affine variety from idealViewing a quasi-projective variety as an affine varietyA principal open set of an...
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affine variety from ideal
Viewing a quasi-projective variety as an affine varietyA principal open set of an affine algebraic set is an affine varietyClosed subset of an affine variety… is it affine?Difference between affine variety and graph of a functionIrreducible components of affine varietyEquivalent definitions of a quasi-affine variety?Example of affine variety with infinite dimensionSpectrum of the coordinate ring of an affine varietyClosure of a quasi-affine varietyHow to define affine/projective varieties over fields which are not algebraically closed?
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I have a problem with this excercise:
*Take a field $F$, algebraic closed, $charFneq 2$.
1)Consider the following ideal $I=(xy+yz+zx-frac{1}{2}u^2 , x+y+z+u)$ in $F[x,y,z,u]$.
Is $Z(I)$ affine variety in $A^4$?
2)Consider the following ideal $I=(x^2+y^2+z^2,x+y+z)$ in $F[x,y,z]$. Is $Z(I)$ affine variety in $A^3$*
How can I check, that it is variety?
I think, that my solutions are incorect, or noncomplete. For now I have:
1)$u=-x-y-z$, then $F[x,y,z,u]/I=F[x,y,z]/(x^2+y^2+z^2)$, polynomial $x^2+y^2+z^2$ is irreducible, so $Z(I)$ must be variety.
2)$x=-y-z$, then $F[x,y,z]/I=F[y,z]/(2(y^2+yz+z^2))$
algebraic-geometry affine-varieties
asked Feb 8 at 21:35
enigmaenigma
62
$endgroup$
add a comment |
$begingroup$
I have a problem with this excercise:
*Take a field $F$, algebraic closed, $charFneq 2$.
1)Consider the following ideal $I=(xy+yz+zx-frac{1}{2}u^2 , x+y+z+u)$ in $F[x,y,z,u]$.
Is $Z(I)$ affine variety in $A^4$?
2)Consider the following ideal $I=(x^2+y^2+z^2,x+y+z)$ in $F[x,y,z]$. Is $Z(I)$ affine variety in $A^3$*
How can I check, that it is variety?
I think, that my solutions are incorect, or noncomplete. For now I have:
1)$u=-x-y-z$, then $F[x,y,z,u]/I=F[x,y,z]/(x^2+y^2+z^2)$, polynomial $x^2+y^2+z^2$ is irreducible, so $Z(I)$ must be variety.
2)$x=-y-z$, then $F[x,y,z]/I=F[y,z]/(2(y^2+yz+z^2))$
algebraic-geometry affine-varieties
asked Feb 8 at 21:35
enigmaenigma
62
$endgroup$
2
$begingroup$
What is your definition of a variety? This varies lots in the literature, so including it will help provide a better answer to your question.
$endgroup$
– KReiser
Feb 8 at 23:23
$begingroup$
I use the following definition: affine variety is an irreducible, closed subset of $A^n$
$endgroup$
– enigma
Feb 10 at 0:39
add a comment |
$begingroup$
I have a problem with this excercise:
*Take a field $F$, algebraic closed, $charFneq 2$.
1)Consider the following ideal $I=(xy+yz+zx-frac{1}{2}u^2 , x+y+z+u)$ in $F[x,y,z,u]$.
Is $Z(I)$ affine variety in $A^4$?
2)Consider the following ideal $I=(x^2+y^2+z^2,x+y+z)$ in $F[x,y,z]$. Is $Z(I)$ affine variety in $A^3$*
How can I check, that it is variety?
I think, that my solutions are incorect, or noncomplete. For now I have:
1)$u=-x-y-z$, then $F[x,y,z,u]/I=F[x,y,z]/(x^2+y^2+z^2)$, polynomial $x^2+y^2+z^2$ is irreducible, so $Z(I)$ must be variety.
2)$x=-y-z$, then $F[x,y,z]/I=F[y,z]/(2(y^2+yz+z^2))$
algebraic-geometry affine-varieties
asked Feb 8 at 21:35
enigmaenigma
62
$endgroup$
I have a problem with this excercise:
*Take a field $F$, algebraic closed, $charFneq 2$.
