1

Let $X=\{(t,t^2,t^5): t\in k\}\subset \mathbb{A}^3$.

  1. Show that the projective closure $\bar{X}$ is a projective variety of dimension 1, and say if it is isomorphic to $\mathbb{P}^1$.
  2. Compute $\mathbb{I}_p(\bar{X})$.

What I tried:

  1. We can see that $X$ is an affine variety with corresponding ideal $J=(y-x^2,z-x^5)$. Since $X$ is an affine variety, then its projective closure is $\mathbb{V}_p({}^hJ)$, where ${}^hJ$ is the homogenization of $J$. We have ${}^h(y-x^2)=wy-x^2$ and ${}^h(z-x^5)=w^4z-x^5.$ Then $\bar{X}=\mathbb{V}_p(wy-x^2,w^4z-x^5)\subset \mathbb{P}^3$ is a projective variety.

Question 1: How can I prove that the dimension of $\bar{X}$ is 1 and decide if it's isomorphic to $\mathbb{P}^1$?

  1. To calculate $\mathbb{I}_p(\bar{X})$ I will use that $\mathbb{I}_p(\bar{X})=\mathbb{I}(C(\bar{X}))$, where $C(\bar{X})$ is the cone of $\bar{X}$.

$C(\bar{X})=\{0\} \cup \{x\in\mathbb{A}^4\setminus{\{0\}}: [x]\in\bar{X}\}= \mathbb{V}(wy-x^2,w^4z-x^5)$.

Then $\mathbb{I}(C(\bar{X}))=\sqrt{(wy-x^2,w^4z-x^5)}$.

Question 2. Is correct what I did in 2.?

Question 3. How can I calculate explicitly the radical that I obtained in 2.?

SSG19
  • 53
  • What about $\langle x^2-yw, y^3-xzw, xy^2-zw^2\rangle$ from $\langle w-s^5,x-ts^4,y-t^2s^3,z-t^5\rangle$? – Jan-Magnus Økland Sep 20 '21 at 10:15
  • Also, primaryDecomposition radical I where $I$ is your $\langle wy-x^2,w^4z-x^5\rangle$ yields in M2 {ideal(w,x), ideal(x^2-y*w,y^3-x*z*w,x*y^2-z*w^2)}. – Jan-Magnus Økland Sep 20 '21 at 10:59
  • I don't understand your first comment. Where do you get these ideals from? Where do you get the polynomial $y^3-xzw$ from? – SSG19 Sep 20 '21 at 11:08
  • $y^3=(t^2s^3)^3=(ts^4)(t^5)s^5=xzw.$ But also look at $s=w=1.$ – Jan-Magnus Økland Sep 20 '21 at 11:11
  • I don't understand if the ideals you give me correspond to the radical I am trying to calculate. Also, from the ideals I already calculated, how do I get the idea $(x^2−yw,y^3−xzw,xy^2−zw^2)$ and the ideal $(w−s^5,x−ts^4,y−t^2s^3,z−t^5)$? – SSG19 Sep 20 '21 at 11:16
  • It's instructive to look at the rational normal quintic, where ${\Bbb P}^1$ has coordinates $(s:t):$ $(s^5:s^4t:s^3t^2:s^2t^3:st^4:t^5).$ Project by omitting the fourth and fifth coordinate and set $s=1$ and you have your parametrization. – Jan-Magnus Økland Sep 20 '21 at 11:22
  • If I do that I obtain $(1:t:t^2:t^3)$, but this is different of the parametrization of the $X$ that should be $(1:t:t^2:t^5)$ – SSG19 Sep 20 '21 at 11:32
  • No, you omitted the fifth and sixth. – Jan-Magnus Økland Sep 20 '21 at 11:33
  • 1
    The problem is that the homogenization $I^h$ of an ideal $I = (f_1, \ldots, f_m)$ is not always the same thing as the ideal $(f_1^h, \ldots, f_m^h)$ generated by the homogenization of its generators. In this problem, the two are not the same. See my answer here for another example. – Viktor Vaughn Sep 20 '21 at 14:58

0 Answers0