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.

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    $\begingroup$ This variety is the cone over $\mathbb{P}^1 \times \mathbb{P}^2$. $\endgroup$ – Sasha May 20 '20 at 8:08
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    $\begingroup$ Please use the relevant mathjax commands instead of leaving math mode to write things like $\dim$. I have updated your post with the changes. $\endgroup$ – KReiser May 20 '20 at 8:33
  • $\begingroup$ 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 $2$. $\endgroup$ – user26857 May 20 '20 at 9:51

Expanding on Sasha's comment above.

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 $$\dim X = \dim \sigma(\mathbb{P}^1\times \mathbb{P}^2) +1 = (1+2)+1 = 4.$$


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