Show that this morphism of varieties is not separable

Question

Let $k$ be an algebraically closed field of characteristic $2$, $G = \textrm{SL}_2(k)$, and $z = \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$. Let $\sigma: G \rightarrow G$ be the automorphism $\sigma(x) = zxz^{-1}$ (actually, $z = z^{-1}$). Let $\chi: G \rightarrow G$ be the morphism of varieties given by $\chi(x) := \sigma(x)x^{-1} = zxz^{-1}x^{-1}$. Since $G$ is irreducible, the image of $\chi$ is irreducible, hence so is its closure in $G$. So $\chi$ induces a dominant morphism of varieties $$\chi: G \rightarrow \overline{\chi(G)}$$ Let $K, L$ be the fields of rational functions of $\overline{\chi(G)}, G$ respectively. For example, $L = \textrm{Quot}(k[X_{ij}]_{\textrm{Det}(X_{ij})}) = k(X_{11},X_{12},X_{21},X_{22})$. Since $\chi$ is dominant, it induces an inclusion of fields $K \rightarrow L$. I am trying to

show that this field extension is not separable.

But I am having trouble computing $\overline{\chi(G)}$ as a closed set in $\textrm{SL}_2(k) = \{ x \in k^4 : \textrm{Det}(x) =1 \}$, which I think I need to do to understand what $K$ is. I have computed the image of $\chi$ to be $$Y = \{ \begin{pmatrix} ac + 1 + c^2 & ac + 1 + a^2 \\ c^2 & ac + 1 \end{pmatrix} : a,c \in k \}$$ Now to get the closure of $Y$ in $G$ (or in $k^4$, same thing), I know I need to take the ideal of functions vanishing on $Y$, then compute the zero set of that.

Answer 1

I'm just gonna post what I have. Identify $k^4$ with $4$ by $4$ matrices, as $\begin{pmatrix} x & y \\ z & w \end{pmatrix}$. The map $\tau: k^4 \rightarrow k^4$ which adds the bottom row of a matrix to the top row is an isomorphism of varieties; its inverse is itself, since we are in characteristic $2$. If you follow $\chi$ by composition with $\tau$, then you as mentioned in the comments, you will get the zero set $S$ of $x-w, x+yz-1$. The coordinate ring is $k[X,Y,Z,W]/(X-W,1-X-YZ)$, which is isomorphic to $k[X,Y,Z]/(1-X-YZ)$ via $f(x,y,z,w) \mapsto f(x,y,z,x)$. This last ring is isomorphic to $k[X,Y,Z]/(X-YZ)$, via $f(x,y,z) \mapsto f(x+1,y,z)$.

Now, $\phi = \tau \circ \chi$ be the dominant morphism $G \rightarrow S$, given by the formula $$\begin{pmatrix} x & y \\ z & w \end{pmatrix} \mapsto \begin{pmatrix} xz+1 & x^2 \\ z^2 & xz+1 \end{pmatrix}$$ Since $\tau$ is an isomorphism of varieties, it suffices to show that $\phi$ is separable. If $x,y,z,w$ are the coordinate functions in $k[S]$, we have for $A = \begin{pmatrix} a & b \\ c& d \end{pmatrix} \in G= \textrm{SL}_2(k)$, $$\phi^{\ast}(x)(A) = x(\phi(A)) = ac + 1$$ $$\phi^{\ast}(y)(A) = a^2 $$ $$\phi^{\ast}(z)(A) = c^2$$

Therefore, $\phi^{\ast}: k[S] \rightarrow k[G]$ is defined by $\phi^{\ast}(f(x,y,z,w)) = f(xz+1,x^2,z^2,xz+1)$.

We identify $R := k[X,Y,Z]/(X-YZ)$ with $k[S]$ by sending $f(x,y,z) \in R$ to $f(x+1,y,z) \in k[S] = k[X,Y,Z,W]/(X-W,1-X-YZ)$. Under this identification, $\phi^{\ast}: R \rightarrow k[G]$ is the map $$\phi^{\ast}(f(x,y,z)) = f(xz,x^2,z^2)$$

Let $K$ be the quotient field of $R$, and $L$ the quotient field of $k[G]$. We know that $\phi^{\ast}$ is injective; this extends to an injection of fields $\psi: K \rightarrow L$. Now, $R$ is a two dimensional ring, e.g. $(X-YZ)$ is a height one prime ideal. And $k[G]$ is three dimensional (in general, $\textrm{SL_n}$ has dimension $n^2-1$). So, $L$ must have transcendence degree $1$ over $\psi(K)$.

So, to show that $K \rightarrow L$ is not a separately generated extension I need to show that, for any $\xi \in L$ which is transcendental over $\psi(K)$, there exists an element $\zeta \in L$ which is not separable over $\psi(K)$.