1)Consider the following ideal $I=(xy+yz+zx-frac{1}{2}u^2 , x+y+z+u)$ in $F[x,y,z,u]$.
Is $Z(I)$ affine variety in $A^4$?
2)Consider the following ideal $I=(x^2+y^2+z^2,x+y+z)$ in $F[x,y,z]$. Is $Z(I)$ affine variety in $A^3$*
How can I check, that it is variety?
I think, that my solutions are incorect, or noncomplete. For now I have:
1)$u=-x-y-z$, then $F[x,y,z,u]/I=F[x,y,z]/(x^2+y^2+z^2)$, polynomial $x^2+y^2+z^2$ is irreducible, so $Z(I)$ must be variety.
2)$x=-y-z$, then $F[x,y,z]/I=F[y,z]/(2(y^2+yz+z^2))$
algebraic-geometry affine-varieties
algebraic-geometry affine-varieties
asked Feb 8 at 21:35
enigmaenigma
62
asked Feb 8 at 21:35
enigmaenigma
62
asked Feb 8 at 21:35
enigmaenigma
62
asked Feb 8 at 21:35
enigmaenigma
62
asked Feb 8 at 21:35
enigmaenigma
62
62
2
$begingroup$
What is your definition of a variety? This varies lots in the literature, so including it will help provide a better answer to your question.
$endgroup$
– KReiser
Feb 8 at 23:23
$begingroup$
I use the following definition: affine variety is an irreducible, closed subset of $A^n$
$endgroup$
– enigma
Feb 10 at 0:39
add a comment |
2
$begingroup$
What is your definition of a variety? This varies lots in the literature, so including it will help provide a better answer to your question.
$endgroup$
– KReiser
Feb 8 at 23:23
$begingroup$
I use the following definition: affine variety is an irreducible, closed subset of $A^n$
$endgroup$
– enigma
Feb 10 at 0:39
2
2
$begingroup$
What is your definition of a variety? This varies lots in the literature, so including it will help provide a better answer to your question.
$endgroup$
– KReiser
Feb 8 at 23:23
$begingroup$
What is your definition of a variety? This varies lots in the literature, so including it will help provide a better answer to your question.
$endgroup$
– KReiser
Feb 8 at 23:23
$begingroup$
I use the following definition: affine variety is an irreducible, closed subset of $A^n$
$endgroup$
– enigma
Feb 10 at 0:39
$begingroup$
I use the following definition: affine variety is an irreducible, closed subset of $A^n$
$endgroup$
– enigma
Feb 10 at 0:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
(2)The answer is no in characteristics other than 3,for which we have to show I is radical,but not prime.For this it's enough to show K[x,y,z]/I has no nilpotent ,but is not an Integral Domain.
We observe the ring homomorphism K[x,y,z] →K[x,y] induced by x → x,y → y,z → -x-y has kernel (x,y,z) and thus it induces an isomorphism.
K[x,y,z]/I=K[x,y]/(x^2+y^2+(-x-y)^2) ,Since, z=-x-y
=K[x,y]/(x^2+xy+y^2)
If Characteristic is not 3,the x^2+xy+y^2 is a product of two distinct linear factors,which shows the quotient is reduced but not Integral,thus Z(I) is not variety.
If characteristic is 3,then (x^2+xy+y^2) is equal to (x-y)^2,so the radical of Ideal it generates is generated by x-y .The quotient by this radical is isomorphic to K[x] ,so is an Integral Domain,hence I(Z(I)) is prime and Z(I) is a variety.
answered Mar 14 at 0:06
Abdul HalimAbdul Halim
12
$endgroup$
add a comment |
$begingroup$
(1) We have,
$$K[x,y,z,u]/I=K[x,y,z]/(xy+yz+zx-(1/2)*(x+y+z)^2)$$
Since $$u=-x-y-z
=K[x,y,z]/(x^2+y^2+z^2)$$
and as we have established $x^2+y^2+z^2$ is obviously irreducible in Field theory,so the ideal it generates is prime,showing the quotient is Integral Domain and so $Z(I)$ is a affine variety.
edited Mar 14 at 2:58
dantopa
6,63342245
answered Mar 13 at 23:49
Abdul HalimAbdul Halim
12
$endgroup$
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Mar 14 at 0:02
add a comment |
protected by Saad Mar 14 at 2:31
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
(2)The answer is no in characteristics other than 3,for which we have to show I is radical,but not prime.For this it's enough to show K[x,y,z]/I has no nilpotent ,but is not an Integral Domain.
