Ideal of regular functions in $Z(xy-z^2)$ vanishing on $(x,y)$ is not principal
$begingroup$
Let $H = Z(xy-z^2) subset Bbb A ^ {3}$.
And let $L = Z(x,z)$.
I need to show that $L subset H$ , that $dim(L) = dim (H) -1$ and that the ideal of regular functions on $H$ which vanishing on $L$ is not a principal ideal.
So showing $Lsubset H$ is easy, and as $xy-z^2$ is irreducible, we have $dim(L) = dim(H) -1$ by principal ideal theorem.
Im not sure how to show the last part.
The regular functions on $H$ are $k[x,y,z]/(xy-z^2)$ and the ideal of regular functions vanishing on $L$ is $(x,z)$ so I need to show $k[x,y,z]/(xy-z^2) cap (x,z)$ is not principal.
Any ideas how?
algebraic-geometry affine-geometry
asked Jan 3 at 13:38
user123user123
1,400316
$endgroup$
add a comment |
$begingroup$
Let $H = Z(xy-z^2) subset Bbb A ^ {3}$.
And let $L = Z(x,z)$.
I need to show that $L subset H$ , that $dim(L) = dim (H) -1$ and that the ideal of regular functions on $H$ which vanishing on $L$ is not a principal ideal.
So showing $Lsubset H$ is easy, and as $xy-z^2$ is irreducible, we have $dim(L) = dim(H) -1$ by principal ideal theorem.
Im not sure how to show the last part.
The regular functions on $H$ are $k[x,y,z]/(xy-z^2)$ and the ideal of regular functions vanishing on $L$ is $(x,z)$ so I need to show $k[x,y,z]/(xy-z^2) cap (x,z)$ is not principal.
Any ideas how?
algebraic-geometry affine-geometry
asked Jan 3 at 13:38
user123user123
1,400316
$endgroup$
add a comment |
$begingroup$
Let $H = Z(xy-z^2) subset Bbb A ^ {3}$.
And let $L = Z(x,z)$.
I need to show that $L subset H$ , that $dim(L) = dim (H) -1$ and that the ideal of regular functions on $H$ which vanishing on $L$ is not a principal ideal.
So showing $Lsubset H$ is easy, and as $xy-z^2$ is irreducible, we have $dim(L) = dim(H) -1$ by principal ideal theorem.
Im not sure how to show the last part.
The regular functions on $H$ are $k[x,y,z]/(xy-z^2)$ and the ideal of regular functions vanishing on $L$ is $(x,z)$ so I need to show $k[x,y,z]/(xy-z^2) cap (x,z)$ is not principal.
Any ideas how?
algebraic-geometry affine-geometry
asked Jan 3 at 13:38
user123user123
1,400316
$endgroup$
Let $H = Z(xy-z^2) subset Bbb A ^ {3}$.
And let $L = Z(x,z)$.
I need to show that $L subset H$ , that $dim(L) = dim (H) -1$ and that the ideal of regular functions on $H$ which vanishing on $L$ is not a principal ideal.
So showing $Lsubset H$ is easy, and as $xy-z^2$ is irreducible, we have $dim(L) = dim(H) -1$ by principal ideal theorem.
Im not sure how to show the last part.
The regular functions on $H$ are $k[x,y,z]/(xy-z^2)$ and the ideal of regular functions vanishing on $L$ is $(x,z)$ so I need to show $k[x,y,z]/(xy-z^2) cap (x,z)$ is not principal.
Any ideas how?
algebraic-geometry affine-geometry
algebraic-geometry affine-geometry
asked Jan 3 at 13:38
user123user123
1,400316
asked Jan 3 at 13:38
user123user123
1,400316
asked Jan 3 at 13:38
user123user123
1,400316
asked Jan 3 at 13:38
user123user123
1,400316
asked Jan 3 at 13:38
user123user123
1,400316
1,400316
add a comment |
add a comment |
1 Answer
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$begingroup$
One approach is to consider the grading.
$$require{cancel}(x,z)/(xy - z^2) = k{x,z} + k{x^2,xy,xz,yz} + k{x^2,x^2y,x^2z,xyz,xy^2,y^2z} +cdots$$
So the dimensions are $2,4,6,dots$.
