4

The following is exercise 3.7 from Undergraduate Algebraic Geometry by Reid.

Let $J=\langle x^2+y^2+z^2, xy+yz+xz\rangle$; identify $V(J)$ and $I(V(J))$.

The question did not specify the field. It is easy to see that in $\mathbb{R}[x,y,z]$, $V(J)=\{(0,0,0)\}$ and $I(V(J))=\langle x,y,z\rangle$.

Consider it in $\mathbb{C}[x,y,z]$. From the defining equations, I got $x+y+z=0$. Substituting $x=-y-z$ into $x^2+y^2+z^2$, we get $y^2+yz+z^2=0$, this is equivalent as $(y+\frac{1}{2}z)^2+\frac{3}{4}z^2=0$. This gives $$y=\left(-\frac{1}{2}\pm\frac{\sqrt{3}}{2}i\right)z$$

So $$I(V(J))=\left\langle y-\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)z, x-\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)z\right\rangle \\\cap \left\langle y-\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)z, x-\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)z \right\rangle$$

Since $\mathbb{C}$ is algebraically closed, this should also be $\sqrt{J}$.

My question is,

Is it possible to find the generators so we don't have to write it as an intersection?

Thank you for your help.

user26857
  • 53,190
KittyL
  • 17,275

2 Answers2

2

I think Proposition $30$ of $\S9.6$ of Dummit and Foote answers your question. (This is also Theorem 11 of Ch. 4, $\S3$ of Cox, Little, and O'Shea.)

Proposition 30. If $I$ and $J$ are any two ideals in $F[x_1, \ldots, x_n]$ then $tI + (1 - t)J$ is an ideal in $F[t, x_1, \ldots, x_n]$ and $I \cap J = (tI + (1 - t)J) \cap F[x_1, \ldots, x_n]$. In particular, $I \cap J$ is the first elimination ideal of $t I + (1 - t) J$ with respect to the ordering $t > x_1 > \cdots > x_n$.

So if you have Gröbner bases for $I$ and $J$, you can compute a Gröbner basis for $I \cap J$.

Viktor Vaughn
  • 20,897
  • Thank you so much! I got $\langle x+y+z, y^2+yz+z^2\rangle$ which verifies my guess! But I didn't know how to show it. – KittyL Jan 24 '16 at 18:51
  • Don't know how to show what? Are you saying you don't know how to prove the proposition? – Viktor Vaughn Jan 24 '16 at 22:12
  • 1
    I meant I guessed that the radical ideal might be this, but didn't know how to get it. The proposition is just what I needed. Thank you. – KittyL Jan 25 '16 at 00:53
1

Let's compute the radical of $J$ over $\mathbb R$. Let $P$ be a minimal prime over $J$. Since $(x+y+z)^2\in J$ we get $x+y+z\in P$. On the other side, $xy+yz+zx\in P$, so $xy-z^2\in P$. Then $(x+y+z,yz-x^2)\subseteq P$. But $(x+y+z,yz-x^2)$ is a prime ideal in $\mathbb R[x,y,z]$ since $$\mathbb R[x,y,z]/(x+y+z,yz-x^2)\simeq\mathbb R[y,z]/(y^2+yz+z^2),$$ and $y^2+yz+z^2$ is an irreducible polynomial in $\mathbb R[y,z]$ (why?). Then $P=(x+y+z,yz-x^2)$ (or $P=(x+y+z,y^2+yz+z^2)$ if you like) and this shows that $$\sqrt J=(x+y+z,y^2+yz+z^2).$$

There is no problem to see that this result is "coefficient field free": see here.

Remark. If you want to recover the above result from what you got, write $$\sqrt J=(y-az,x-\bar az)\cap(y-\bar az,x-az),$$ where $a=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$. Now notice that $$(y-az,x-\bar az)=(x+y+z,y-az),$$ and $$(y-\bar az,x-az)=(x+y+z,y-\bar az),$$ so $$\sqrt J=(x+y+z,y-az)\cap(x+y+z,y-\bar az)=(x+y+z,(y-az)(y-\bar az))=(x+y+z,y^2+yx+z^2).$$

Community
  • 1
user26857
  • 53,190
  • Could you elaborate a little bit more on how did we get from $(x+y+z,y-az)\cap (x+y+z,y-\bar{a}z)$ to $(x+y+z, (y-az)(y-\bar{a}z))$? – Xipan Xiao Jun 30 '20 at 18:58
  • oh, is it because $y-az, y-\bar{a}z$ are irreducible, so $(y-az)\mid (y-\bar{a}z) \cdot f$ means that $y-az\mid f$, hence $(y-az)\cap (y-\bar{a}z) = ((y-az)(y-\bar{a}z))$? – Xipan Xiao Jun 30 '20 at 19:11