A problem about the twisted cubic
$begingroup$
I have some difficulty with the following problem:
Let $f : k → k^3$ be the map which associates $(t, t^2, t^3)$ to $t$ and let $C$ be the image of $f$ (the twisted cubic). Show that $C$ is an affine algebraic set and calculate $I(C)$. Show that the affine algebra $Γ(C)=k[X,Y,Z]/I(C)$ is isomorphic to the ring of polynomials $k[T]$.
I found a solution in the case when $k$ is an infinite field:
Clearly $C$ is the affine algebraic set $C=big<X^3-Z,;X^2-Ybig>$ and moreover $big<X^3-Z,;X^2-Ybig>subseteq I(C) $. To prove the inclusion "$supseteq$", let $F$ be a polynomial with $Fin I(C)$. Then we can write (thanks to successive divisions):
$$F(X,Y,Z)=(X^3-Z)cdot Q(X,Y,Z)+(X^2-Y)cdot P(X,Y)+R(X).$$
For all $tin k$ we have
$$0=(t,t^2,t^3)=R(t)$$
and since $k$ is infinite then $R$ is the zero polynomial and $I(C)=big<X^3-Z,;X^2-Ybig>$.
Now since the function $f$ is an isomorfism between $C$ and $k$ follows that $Gamma(C)cong Gamma(k)=k[T]$.
I can not find the ideal $I(C)$ if the field $k$ is finite.
algebraic-geometry commutative-algebra algebraic-curves
asked May 3 '12 at 8:18
DubiousDubious
3,27452675
$endgroup$
add a comment |
$begingroup$
I have some difficulty with the following problem:
Let $f : k → k^3$ be the map which associates $(t, t^2, t^3)$ to $t$ and let $C$ be the image of $f$ (the twisted cubic). Show that $C$ is an affine algebraic set and calculate $I(C)$. Show that the affine algebra $Γ(C)=k[X,Y,Z]/I(C)$ is isomorphic to the ring of polynomials $k[T]$.
I found a solution in the case when $k$ is an infinite field:
Clearly $C$ is the affine algebraic set $C=big<X^3-Z,;X^2-Ybig>$ and moreover $big<X^3-Z,;X^2-Ybig>subseteq I(C) $. To prove the inclusion "$supseteq$", let $F$ be a polynomial with $Fin I(C)$. Then we can write (thanks to successive divisions):
$$F(X,Y,Z)=(X^3-Z)cdot Q(X,Y,Z)+(X^2-Y)cdot P(X,Y)+R(X).$$
For all $tin k$ we have
$$0=(t,t^2,t^3)=R(t)$$
and since $k$ is infinite then $R$ is the zero polynomial and $I(C)=big<X^3-Z,;X^2-Ybig>$.
Now since the function $f$ is an isomorfism between $C$ and $k$ follows that $Gamma(C)cong Gamma(k)=k[T]$.
I can not find the ideal $I(C)$ if the field $k$ is finite.
algebraic-geometry commutative-algebra algebraic-curves
asked May 3 '12 at 8:18
DubiousDubious
3,27452675
$endgroup$
2
$begingroup$
If $k$ is finite then $C$ is finite and so $Gamma(C)$ is a finite-dimensional vector space over $k$, so it is certainly not isomorphic to $k[T]$.
$endgroup$
– user8268
May 3 '12 at 12:05
$begingroup$
Thanks! your comment helped me a lot. So the statement of the problem is wrong.
$endgroup$
– Dubious
May 3 '12 at 13:03
$begingroup$
Algebraic closed field is always infinite. And when we are talking about (most part of) algebraic geometry, $k$ is implicitly assumed to be algebraic closed.
$endgroup$
– asatzhh
Oct 28 '13 at 6:25
$begingroup$
Aha, I always forget that the affine twisted cubic is actually a complete intersection (as opposed to the projective case)
$endgroup$
– Theo
Mar 30 '15 at 16:00
add a comment |
$begingroup$
I have some difficulty with the following problem:
Let $f : k → k^3$ be the map which associates $(t, t^2, t^3)$ to $t$ and let $C$ be the image of $f$ (the twisted cubic). Show that $C$ is an affine algebraic set and calculate $I(C)$. Show that the affine algebra $Γ(C)=k[X,Y,Z]/I(C)$ is isomorphic to the ring of polynomials $k[T]$.
