Zariski closure of $T:= \{(t_1,t_2,t_3,t_1t_2t_3^{-1})|t_i\in \mathbb{C}^*\}\subseteq \mathbb{C}^4$?

Question

Let $V= \mathcal{V}(\langle xy-zw\rangle)\subseteq \mathbb{C}^4$ be an affine variety. The set $T:= \{(t_1,t_2,t_3,t_1t_2t_3^{-1})|t_i\in \mathbb{C}^*\}$ is a torus contained in $V$. I am trying to understand why $\overline{T}=V$, where $\overline{T}$ is the Zariski closure of $T$ in $\mathbb{C}^4$.

It is clear that $\overline{T}\subseteq V$, but I have no idea how to proceed in the other direction.

Answer 1

Let $k=\mathbb{C}$ and $R=k[x,y,z,w]$. Furthermore let $S=k[t_1,t_2,t_3,t_1^{-1}, t_2^{-1}, t_3^{-1}]$.

Call $f$ the map $f:R \to S$ defined by $f(x) = t_1$, $f(y) = t_2$, $f(z) = t_3$ and $f(w) = t_1 t_2 t_3^{-1}$. Let $I \subseteq R$ be the kernel ideal of $f$. By a Gröbner base calculation with a suitable CAS (e.g. Macaulay 2) it can be proved that $I = \ker f = (x y - z w)$, at least for $k=\mathbb{Q}$.

So we have an exact sequence

$$R \twoheadrightarrow R/I \hookrightarrow S$$

As $R/I \hookrightarrow S$ we have $\mathrm{Spec}(S) \to \mathrm{Spec}(R/I)$ is a dominant map, that is with dense image in $\mathrm{Spec}(R/I)$. This proves the sought for theorem.