We observe the ring homomorphism K[x,y,z] →K[x,y] induced by x → x,y → y,z → -x-y has kernel (x,y,z) and thus it induces an isomorphism.
K[x,y,z]/I=K[x,y]/(x^2+y^2+(-x-y)^2) ,Since, z=-x-y
=K[x,y]/(x^2+xy+y^2)
If Characteristic is not 3,the x^2+xy+y^2 is a product of two distinct linear factors,which shows the quotient is reduced but not Integral,thus Z(I) is not variety.
If characteristic is 3,then (x^2+xy+y^2) is equal to (x-y)^2,so the radical of Ideal it generates is generated by x-y .The quotient by this radical is isomorphic to K[x] ,so is an Integral Domain,hence I(Z(I)) is prime and Z(I) is a variety.
answered Mar 14 at 0:06
Abdul HalimAbdul Halim
12
$endgroup$
add a comment |
$begingroup$
(2)The answer is no in characteristics other than 3,for which we have to show I is radical,but not prime.For this it's enough to show K[x,y,z]/I has no nilpotent ,but is not an Integral Domain.
We observe the ring homomorphism K[x,y,z] →K[x,y] induced by x → x,y → y,z → -x-y has kernel (x,y,z) and thus it induces an isomorphism.
K[x,y,z]/I=K[x,y]/(x^2+y^2+(-x-y)^2) ,Since, z=-x-y
=K[x,y]/(x^2+xy+y^2)
If Characteristic is not 3,the x^2+xy+y^2 is a product of two distinct linear factors,which shows the quotient is reduced but not Integral,thus Z(I) is not variety.
If characteristic is 3,then (x^2+xy+y^2) is equal to (x-y)^2,so the radical of Ideal it generates is generated by x-y .The quotient by this radical is isomorphic to K[x] ,so is an Integral Domain,hence I(Z(I)) is prime and Z(I) is a variety.
answered Mar 14 at 0:06
Abdul HalimAbdul Halim
12
$endgroup$
add a comment |
$begingroup$
(2)The answer is no in characteristics other than 3,for which we have to show I is radical,but not prime.For this it's enough to show K[x,y,z]/I has no nilpotent ,but is not an Integral Domain.
We observe the ring homomorphism K[x,y,z] →K[x,y] induced by x → x,y → y,z → -x-y has kernel (x,y,z) and thus it induces an isomorphism.
K[x,y,z]/I=K[x,y]/(x^2+y^2+(-x-y)^2) ,Since, z=-x-y
=K[x,y]/(x^2+xy+y^2)
If Characteristic is not 3,the x^2+xy+y^2 is a product of two distinct linear factors,which shows the quotient is reduced but not Integral,thus Z(I) is not variety.
If characteristic is 3,then (x^2+xy+y^2) is equal to (x-y)^2,so the radical of Ideal it generates is generated by x-y .The quotient by this radical is isomorphic to K[x] ,so is an Integral Domain,hence I(Z(I)) is prime and Z(I) is a variety.
answered Mar 14 at 0:06
Abdul HalimAbdul Halim
12
$endgroup$
(2)The answer is no in characteristics other than 3,for which we have to show I is radical,but not prime.For this it's enough to show K[x,y,z]/I has no nilpotent ,but is not an Integral Domain.
We observe the ring homomorphism K[x,y,z] →K[x,y] induced by x → x,y → y,z → -x-y has kernel (x,y,z) and thus it induces an isomorphism.
K[x,y,z]/I=K[x,y]/(x^2+y^2+(-x-y)^2) ,Since, z=-x-y
=K[x,y]/(x^2+xy+y^2)
If Characteristic is not 3,the x^2+xy+y^2 is a product of two distinct linear factors,which shows the quotient is reduced but not Integral,thus Z(I) is not variety.
