# Does $k[B]^{B'}=k[\det(b),\det(b)^{-1}]$?

Let $k=\overline{k}$.

Suppose $GL_2(k)$ acts on $GL_2(k)$ by left multiplication. Then $k[GL_2]^{SL_2} = k[\det(g),\det(g)^{-1}]$.

Now for $B=\left\{ \left( \begin{array}{cc} b_{11} & b_{12} \\ 0 & b_{22} \\ \end{array}\right) : b_{ii}\not=0, b_{12}\in k \right\}$, let $B'=\{ b\in B: \det(B)=1 \}$. Suppose $B$ acts on $B$ by left multiplication.

Then does $k[B]^{B'}=k[b_{11}, b_{22},b_{11}^{-1},b_{22}^{-1}]$ or does $k[B]^{B'}=k[\det(b),\det(b)^{-1}]$?

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