Is $X=(x^2-yx, y^2-xz, z^2-x^2y) $ isomorphic to $ \mathbb A_{K}^1$?

Question

Is the algebraic affine set $X=(x^2-yx, y^2-xz, z^2-x^2y) \subset \mathbb A_{K}^3$ isomorphic to $ \mathbb A_{K}^1$?

Any hints and comments would be highly appreciated.

Answer 1

You can just compute what this is: the first two generators $x^2-yx=y^2-xz=0$ define the intersection between the union of two planes and a cone. The first condition $x(x-y)=0$ tells you $x=0$ or $x=y$ and the second condition tells you that if $x=0$ then $y=0$ and $z$ can be anything so you get the $z$-axis. In the second case $x\neq 0$ implies $x=y$ which therefore implies $x=y=z$ (look at the second condition), and so far we have the union of the $z$-axis and the line ${x=y=z}$. Now impose final condition $z^2-x^2y=0$. If $x=0$ then $y=0$ and hence $z=0$ as well. If $x\neq 0$ then $x=y=z$ and the third condition says (by plugging in $x$ for $z$ and $y$) that $x^2-x^3=0$ which means $x=1$ giving us another point $(1,1,1)$. So this variety is just two points $(0,0,0),(1,1,1)$.

Edit: to address the comment: There are several ways to see this:

First way: You can basically perform the same argument/computation in the context of the coordinate rings by duality. For example, $K[x,y,z]/(f_1,f_2,f_3)$ can be thought of as the composition of the maps $$K[x,y,z] \to K[x,y,z]/(f_1) \to (K[x,y,z]/(f_1))/(\bar{f_2}) \cong K[x,y,z]/(f_1,f_2)\to K[x,y,z]/I \cong (K[x,y,z]/(f_1,f_2))/(\bar{f_3})$$ where $\bar{f}$ is understood to be the image in the appropriate quotient ring.

As before we begin $K[x,y,z] \to K[x,y,z]/(x(x-y)) \hookrightarrow K[x,y,z]/(x) \times K[x,y,z]/(x-y)$ identifies our first quotient with a subring of the coordinate rings of the two planes. Composing each of the factors with the canonical projection onto the quotient by $(\bar{y}^2-\bar{x}\bar{z})$ (Image taken in the respective factors) $$\\ \displaystyle \to (K[x,y,z]/(x))/(\bar{y}^2-\bar{x}\bar{z}) \times (K[x,y,z]/(x-y))/(\bar{y}^2-\bar{x}\bar{z}) \\ \cong K[y,z]/(y^2) \times K[y,z]/(y^2-yz)$$ Again the second factor embeds into the product of two rings and taking a further quotient by the image of $(z^2-x^2y)$ in each factor we can see that our original ring can be realized as a (subring of) something that looks $\cong K\times K \times K $.

Another (easier) way: Supposing $K[x,y,z]/I$ is isomorphic to affine space, then it must be smooth of dimension 1. Hence the Jacobian matrix $(\frac{\partial f_i}{\partial x_j})_{ij}$ should have rank $3-1=2$ at all points of the variety, but it is a quick computation to see that the rank is zero at the origin, contradiction.