This is an exercise from Chapter 8 of Ideals, Varieties and Algorithms by Cox et al.

The projective Veronese surface in $\mathbb{P}^5$ is defined as the projective closure of the surface $S$ which is the image of the map $$\phi(x_1,x_2)=(x_1,x_2,x_1^2,x_1x_2,x_2^2)$$ I will denote it by $V_V$.

The Segre variety is defined as the image of $\sigma: \mathbb{P}^2\times \mathbb{P}^1\rightarrow \mathbb{P}^5$ where $$\sigma(x_0,x_1,x_2,y_0,y_1)=(x_0y_0,x_0y_1,x_1y_0,x_1y_1,x_2y_0,x_2y_1)$$ I will denote it by $V_S$.

The question is what is the intersection of $V_V$ and $V_S$.

My attempt:

I calculated the variety by using Groebner basis. This is what I got: $$V_V=V(x_1^2-x_0x_3,x_1x_2-x_4x_0,x_1x_4-x_2x_3,x_1x_5-x_2x_4,x_2^2-x_5x_0,x_3x_5-x_4^2)\\ V_S=(x_0x_3-x_1x_2,x_0x_5-x_1x_4,x_2x_5-x_3x_4)$$

I see that $V_S$ cannot be transformed to the equations in $V_V$ by change of variables.

So I am not sure how to proceed from here.

Thank you for any help!


The intersection of two varieties is given by the sum of their ideals. So you're looking at the ideal generated by the union of the generators of $V_V$ and $V_S$.

Immediately we see that both $x_1^2-x_0x_3$ and $x_0x_3-x_1x_2$ are generators of $I=I_{S \cap V}$. But this implies that $x_1^2-x_1x_2=x_1(x_1-x_2)$ is a generator of $I$. Thus $I$ is reducible.

We can try to find the intersection directly. A point on the intersection must satisfy $x_1(x_1-x_2)=0$. So assume $x_1 \neq x_ 2$. Then we must have $x_1=0$. Then the ideal reduces to $$ I = (x_0x_3,x_4x_0,x_2x_3,x_2x_4,x_2^2-x_5x_0,x_3x_5-x_4^2,x_0x_5,x_2x_5-x_3x_4) $$ Removing duplicates and simplifying, this is $$ I = (x_0x_3,x_0x_4,x_0x_5,x_2x_3,x_2x_4,x_2^2,x_3x_5-x_4^2,x_2x_5-x_3x_4) $$ This is clearly reducible. For example, $x_0x_3=0$. Assume first $x_0=0$. Then the (radical of the ideal is) $$ (x_2,x_3x_5-x_4^2,x_3x_4) $$ This decomposes as well. Assume first $x_3=0$. Then the ideal is $(x_2,x_4^2)$. Hence we have found a point on the intersection, namely the one given by ideal $(x_1,x_0,x_3,x_2,x_4)$. This is the point $(0:0:0:0:1)$ in $\mathbb P^5$.

If we assume $x_3=0$ above, we get by the same procedure the ideal $(x_0x_5,x_2,x_4)$. This gives us two more points, namely $(0:0:0:0:0:1)$ and $(1:0:0:0:0:0)$.

Lastly, assume $x_1=x_2$. Then the equations simplifly to $$ (x_1^2-x_0x_3,x_1^2-x_0x_4,x_1x_4-x_1x_3,x_1x_4-x_1x_4,x_1^2-x_0x_5,x_3x_5-x_4^2,x_0x_3-x_1^2,x_0x_5-x_1x_4,x_1x_5-x_3x_5) $$ We can assume $x_1 \neq 0$, since we have treated that case already. Hence we can divide by $x_1$ everywhere. $$ (x_1^2-x_0x_3,x_1^2-x_0x_4,x_3-x_4,x_1^2-x_0x_5,x_3x_5-x_4^2,x_0 x_3-x_1^2,x_0x_5-x_1x_4,x_1x_5-x_3x_5). $$ Not too big a simplification, but now we see that $x_3=x_4$. This again simplifies a lot, since we get that $x_3x_5-x_3^2=0$. This implies either $x_3=0$ or $x_5=x_3$. First assume $x_3=0$. Then a similar calculation gives the point $(0:0:0:0:0:1)$ which we have already found.

Assume now $x_3 \neq 0$. Then $x_3=x_5$, so we get $$ (x_1^2-x_0x_3,x_0x_3-x_1^2,x_0x_3-x_1x_3,x_1x_3-x_3^2) $$ Since $x_3 \neq 0$, we get $x_0=x_1$. And $x_1=x_3$. All in all, we get the point $(1:1:1:1:1:1)$.

In conclusion: the intersection is a union of 4 points.

If you have Macaulay2, all this could have been done as follows:

i23 : decompose ideal mingens (I1 + I2)

o23 = {ideal (x  - x , x  - x , - x  + x , x  - x , x  - x ), ideal (x , x , x , x , x ), ideal (x ,
               2    4   1    4     4    5   0    4   3    4           4   2   1   0   3           4 
      x , x , x , x ), ideal (x , x , x , x , x )}
       2   1   5   0           4   2   1   5   3

Here $I1,I2$ are the ideals of the Veronese and Segre variety resepctively. We see that the ideals we get correspond to the points found above.

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