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Let $X=\{(r^2,r^3,r^4) : r\in\Bbb R\}\subset \Bbb R^3$. Show that

1) $X$ is an affine variety.

2) Determine the ideal of $X$. Every $f\in\Bbb R[x,y,z]$, can we write $f$ in the form $f=p(xz-y^2)+q(z-x^2) +r$ , where $p,q\in\Bbb R[x,y,z]$?

i) Affine variety is defined as the solution set to some polynomials in the affine space $k^n$.

ii) The ideal of $X$ is defined as

$$I(X)=\{f\in \Bbb R[x,y,z]:f(x,y,z)=0, \forall (x,y,z)\in X\}$$

user26857
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ziang chen
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3 Answers3

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1) It is immediate that $X$ is included in the variety $V:=V(xz-y^2,z-x^2)$.

2) To prove the converse inclusion $V\subset X$, we will take some point $P=(a,b,c)\in V$, i.e. a point satisfying $ac=b^2 $ and $c=a^2$, and show that $P\in X$.
a) If $a=0$, then $ac-b^2=0$ forces $b=0$ and $c-a^2=0$ forces $c=0$.
So actually $P=(0,0,0)$ and $P\in X $, corresponding to $r=0$.
b) If $a\neq0$, a few little calculations (using $ac=b^2, c=a^2$) show that $P=(a,b,c)=(r^2,r^3,r^4)$ with $r=\frac ba$
[For example $r^2=\frac {b^2}{a^2}=\frac {ac}{a^2}=\frac {ac}{c}=a$ ]

3) Hence $X=V(I)$ with $I=(xz-y^2,z-x^2)$, proving that $X$ is an affine variety.

4) The calculations above are valid over any field and the Nullstellensatz implies that $I (X)=\sqrt I$.
Now $I$ is prime since $k(x,y,z]/I=k[x,y]/(y^2-x^3)$ and $y^2-x^3$ is irrreducible. So finally $$I(X)=I=(xz-y^2,z-x^2)$$

Georges Elencwajg
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For this simple example, you do not need to use computer algebra software to compute the ideal. The ideal $I$ is the kernel of the homomorphmism $$ k[x,y,z] \to k[r],$$ defined by $x \mapsto r^2, y \mapsto r^3, z \mapsto r^4$. Your task task is to find relations among the images of the homomorphism.

By inspection you see that $x^2$ and $z$ are sent to the same element, and also that $xz$ and $y^2$ are sent to the same element. Thus $I=(x^2-z,xz-y^2) \subseteq I(X)$. To show that this is an equality, you need to show that any point $(x,y,z) \in \mathbb A^3$ that satisfies these polynomial relations actually lie on the curve that any polynomial .

Hint: start with $(x,y,z) \in \mathbb A^3$. By the second equation, this is equal to $(x,y,x^2)$. Now, if $y$ is zero, then both $x$ and $z$ are. So assume $y > 0$. Then $y=\sqrt{x^3}$ (we take the positive square root always). The variables $x$ and $z$ are always positive because $xz > 0$ and $x^2=z$. Then $\sqrt{x}^2$ makes sense and is equal to $x$, and also $z=\sqrt{x}^4$. This means that $(x,y,z)=(\sqrt{x}^2,\sqrt{x}^3,\sqrt{x}^4)$, which have the required form. Now assume $y <0$. Then the same arguments give $(x,y,z)=(\sqrt{x}^2,-\sqrt{x}^3,\sqrt{x}^4)$. So $Z(I) \subset Z(I(X))$ also.

Fredrik Meyer
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  • http://en.wiktionary.org/wiki/give_a_man_a_fish_and_you_feed_him_for_a_day._Teach_a_man_to_fish_and_you_feed_him_for_a_lifetime – OR. Oct 16 '13 at 11:22
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    @ABC: I agree that I perhaps gave away a bit too much, but there are still things to check in my argument. For example, if some of the variables are negative, then the square roots doesn't make sense. That being said, I believe my method is the right one for this example - there's no need to introduce the machinery of Gröbner bases and computer algebra software when the problem can easily be done hands on. – Fredrik Meyer Oct 16 '13 at 11:28
  • No, I am not saying anything about giving too much. The claim "the right one" simply doesn't make sense. There is no such thing as the right method among methods that work. The proverb "give a man a fish ..." means only that this method solves this problem and this problem only. Grobner basis "is not necessary" but will "feed him for a lifetime". – OR. Oct 16 '13 at 11:40
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    @FredrikMeyer Your second paragraph is incorrect. What you need to show is that any polynomial that vanishes on $X$ is in the given ideal. – Zhen Lin Oct 16 '13 at 12:31
  • This answer refers to field elements like $x$ and $z$ being positive. The notion positive is not defined for a general field $k$. – John Gowers May 09 '14 at 20:13
  • @Donkey_2009 The OP asks for the case $k=\mathbb R$, so that is what I'm assuming here. If you like, choose $k$ to be your favourite ordered algebraically closed field. – Fredrik Meyer May 09 '14 at 20:25
  • Yes, sorry you're right. I just saw the $k$ in your answer and didn't bother to scroll up to see what had actually been asked in the question. – John Gowers May 09 '14 at 20:32
  • Another, more important point - you haven't actually shown that $I(X)=(x^2-z, xz-y^2)$. Instead, you've shown that $X=I(x^2-z, xz-y^2)$. In other words, $(x^2-z, xz-y^2)$ is an ideal such that every polynomial in it vanishes on $X$ and every point on $X$ satisfies those polynomial equations - but that does not mean that every polynomial that vanishes on $X$ is an element of $(x^2-z, xz-y^2)$. – John Gowers May 10 '14 at 09:21
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  1. Write down the ideal $J:=(x-r^2,y-r^3,z-r^4)$.
  2. Find a Groebner basis, using Buchberger's algorithm for example. Use the monomial order $r>x>y>z$ but such that the $r$ parts of the monomials are compared first.
  3. The elements of the basis that do not depend on $r$ give you a Groebner basis (a set of generators) of the ideal of $X$.

When we compute we get $J=(y^2-xz,x^2-z,rz-xy,ry-x^2,rx-y,r^2-x)$.

So, $I(X)=(y^2-xz,x^2-z)$.

Code in Singular:

 > ring R=0,(r,x,y,z),(dp(1),dp(3));
 > ideal J=(x-r^2,y-r^3,z-r^4);
 > J=groebner(J);
 > J;
 J[1]=y2-xz
 J[2]=x2-z
 J[3]=rz-xy
 J[4]=ry-x2
 J[5]=rx-y
 J[6]=r2-x
 >

Alternatively Singular has a command to do elimination directly.

> eliminate(J,r);
_[1]=y2-xz
_[2]=x2-z

What we did is one of the applications of Groebner bases. In this case, Implicitization.

user26857
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OR.
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