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These are problems from Miles Reid's book "Undergraduate algebraic geometry"

Let $k$ be an algebraically closed field.

Question 1.

Let $I= (xy,xz,yz) \subset k[x,y,z]$. I want to find $Z(I)$ ok it is clear that $I$ is the union of the three coordinate axes, now I want to prove rigorously each coordinate axis is closed and irreducible. Can we simply say:

For example the z-axis is homeomorphic to $\mathbb{A}^{1}$. But any homeomorphism preserves closed sets and irreducible sets so since $\mathbb{A}^{1}$ is algebraic (i.e closed) and irreducible we are done.

Question 2:

Let $I=(x^{2}+y^{2}+z^{2},xy+xz+yz)$. Identify $Z(I)$ and $I(Z(I))$.

I'm confused with this one because here in case the field is $\mathbb{R}$ then we get $x=y=z$ however $\mathbb{R}$ is not algebraically closed so how do we proceed?

Now I was thinking in using the identity $(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(xy+xz+yz)$ from here we get that $x+y+z=0$. So it seems $Z(I)$ is the set of all points in the plane $x+y+z=0$. Is this wrong? But what if the underlying field has characteristic equal $2$?

Finally how do we find $I(Z(I))$?

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Dear user6495, it is not true that for $k=\mathbb R$ you get $x=y=z$ as the set of zeros $Z(I)$ in Question 2 since $x^2+y^2+z^2$ takes the value $3a^2$ instead of $0$ at $(a,a,a)$. –  Georges Elencwajg Feb 14 '12 at 19:10
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1 Answer

up vote 4 down vote accepted

Question 1: The $z$-axis is closed and irreducible because it is the locus of zeros $Z(x,y )$ of the prime ideal $(x,y)$.

Question 2: No, the locus of zeros is not the plane $x+y+z=0$.
Your identity $(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(xy+xz+yz)$ proves that you can write $I=(x^{2}+y^{2}+z^{2},xy+xz+yz)=((x+y+z)^{2}-2(xy+xz+yz),xy+xz+yz)$
$=((x+y+z)^{2},xy+xz+yz)$.
Since the set of zeros of a polynomial is the same as the set of zeros of its square, you see that $Z(I)=Z(I')$ with $I'=(x+y+z,xy+xz+yz)=(x+y+z,xy+(x+y)z)$
$=(x+y+z,xy-(x+y)^2)=(x+y+z,-(xy+x^2+y^2))$

Hence $Z(I)=Z(I')$ is the intersection of the plane $x+y+z=0$ with the quadratic cylinder$Q$ given by $x^2+xy+y^2=0$ .
That cylinder is a single plane (with multiplicity 2) or the union of two planes according as $char(k)=3$ or $char (k)\neq3$ .
So $Z(I)$ is a single (double) line if $char(k)=3$ or the union of two lines if $char (k)\neq3$.

The Nullstellensatz tells you that $I(Z(I))=\sqrt I$ where $I= ((x+y+z)^{2},xy+xz+yz)$.
$\bullet $ If $char(k)\neq 3$, then $\sqrt I=(x+y+z ,\: x^2+y^2+xy) =I'$
$\bullet \bullet$ If $char(k)= 3$, then $ I=((x+y+z)^2 ,\:(x-y)^2)$ [because $x^2+y^2+xy=(x-y)^2$]
so that $\sqrt I=(x+y+z, x-y)$

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thanks, is it possible that you can please explain how did you obtain the locus of zeros? –  user6495 Feb 14 '12 at 1:23
    
Dear user6495, I have expanded the calculation of the locus of zeros $Z(I)$ by showing that it is equal to the locus of zeros $Z(I')$ of a bigger ideal $I'$, which turns out (in characteristic not 3) to be the radical of $I$, that is $I'=\sqrt I$. I have also added an explicit description of $I(Z(I)))$ in characteristic $3$. –  Georges Elencwajg Feb 14 '12 at 20:08
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