If characteristic is 3,then (x^2+xy+y^2) is equal to (x-y)^2,so the radical of Ideal it generates is generated by x-y .The quotient by this radical is isomorphic to K[x] ,so is an Integral Domain,hence I(Z(I)) is prime and Z(I) is a variety.
answered Mar 14 at 0:06
Abdul HalimAbdul Halim
12
answered Mar 14 at 0:06
Abdul HalimAbdul Halim
12
answered Mar 14 at 0:06
Abdul HalimAbdul Halim
12
answered Mar 14 at 0:06
Abdul HalimAbdul Halim
12
12
add a comment |
add a comment |
$begingroup$
(1) We have,
$$K[x,y,z,u]/I=K[x,y,z]/(xy+yz+zx-(1/2)*(x+y+z)^2)$$
Since $$u=-x-y-z
=K[x,y,z]/(x^2+y^2+z^2)$$
and as we have established $x^2+y^2+z^2$ is obviously irreducible in Field theory,so the ideal it generates is prime,showing the quotient is Integral Domain and so $Z(I)$ is a affine variety.
edited Mar 14 at 2:58
dantopa
6,63342245
answered Mar 13 at 23:49
Abdul HalimAbdul Halim
12
$endgroup$
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Mar 14 at 0:02
add a comment |
$begingroup$
(1) We have,
$$K[x,y,z,u]/I=K[x,y,z]/(xy+yz+zx-(1/2)*(x+y+z)^2)$$
Since $$u=-x-y-z
=K[x,y,z]/(x^2+y^2+z^2)$$
and as we have established $x^2+y^2+z^2$ is obviously irreducible in Field theory,so the ideal it generates is prime,showing the quotient is Integral Domain and so $Z(I)$ is a affine variety.
edited Mar 14 at 2:58
dantopa
6,63342245
answered Mar 13 at 23:49
Abdul HalimAbdul Halim
12
$endgroup$
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Mar 14 at 0:02
add a comment |
$begingroup$
(1) We have,
$$K[x,y,z,u]/I=K[x,y,z]/(xy+yz+zx-(1/2)*(x+y+z)^2)$$
Since $$u=-x-y-z
=K[x,y,z]/(x^2+y^2+z^2)$$
and as we have established $x^2+y^2+z^2$ is obviously irreducible in Field theory,so the ideal it generates is prime,showing the quotient is Integral Domain and so $Z(I)$ is a affine variety.
edited Mar 14 at 2:58
dantopa
6,63342245
answered Mar 13 at 23:49
Abdul HalimAbdul Halim
12
$endgroup$
(1) We have,
$$K[x,y,z,u]/I=K[x,y,z]/(xy+yz+zx-(1/2)*(x+y+z)^2)$$
Since $$u=-x-y-z
=K[x,y,z]/(x^2+y^2+z^2)$$
and as we have established $x^2+y^2+z^2$ is obviously irreducible in Field theory,so the ideal it generates is prime,showing the quotient is Integral Domain and so $Z(I)$ is a affine variety.
edited Mar 14 at 2:58
dantopa
6,63342245
answered Mar 13 at 23:49
Abdul HalimAbdul Halim
12
edited Mar 14 at 2:58
dantopa
6,63342245
edited Mar 14 at 2:58
dantopa
6,63342245
edited Mar 14 at 2:58
dantopa
6,63342245
6,63342245
answered Mar 13 at 23:49
Abdul HalimAbdul Halim
12
answered Mar 13 at 23:49
Abdul HalimAbdul Halim
12
answered Mar 13 at 23:49
Abdul HalimAbdul Halim
12
12
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Mar 14 at 0:02
add a comment |
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Mar 14 at 0:02
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Mar 14 at 0:02
$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
Mar 14 at 0:02
add a comment |
protected by Saad Mar 14 at 2:31
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
2
$begingroup$
What is your definition of a variety? This varies lots in the literature, so including it will help provide a better answer to your question.
$endgroup$
– KReiser
Feb 8 at 23:23
$begingroup$
I use the following definition: affine variety is an irreducible, closed subset of $A^n$
$endgroup$
– enigma
Feb 10 at 0:39