I found a solution in the case when $k$ is an infinite field:
Clearly $C$ is the affine algebraic set $C=big<X^3-Z,;X^2-Ybig>$ and moreover $big<X^3-Z,;X^2-Ybig>subseteq I(C) $. To prove the inclusion "$supseteq$", let $F$ be a polynomial with $Fin I(C)$. Then we can write (thanks to successive divisions):
$$F(X,Y,Z)=(X^3-Z)cdot Q(X,Y,Z)+(X^2-Y)cdot P(X,Y)+R(X).$$
For all $tin k$ we have
$$0=(t,t^2,t^3)=R(t)$$
and since $k$ is infinite then $R$ is the zero polynomial and $I(C)=big<X^3-Z,;X^2-Ybig>$.
Now since the function $f$ is an isomorfism between $C$ and $k$ follows that $Gamma(C)cong Gamma(k)=k[T]$.
I can not find the ideal $I(C)$ if the field $k$ is finite.
algebraic-geometry commutative-algebra algebraic-curves
asked May 3 '12 at 8:18
DubiousDubious
3,27452675
$endgroup$
I have some difficulty with the following problem:
Let $f : k → k^3$ be the map which associates $(t, t^2, t^3)$ to $t$ and let $C$ be the image of $f$ (the twisted cubic). Show that $C$ is an affine algebraic set and calculate $I(C)$. Show that the affine algebra $Γ(C)=k[X,Y,Z]/I(C)$ is isomorphic to the ring of polynomials $k[T]$.
I found a solution in the case when $k$ is an infinite field:
Clearly $C$ is the affine algebraic set $C=big<X^3-Z,;X^2-Ybig>$ and moreover $big<X^3-Z,;X^2-Ybig>subseteq I(C) $. To prove the inclusion "$supseteq$", let $F$ be a polynomial with $Fin I(C)$. Then we can write (thanks to successive divisions):
$$F(X,Y,Z)=(X^3-Z)cdot Q(X,Y,Z)+(X^2-Y)cdot P(X,Y)+R(X).$$
For all $tin k$ we have
$$0=(t,t^2,t^3)=R(t)$$
and since $k$ is infinite then $R$ is the zero polynomial and $I(C)=big<X^3-Z,;X^2-Ybig>$.
Now since the function $f$ is an isomorfism between $C$ and $k$ follows that $Gamma(C)cong Gamma(k)=k[T]$.
I can not find the ideal $I(C)$ if the field $k$ is finite.
algebraic-geometry commutative-algebra algebraic-curves
algebraic-geometry commutative-algebra algebraic-curves
asked May 3 '12 at 8:18
DubiousDubious
3,27452675
asked May 3 '12 at 8:18
DubiousDubious
3,27452675
edited Dec 6 '13 at 16:43
user26857
asked May 3 '12 at 8:18
DubiousDubious
3,27452675
asked May 3 '12 at 8:18
DubiousDubious
3,27452675
asked May 3 '12 at 8:18
DubiousDubious
3,27452675
3,27452675
2
$begingroup$
If $k$ is finite then $C$ is finite and so $Gamma(C)$ is a finite-dimensional vector space over $k$, so it is certainly not isomorphic to $k[T]$.
$endgroup$
– user8268
May 3 '12 at 12:05
$begingroup$
Thanks! your comment helped me a lot. So the statement of the problem is wrong.
$endgroup$
– Dubious
May 3 '12 at 13:03
$begingroup$
Algebraic closed field is always infinite. And when we are talking about (most part of) algebraic geometry, $k$ is implicitly assumed to be algebraic closed.
