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A map from $\mathbb{P^2} \rightarrow \mathbb{P^3}$ is given by: $$f(x:y:z) = (-xyz:xy(x+y+z):xz(x+y+z):yz(x+y+z))$$

How might one find the equations of it's image?

baltazar
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    How do we find the equations of the image of $t\mapsto(t^2,t^3)$? We consider the ideal of the graph of the function $I=(t^2-x,t^3-y)$ and then find an special set of generators, $$I=(x^3-y^2,ty-x^2,tx-y,t^2-x)$$ that allows us to compute $I\cap k[x,y]=(x^3-y^2)$. The set of generators is called a Groebner basis with respect to convenient monomial order called elimination monomial order. – Pp.. Jan 16 '15 at 16:13
  • @Pp.. we didn't deal with the Groebner basis so far, but I know it's important and I'll look into applying your method for this exercise. – baltazar Jan 16 '15 at 16:26
  • @Pp.. if you know a way without using it, or at least have an idea, I would appreciate it, since I am kind of stuck trying to find relations by hand. – baltazar Jan 16 '15 at 16:29
  • Some sort-of-hints: 1. Do you know how to find the dimension and degree of the image? 2. How many equations do you need to cut out a surface in $\mathbf P^3$? –  Jan 16 '15 at 16:41
  • @AsalBeagDubh 1. no, not without the equations. Is there a way? 2. I was under the impression that one equation defines a surface in any $\mathbb{P}^n$. Is this wrong? – baltazar Jan 16 '15 at 16:46
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    Yes, you are basically right: a hypersurface in $\mathbf P^n$ (meaning a variety of dimension $n-1$) is always cut out by 1 equation. So here, if the image of your map is 2-dimensional, you only need 1 equation. You can check if the image is 2-dimensional by calculating the differential $df$ at a general point of $\mathbf P^2$. As for the degree: it is the number of intersection points of the image with a general line in $\mathbf P^3$. Can you see how to find that number just using the formula for the map? (If not I will try to answer later when I have time...) –  Jan 16 '15 at 16:55
  • @AsalBeagDubh I tried some calculations but I am not getting anywhere. If it's not too much trouble I would appreciate a demonstration. Thank you in any case. – baltazar Jan 16 '15 at 19:55

1 Answers1

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The points where at least one of $x,y,z$ is zero get mapped to one point. For example $$f(0:y:z)=(0:0:0:yz(y+z))=(0:0:0:1)$$

Let us look at $x,y,z\neq0$ where the bulk of the action is happening. There $$\begin{align}f(x:y:z)&=f(u:v:1)\\&=(-1:(u+v+1):(u+v+1)/v:(u+v+1)/u)\end{align}$$

We consider $$\begin{align}I&=(u+v+1-X,u+v+1-vY,u+v+1-uZ)\\&=(XYZ-XY-XZ-YZ,vZ-XZ+X+Z,vY-X,u+v-X+1)\end{align}$$

The code in Singular

ring R=0,(u,v,X,Y,Z),(dp(2),lp);

ideal I=(u+v+1-X,u+v+1-v*Y,u+v+1-u*Z);

I=groebner(I);

I;

computes it. The monomial order here compares the part formed by $u,v$ first, only if that is equal it compares the $X,Y,Z$ part of the monomial.

The computation by hand is not too bad, but I never trust my hand with something my computer does better. You can take a look at Buchberger's algorithm. It is not complicated (and is useful to learn), the annoying thing can be running it by hand. Alternatively, use the computer, and then check that it works.

OK. So, now we have that $I\cap k[X,Y,Z]=(XYZ-XY-XZ-YZ)$. Now we homogenize to get the equation in $\mathbb{P}^3$. We put $X=-b/a,Y=-c/a,Z=-d/a$ and multiply by $-a^3$. The $-$ signs is because I left a $-1$ up there when I went to the chart in $\mathbb{P}^3$. We get

$$bcd+abc+abd+acd=0\ \ \ \ \text{ as the equation.}$$

Pp..
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  • It's a helpful answer and a very practical one, but unfortunately this is an example of an exercise that I need to be able to do by hand. And without using the methods you used, at least for now. Upvoted. – baltazar Jan 16 '15 at 20:42
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    @baltazar You can certainly do it by hand. The same way you do the $t\mapsto(t^2,t^3)$. You guess the equation $x^3-y^2=0$ by playing with the formula and then follow Asal's comment to show you only need one equation. Computing the degree tells you the degree you need for guessing the equation. – Pp.. Jan 16 '15 at 21:10
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    @baltazar And once you have seen the equation we got you can't un-see it. The computation gives us $$x^2y^2z^2(x+y+z)^3-x^3y^2z^2(x+y+z)^2-x^2y^3^2(x+y+z)^2-z^2y^2z^3(x+y+z)^2=x^2y^2z^2[(x+y+z)-(x+y+z)]=0$$ – Pp.. Jan 16 '15 at 21:13