A problem about the twisted cubic












9












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















9












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














9












9








9


3



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '13 at 16:43







user26857

















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














  • 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










1 Answer
1






active

oldest

votes


















0












$begingroup$

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 $C$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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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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active

oldest

votes









0












$begingroup$

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 $C$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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 $C$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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 $C$.






share|cite|improve this answer











$endgroup$



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 $C$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 17 '18 at 16:45









user26857

39.3k124183




39.3k124183










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


















  • $begingroup$
    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


















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