Describing the orbits of an algebraic action on $\mathbb{A}^4$ I have an algebraic action of $(k^*)^2$ (an algebraic group) on the variety $\mathbb{A}^4$ by: $(t,u) \cdot (x,y,z,v) = (tx, uy, t^{-1}z, u^{-1}v)$. What are the orbits of this action?
I am not sure exactly how I am supposed to describe these orbits. For example, it is clear to me that $(0,0,0,0)$ is in its own orbit, and there is an orbit where $x=0$ with the rest nonzero, $y=0$ with the rest nonzero, and so on. And in some cases, like if $z=v=0$ and $x$ and $y$ are nonzero, then you can get any nonzero values in the first two slots, so the orbit "looks like" $(k^*)^2$. Isn't there a better way of describing these orbits?
It seems clear to me that the quotient by the action is given by $(x,y,z,v) \mapsto (xz, yv)$. This is very obviously constant on orbits, but doesn't seem to actually tell me much about the orbits?
Any insight into how to describe these orbits would be appreciated. Is it just a big list of all of the cases where various coordinates are zero? What about when they are all nonzero?

Yes, it is just a big list.


*

*If none of $x,y,z,v$ are zero, then the orbit is a copy of $\mathbb{G}_m^2$: the intersection of two quadrics $x_1x_3=xz$ and $x_2x_4=yv$ in $\mathbb{A}^4$.

*If $x=0$ but $y,z,v$ are nonzero, then you also have a copy of $\mathbb{G}_m^2$ as the orbit: $x_1=0, x_3\neq 0, x_2x_4=yv$.  Similarly the other three cases of one zero.

*If exactly one from each of $\{x,z\}, \{y,v\}$ are zero, then we have one orbit, a copy of $\mathbb{G}_m^2$: e.g., $x=y=0$ given by $x_1=x_2=0,x_3x_4\neq 0$.
On the other hand, if $x=z=0$, $yv\neq 0$, then we get orbits which are a copy of $\mathbb{G}_m$: $x_1=x_3=0$, $x_2x_4=yv$.  Similarly for $y=v=0$, $xz\neq 0$.

*If $x=y=z=0$, then $v\neq 0$ is an orbit which is a copy of $\mathbb{G}_m$.  Similarly the other three cases of three zeros.

*Finally, the origin $(0,0,0,0)$ is a fixed point of the action.