$endgroup$
– asatzhh
Oct 28 '13 at 6:25
$begingroup$
Aha, I always forget that the affine twisted cubic is actually a complete intersection (as opposed to the projective case)
$endgroup$
– Theo
Mar 30 '15 at 16:00
add a comment |
2
$begingroup$
If $k$ is finite then $C$ is finite and so $Gamma(C)$ is a finite-dimensional vector space over $k$, so it is certainly not isomorphic to $k[T]$.
$endgroup$
– user8268
May 3 '12 at 12:05
$begingroup$
Thanks! your comment helped me a lot. So the statement of the problem is wrong.
$endgroup$
– Dubious
May 3 '12 at 13:03
$begingroup$
Algebraic closed field is always infinite. And when we are talking about (most part of) algebraic geometry, $k$ is implicitly assumed to be algebraic closed.
$endgroup$
– asatzhh
Oct 28 '13 at 6:25
$begingroup$
Aha, I always forget that the affine twisted cubic is actually a complete intersection (as opposed to the projective case)
$endgroup$
– Theo
Mar 30 '15 at 16:00
2
2
$begingroup$
If $k$ is finite then $C$ is finite and so $Gamma(C)$ is a finite-dimensional vector space over $k$, so it is certainly not isomorphic to $k[T]$.
$endgroup$
– user8268
May 3 '12 at 12:05
$begingroup$
If $k$ is finite then $C$ is finite and so $Gamma(C)$ is a finite-dimensional vector space over $k$, so it is certainly not isomorphic to $k[T]$.
$endgroup$
– user8268
May 3 '12 at 12:05
$begingroup$
Thanks! your comment helped me a lot. So the statement of the problem is wrong.
$endgroup$
– Dubious
May 3 '12 at 13:03
$begingroup$
Thanks! your comment helped me a lot. So the statement of the problem is wrong.
$endgroup$
– Dubious
May 3 '12 at 13:03
$begingroup$
Algebraic closed field is always infinite. And when we are talking about (most part of) algebraic geometry, $k$ is implicitly assumed to be algebraic closed.
$endgroup$
– asatzhh
Oct 28 '13 at 6:25
$begingroup$
Algebraic closed field is always infinite. And when we are talking about (most part of) algebraic geometry, $k$ is implicitly assumed to be algebraic closed.
$endgroup$
– asatzhh
Oct 28 '13 at 6:25
$begingroup$
Aha, I always forget that the affine twisted cubic is actually a complete intersection (as opposed to the projective case)
$endgroup$
– Theo
Mar 30 '15 at 16:00
$begingroup$
Aha, I always forget that the affine twisted cubic is actually a complete intersection (as opposed to the projective case)
$endgroup$
– Theo
Mar 30 '15 at 16:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Just to remove this question from the list of unanswered questions. It was found in comments that the statement of the problem is wrong if $k$ is finite. In this case the ideal $I(C)$ is the product of maximal ideals corresponding to the points of $C$.
edited Jan 17 '18 at 16:45
user26857
39.3k124183
answered Jan 17 '18 at 13:09
danneksdanneks
517211
$endgroup$
$begingroup$
I don't get the reasoning in the comments: Why does the finiteness of $C$ imply that $Gamma(C)$ is finite-dimensional? I mean $mathbb{A}^n_k$ is finite if $k$ is a finite field, but $Gamma(mathbb{A}^n_k) cong k[x_0,dots,x_n]$.
$endgroup$
– red_trumpet
Aug 23 '18 at 17:23
$begingroup$
By definition, $k[x_0,ldots,x_n]$ is a finitely-generated $k$-algebra (also known as a $k$-algebra of finite type). However, it is not a finite-dimensional vector space over $k$. Indeed, a basis for $k[x_0,ldots,x_n]$ is ${1,x_0,x_0^2,ldots}cup{x_1,x_1^2,ldots}cupldotscup{x_n,x_n^2,ldots}$, which is clearly not a finite set.
