# Dimension of affine algebraic set in $\mathbb{A^6}$

If $$X=V(f_1, f_2, f_3) \subseteq \mathbb{A}^6$$, with $$f_1=x_1x_5-x_4x_2, \qquad f_2=x_1x_6-x_4x_3, \qquad f_3=x_2x_6-x_5x_3,$$

how can I show $$\dim X=4$$? I was trying to find $$\operatorname{ht} I(X)$$, since $$\dim X=6- \operatorname{ht} I(X)$$ or construct explicitely an isomorphism between $$A(X)$$ and a polynomial ring with $$4$$ variables, but I haven't succeeded.

• This variety is the cone over $\mathbb{P}^1 \times \mathbb{P}^2$. – Sasha May 20 '20 at 8:08
• Please use the relevant mathjax commands instead of leaving math mode to write things like $\dim$. I have updated your post with the changes. – KReiser May 20 '20 at 8:33
• Note that $f_3=x_2f_2-x_3f_1$, so one generator of $I(X)$ is superflous. Now it is clear that the height is $2$. – user26857 May 20 '20 at 9:51

Consider the Segre embedding $$\sigma:\mathbb{P}^1\times \mathbb{P}^2 \to \mathbb{P}^5$$ given by $$\sigma:((a_0:a_1),(b_0:b_1:b_2)) \longmapsto (a_0b_0:a_0b_1:a_0b_2:a_1b_0:a_1b_1:a_1b_2).$$ Letting $$x_1,\ldots,x_6$$ be the coordinate functions on $$\mathbb{P}^5,$$ we see that $$\sigma(\mathbb{P}^1\times\mathbb{P}^2)=V(x_1x_5-x_2x_4,\,x_1x_6-x_3x_4,\,x_2x_6-x_3x_5)\subseteq \mathbb{P}^5.$$ Hence your variety $$X$$ is the affine cone in $$\mathbb{A}^6$$ over $$\sigma(\mathbb{P}^1\times \mathbb{P}^2)$$ and so we have $$\dim X = \dim \sigma(\mathbb{P}^1\times \mathbb{P}^2) +1 = (1+2)+1 = 4.$$