny T.
    Dec 22 '18 at 20:34






  • 1




    $begingroup$
    It is a morphism of varieties :)
    $endgroup$
    – Levent
    Dec 22 '18 at 20:38










  • $begingroup$
    Thank you very much. I was wondering if I could ask you just one more question: is this (this composed map) a finite morphism?
    $endgroup$
    – Johnny T.
    Dec 22 '18 at 20:46








  • 1




    $begingroup$
    No it is not. Note that finite morphisms necessarily have finite fibers, where as for example when $k_0=k_1=k_2=1$ all fibers are infinite.
    $endgroup$
    – Levent
    Dec 22 '18 at 20:51






  • 1




    $begingroup$
    The key concept here, as Mohan mentions above, is weighted projective space.
    $endgroup$
    – user347489
    Dec 23 '18 at 2:10
















$begingroup$
Is there a terminology for this composed map? (as in it's not a morphism between affine varieties or morphism between projective varieties. So do we still say this is a morphism between something?)
$endgroup$
– Johnny T.
Dec 22 '18 at 20:34




$begingroup$
Is there a terminology for this composed map? (as in it's not a morphism between affine varieties or morphism between projective varieties. So do we still say this is a morphism between something?)
$endgroup$
– Johnny T.
Dec 22 '18 at 20:34




1




1




$begingroup$
It is a morphism of varieties :)
$endgroup$
– Levent
Dec 22 '18 at 20:38




$begingroup$
It is a morphism of varieties :)
$endgroup$
– Levent
Dec 22 '18 at 20:38












$begingroup$
Thank you very much. I was wondering if I could ask you just one more question: is this (this composed map) a finite morphism?
$endgroup$
– Johnny T.
Dec 22 '18 at 20:46






$begingroup$
Thank you very much. I was wondering if I could ask you just one more question: is this (this composed map) a finite morphism?
$endgroup$
– Johnny T.
Dec 22 '18 at 20:46






1




1




$begingroup$
No it is not. Note that finite morphisms necessarily have finite fibers, where as for example when $k_0=k_1=k_2=1$ all fibers are infinite.
$endgroup$
– Levent
Dec 22 '18 at 20:51




$begingroup$
No it is not. Note that finite morphisms necessarily have finite fibers, where as for example when $k_0=k_1=k_2=1$ all fibers are infinite.
$endgroup$
– Levent
Dec 22 '18 at 20:51




1




1




$begingroup$
The key concept here, as Mohan mentions above, is weighted projective space.
$endgroup$
– user347489
Dec 23 '18 at 2:10




$begingroup$
The key concept here, as Mohan mentions above, is weighted projective space.
$endgroup$
– user347489
Dec 23 '18 at 2:10


















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