$endgroup$
– eloiprime
Sep 12 '18 at 11:24
add a comment |
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1 Answer
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$begingroup$
Just to remove this question from the list of unanswered questions. It was found in comments that the statement of the problem is wrong if $k$ is finite. In this case the ideal $I(C)$ is the product of maximal ideals corresponding to the points of $C$.
edited Jan 17 '18 at 16:45
user26857
39.3k124183
answered Jan 17 '18 at 13:09
danneksdanneks
517211
$endgroup$
$begingroup$
I don't get the reasoning in the comments: Why does the finiteness of $C$ imply that $Gamma(C)$ is finite-dimensional? I mean $mathbb{A}^n_k$ is finite if $k$ is a finite field, but $Gamma(mathbb{A}^n_k) cong k[x_0,dots,x_n]$.
$endgroup$
– red_trumpet
Aug 23 '18 at 17:23
$begingroup$
By definition, $k[x_0,ldots,x_n]$ is a finitely-generated $k$-algebra (also known as a $k$-algebra of finite type). However, it is not a finite-dimensional vector space over $k$. Indeed, a basis for $k[x_0,ldots,x_n]$ is ${1,x_0,x_0^2,ldots}cup{x_1,x_1^2,ldots}cupldotscup{x_n,x_n^2,ldots}$, which is clearly not a finite set.
$endgroup$
– eloiprime
Sep 12 '18 at 11:24
add a comment |
$begingroup$
Just to remove this question from the list of unanswered questions. It was found in comments that the statement of the problem is wrong if $k$ is finite. In this case the ideal $I(C)$ is the product of maximal ideals corresponding to the points of $C$.
edited Jan 17 '18 at 16:45
user26857
39.3k124183
answered Jan 17 '18 at 13:09
danneksdanneks
517211
$endgroup$
$begingroup$
I don't get the reasoning in the comments: Why does the finiteness of $C$ imply that $Gamma(C)$ is finite-dimensional? I mean $mathbb{A}^n_k$ is finite if $k$ is a finite field, but $Gamma(mathbb{A}^n_k) cong k[x_0,dots,x_n]$.
$endgroup$
– red_trumpet
Aug 23 '18 at 17:23
$begingroup$
By definition, $k[x_0,ldots,x_n]$ is a finitely-generated $k$-algebra (also known as a $k$-algebra of finite type). However, it is not a finite-dimensional vector space over $k$. Indeed, a basis for $k[x_0,ldots,x_n]$ is ${1,x_0,x_0^2,ldots}cup{x_1,x_1^2,ldots}cupldotscup{x_n,x_n^2,ldots}$, which is clearly not a finite set.
$endgroup$
– eloiprime
Sep 12 '18 at 11:24
add a comment |
$begingroup$
Just to remove this question from the list of unanswered questions. It was found in comments that the statement of the problem is wrong if $k$ is finite. In this case the ideal $I(C)$ is the product of maximal ideals corresponding to the points of $C$.
edited Jan 17 '18 at 16:45
user26857
39.3k124183
answered Jan 17 '18 at 13:09
danneksdanneks
517211
$endgroup$
Just to remove this question from the list of unanswered questions. It was found in comments that the statement of the problem is wrong if $k$ is finite. In this case the ideal $I(C)$ is the product of maximal ideals corresponding to the points of $C$.
edited Jan 17 '18 at 16:45
user26857
39.3k124183
answered Jan 17 '18 at 13:09
danneksdanneks
517211
edited Jan 17 '18 at 16:45
user26857
39.3k124183
edited Jan 17 '18 at 16:45
user26857
39.3k124183
edited Jan 17 '18 at 16:45
user26857
39.3k124183
39.3k124183
answered Jan 17 '18 at 13:09
danneksdanneks
517211
answered Jan 17 '18 at 13:09
danneksdanneks
517211
answered Jan 17 '18 at 13:09
danneksdanneks
517211
517211
$begingroup$
I don't get the reasoning in the comments: Why does the finiteness of $C$ imply that $Gamma(C)$ is finite-dimensional? I mean $mathbb{A}^n_k$ is finite if $k$ is a finite field, but $Gamma(mathbb{A}^n_k) cong k[x_0,dots,x_n]$.
