A torus splits over a finite Galois extension Let $k$ be a field.  A torus $T$ is an affine group scheme over $k$ such that $T \times_k \overline{k}$ is isomorphic as $\overline{k}$-schemes to $\textrm{Spec }\overline{k}[T_1^{\pm1}, ... , T_n^{\pm1}]$.  If $E$ is a field containing $k$, we say that $T$ splits over $E$ if $T \times_k E$ is isomorphic as $E$-schemes to $\textrm{Spec }E[T_1^{\pm1}, ... , T_n^{\pm1}]$
Proposition: If $T$ splits over a finite, purely inseparable extension of $k$, then $T$ splits over $k$.
In particular, $T$ must split over the separable closure of $k$ in $\overline{k}$, hence over a finite Galois extension of $k$.
I found this proposition to be particularly interesting, so I wanted to share a method of proof suggested in a homework set I found online.
 A: Lemma 1: Let $k'$ be a finite extension field of $k$, $V$ a $k$-vector space, and $R = k' \otimes_k k'$.  Let $V_{k'} = V \otimes_k k'$, $V_R = V \otimes_k R$.  There are two ways to regard $R$ as a $k$'-vector space, hence two $k$-linear maps $V_{k'} \rightarrow V_R$ given on generators by $v \otimes c \mapsto v \otimes c \otimes 1$ and $v \otimes c \mapsto v \otimes 1 \otimes c$, each being $k$'-linear in their own way.  Then the set of $v \in V_{k'}$ on which both these maps agree is exactly $V$.
Proof: Let $e_i : i \in I$ be a basis for $V$ over $k$, so $e_i \otimes 1$ is one for $V_{k'}$, and $e_i \otimes 1 \otimes 1$ is one for $V_R$ as an $R$-module.  Since we can take the direct sum out of the tensor product, we see that the $k$-linear maps $V_{k'} \rightarrow V_R$ are direct sums of maps $k' \rightarrow k' \otimes_k k'$.  So we may assume $V = k$, and we are done once we show that $k = \{ x \in k' : x \otimes 1 = 1 \otimes x \textrm{ in } k' \otimes_k k' \}$.  Let $v_1, ... , v_n$ be a $k$-basis for $k'$, and choose $v_1$ to be $1$.  Then $c_1v_1 + \cdots + c_nv_n, c_i \in k$ lies in $k$ if and only if $c_2 = \cdots = c_n = 0$.  Since $v_i \otimes v_j$ is a basis for $k' \otimes_k k'$ over $k$, the result follows.  $\blacksquare$
Now let $k'$ be a finite purely inseparable extension of $k$, and assume that $T$ splits over $k'$.  Let $R = k' \otimes_k k'$.  Let $A$ be the coordinate ring of $T$, $B = k[T_1^{\pm 1}, ... , T_n^{\pm 1}]$, $A_{k'} = A \otimes_k k'$, $B_{k'} = B \otimes_k k'$, and similarly $A_R, B_R$.
Our assumption is that there exists some $k'$-algebra isomorphism $f: B_{k'} \rightarrow A_{k'}$.  We are done if we can show that the restriction of $f$ to $B$ is mapped onto $A$.
By definition, $A_R = A \otimes_k R$.  Choosing the right $k'$-algebra structure on $R$ gives $A_R$ the structure of a $k'$-algebra, as $A_R = A \otimes_k k' \otimes_k k' = A_{k'} \otimes_{k'} k'$.  Define $i_1: A_{k'} \rightarrow A_R$ on generators by $x \otimes c \mapsto x \otimes c \otimes 1$, and $i_2: A_{k'} \rightarrow A_R$ by $x \otimes c \mapsto x \otimes 1 \otimes c$.  These are $k'$-algebra homomorphisms when $R$ is given the left (resp. right) $k'$-algebra structure.  
Tensoring the product mapping $R \rightarrow k', x \otimes y \mapsto xy$ with the identity $A_{k'} \rightarrow A_{k'}$ gives a $k'$-algebra homomorphism $\pi: A_R \rightarrow A_{k'}$ (with either $k'$-algebra structure on $A_R$) such that $\pi \circ i_1 = \pi \circ i_2 = 1_{A_{k'}}$.
We can do the same with $B, B_{k'}, B_R$, obtaining $k'$-algebra homomorphisms $j_1, j_2:B_{k'} \rightarrow B_R$ and $\rho: B_R \rightarrow B_{k'}$ such that $\rho \circ j_1 = \rho \circ j_2 = B_{k'}$.
At the same time, we can tensor the $k'$-isomorphism $f: B_{k'} \rightarrow A_{k'}$ with the identity $k'$-algebra map $R \rightarrow R$ to obtain two $k$'-algebra isomorphisms $f_1, f_2: B_R \rightarrow A_R$, one for each pair of left (resp. right) $k'$-algebra structures on $R$ inducing such $k'$-algebra structures on $A_R$ and $B_R$.
We obtain a large commutative diagram:

where $\rho \circ j_1, \rho \circ j_2, \pi \circ i_1, \pi \circ i_2$ are the identity maps.  By the lemma, we know that $A$ is the set of points in $A_{k'}$ on which $i_1$ and $i_2$ coincide, and $B$ is the set of points in $B_{k'}$ on which $j_1$ and $j_2$ coincide.  If we can show that $f_1 = f_2$ as functions, it will follow easily that $f$ maps $A$ onto $B$.
We have $B_R = R[T_1^{\pm1 }, ... , T_n^{\pm 1}]$.  Actually, $f_1$ and $f_2$ are $R$-algebra isomorphisms, so it is enough to show that they send $T_1, ... , T_n$ to the same places.  Let's write $f(T_l) = \sum\limits_i a_i \otimes c_i$ for $a_i \in A, c_i \in k'$.  Now $i_1 \circ f = f_1 \circ j_1$, as can easily be deduced from the diagram, so we can see that
$$f_1(T_l) = \sum\limits_i a_i \otimes c_i \otimes 1$$
and similarly,
$$f_2(T_l) = \sum\limits_i a_i \otimes 1 \otimes c_i$$
Now we cannot ordinarily transfer $c_i$ over, but since $k'/k$ is purely inseparable, we can see that for a large integer $n$, we have $f_1(T_l^{p^n}) = f_2(T_l^{p^n})$ for all $l$.  
In general, if $R$ is a ring, and $\phi$ is an $R$-algebra automorphism of $R[T_1^{\pm 1}, ... , T_n^{\pm 1}]$, and there is a positive integer $m$ such that $\phi(T_l^m) = T_l^m$ for all $l$, then $\phi$ has the be the identity. This is clear, because $\phi$ has to permute the set $\{T_1^{\pm 1}, ... , T_n^{\pm 1} \}$.  So $f_1 = f_2$, as required.   