On the other hand, what does the grading look like for $(f)/(xy-z^2)$? It would be
$$ k{f}+ frac{k{xf,yf,zf}}{z^2 sim xy} + cdots $$
Ok, so we can't really get an explicit basis for the higher degree terms but we can see an issue with $k{f}$ as this is one dimensional.
answered Jan 3 at 14:17
Trevor GunnTrevor Gunn
14.8k32047
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
One approach is to consider the grading.
$$require{cancel}(x,z)/(xy - z^2) = k{x,z} + k{x^2,xy,xz,yz} + k{x^2,x^2y,x^2z,xyz,xy^2,y^2z} +cdots$$
So the dimensions are $2,4,6,dots$.
On the other hand, what does the grading look like for $(f)/(xy-z^2)$? It would be
$$ k{f}+ frac{k{xf,yf,zf}}{z^2 sim xy} + cdots $$
Ok, so we can't really get an explicit basis for the higher degree terms but we can see an issue with $k{f}$ as this is one dimensional.
answered Jan 3 at 14:17
Trevor GunnTrevor Gunn
14.8k32047
$endgroup$
add a comment |
$begingroup$
One approach is to consider the grading.
$$require{cancel}(x,z)/(xy - z^2) = k{x,z} + k{x^2,xy,xz,yz} + k{x^2,x^2y,x^2z,xyz,xy^2,y^2z} +cdots$$
So the dimensions are $2,4,6,dots$.
On the other hand, what does the grading look like for $(f)/(xy-z^2)$? It would be
$$ k{f}+ frac{k{xf,yf,zf}}{z^2 sim xy} + cdots $$
Ok, so we can't really get an explicit basis for the higher degree terms but we can see an issue with $k{f}$ as this is one dimensional.
answered Jan 3 at 14:17
Trevor GunnTrevor Gunn
14.8k32047
$endgroup$
add a comment |
$begingroup$
One approach is to consider the grading.
$$require{cancel}(x,z)/(xy - z^2) = k{x,z} + k{x^2,xy,xz,yz} + k{x^2,x^2y,x^2z,xyz,xy^2,y^2z} +cdots$$
So the dimensions are $2,4,6,dots$.
On the other hand, what does the grading look like for $(f)/(xy-z^2)$? It would be
$$ k{f}+ frac{k{xf,yf,zf}}{z^2 sim xy} + cdots $$
Ok, so we can't really get an explicit basis for the higher degree terms but we can see an issue with $k{f}$ as this is one dimensional.
answered Jan 3 at 14:17
Trevor GunnTrevor Gunn
14.8k32047
$endgroup$
One approach is to consider the grading.
$$require{cancel}(x,z)/(xy - z^2) = k{x,z} + k{x^2,xy,xz,yz} + k{x^2,x^2y,x^2z,xyz,xy^2,y^2z} +cdots$$
So the dimensions are $2,4,6,dots$.
On the other hand, what does the grading look like for $(f)/(xy-z^2)$? It would be
$$ k{f}+ frac{k{xf,yf,zf}}{z^2 sim xy} + cdots $$
Ok, so we can't really get an explicit basis for the higher degree terms but we can see an issue with $k{f}$ as this is one dimensional.
answered Jan 3 at 14:17
Trevor GunnTrevor Gunn
14.8k32047
answered Jan 3 at 14:17
Trevor GunnTrevor Gunn
14.8k32047
answered Jan 3 at 14:17
Trevor GunnTrevor Gunn
14.8k32047
answered Jan 3 at 14:17
Trevor GunnTrevor Gunn
14.8k32047
14.8k32047
add a comment |
add a comment |
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