$endgroup$
– red_trumpet
Aug 23 '18 at 17:23
$begingroup$
By definition, $k[x_0,ldots,x_n]$ is a finitely-generated $k$-algebra (also known as a $k$-algebra of finite type). However, it is not a finite-dimensional vector space over $k$. Indeed, a basis for $k[x_0,ldots,x_n]$ is ${1,x_0,x_0^2,ldots}cup{x_1,x_1^2,ldots}cupldotscup{x_n,x_n^2,ldots}$, which is clearly not a finite set.
$endgroup$
– eloiprime
Sep 12 '18 at 11:24
add a comment |
$begingroup$
I don't get the reasoning in the comments: Why does the finiteness of $C$ imply that $Gamma(C)$ is finite-dimensional? I mean $mathbb{A}^n_k$ is finite if $k$ is a finite field, but $Gamma(mathbb{A}^n_k) cong k[x_0,dots,x_n]$.
$endgroup$
– red_trumpet
Aug 23 '18 at 17:23
$begingroup$
By definition, $k[x_0,ldots,x_n]$ is a finitely-generated $k$-algebra (also known as a $k$-algebra of finite type). However, it is not a finite-dimensional vector space over $k$. Indeed, a basis for $k[x_0,ldots,x_n]$ is ${1,x_0,x_0^2,ldots}cup{x_1,x_1^2,ldots}cupldotscup{x_n,x_n^2,ldots}$, which is clearly not a finite set.
$endgroup$
– eloiprime
Sep 12 '18 at 11:24
$begingroup$
I don't get the reasoning in the comments: Why does the finiteness of $C$ imply that $Gamma(C)$ is finite-dimensional? I mean $mathbb{A}^n_k$ is finite if $k$ is a finite field, but $Gamma(mathbb{A}^n_k) cong k[x_0,dots,x_n]$.
$endgroup$
– red_trumpet
Aug 23 '18 at 17:23
$begingroup$
I don't get the reasoning in the comments: Why does the finiteness of $C$ imply that $Gamma(C)$ is finite-dimensional? I mean $mathbb{A}^n_k$ is finite if $k$ is a finite field, but $Gamma(mathbb{A}^n_k) cong k[x_0,dots,x_n]$.
$endgroup$
– red_trumpet
Aug 23 '18 at 17:23
$begingroup$
By definition, $k[x_0,ldots,x_n]$ is a finitely-generated $k$-algebra (also known as a $k$-algebra of finite type). However, it is not a finite-dimensional vector space over $k$. Indeed, a basis for $k[x_0,ldots,x_n]$ is ${1,x_0,x_0^2,ldots}cup{x_1,x_1^2,ldots}cupldotscup{x_n,x_n^2,ldots}$, which is clearly not a finite set.
$endgroup$
– eloiprime
Sep 12 '18 at 11:24
$begingroup$
By definition, $k[x_0,ldots,x_n]$ is a finitely-generated $k$-algebra (also known as a $k$-algebra of finite type). However, it is not a finite-dimensional vector space over $k$. Indeed, a basis for $k[x_0,ldots,x_n]$ is ${1,x_0,x_0^2,ldots}cup{x_1,x_1^2,ldots}cupldotscup{x_n,x_n^2,ldots}$, which is clearly not a finite set.
$endgroup$
– eloiprime
Sep 12 '18 at 11:24
add a comment |
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If $k$ is finite then $C$ is finite and so $Gamma(C)$ is a finite-dimensional vector space over $k$, so it is certainly not isomorphic to $k[T]$.
$endgroup$
– user8268
May 3 '12 at 12:05
$begingroup$
Thanks! your comment helped me a lot. So the statement of the problem is wrong.
$endgroup$
– Dubious
May 3 '12 at 13:03
$begingroup$
Algebraic closed field is always infinite. And when we are talking about (most part of) algebraic geometry, $k$ is implicitly assumed to be algebraic closed.
$endgroup$
– asatzhh
Oct 28 '13 at 6:25
$begingroup$
Aha, I always forget that the affine twisted cubic is actually a complete intersection (as opposed to the projective case)
$endgroup$
– Theo
Mar 30 '15 at 16:00