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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!

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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 
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      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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