Borel-Morozov theorem fails for non-algebraically closed field

Statement
It is possible to have an infinite field $$K$$ that is not an algebraically closed field and a connected linear algebraic group $$G$$ such that $$G$$ has more than one conjugacy class of Borel subgroups. In other words, the analogue of the Borel-Morozov theorem fails for $$G$$.

Proof
Since $$K$$ is not algebraically closed, there is a finite-dimensional extension $$L$$ of $$K$$ of degree $$d > 1$$. Consider $$L$$ as $$K^d$$. Let $$G = GL(d,K)$$. We claim that $$G$$ has the following two non-conjugate Borel subgroups:


 * The multiplicative group $$L^\ast$$ with the following embedding: each element of $$L^\ast$$ becomes the corresponding linear transformation of $$L = K^d$$ induced by multiplication by it.
 * The subgroup of upper-triangular matrices with nonzero entries on the diagonal (the standard Borel subgroup).