= Solution
An <affine algebraic group> is a <linearly reductive algebraic group> if every finite-dimensional <rational representation> is completely reducible; equivalently the functor of invariants is exact. It is a <geometrically reductive algebraic group> if, whenever $0\ne v\in V^G$, there is a homogeneous invariant polynomial $F\in k[V]^G$ of positive degree with $F(v)\ne0$. Linear reductivity permits degree one; geometric reductivity allows higher degree.
Over the <algebraically closed field>, an <algebraic torus> is $T\cong(\mathbb G_m)^r$. Its <coordinate ring> is the Laurent <polynomial ring>. A rational action writes the coaction of a vector as a finite sum $\delta(v)=\sum_\chi v_\chi\otimes t^\chi$. The coaction identity and linear independence of Laurent monomials give $\delta(v_\chi)=v_\chi\otimes t^\chi$ and $v=\sum_\chi v_\chi$. Thus
$$
\boxed{V=\bigoplus_{\chi\in\mathbb Z^r}V_\chi.}
$$
Each <weight space> has scalar action, so any vector-space complement within each <weight space> is invariant. This proves complete reducibility, and hence \b[tori are linearly reductive in every characteristic]. The argument uses characters themselves, not their <Lie algebra> differentials, which could coincide modulo $p$.
We now prove geometric reductivity of $G=SL_2(k)$ by <determinant separation for positive-characteristic SL2>. Fix a rational $G$-module $V$ and a nonzero fixed vector $v$. For the diagonal torus, choose a weight-zero functional $l\in V^*$ with $l(v)=1$, using the torus decomposition just proved. Define matrix coefficients
$$
\Phi(w)(g)=l(g^{-1}w).
$$
These are <regular functions> on $G$, and $\Phi(hw)(g)=\Phi(w)(h^{-1}g)$ makes $\Phi$ equivariant for left translations. Moreover $\Phi(w)(gt)=\Phi(w)(g)$ because $l$ is torus-invariant. Thus $\Phi$ maps into $k[G]^T$ for the right torus action, and $\Phi(v)=1$.
Write a matrix's two columns as $a=(a_0,a_1)$ and $b=(b_0,b_1)$, with $\Delta=a_0b_1-a_1b_0=1$. Right torus multiplication scales these columns by $t$ and $t^{-1}$. Every right-torus-invariant <regular function> has a weight-zero polynomial lift, hence a sum of polynomials of bidegrees $(j,j)$ in $(a,b)$. To see that a weight-zero lift exists, take any lift and retain its zero-weight component; the ideal $(\Delta-1)$ is torus-stable and the quotient map preserves weights. Multiply each piece by $\Delta^{n-j}$ to put the entire sum in bidegree $(n,n)$ without changing its restriction to $G$.
Let $P_n$ be the space of degree-$n$ polynomials in one column. Restriction gives an injective equivariant map $P_n\otimes P_n\to k[G]^T$. Indeed if a bidegree-$(n,n)$ polynomial vanishes on <determinant>-one matrices, scale the first column of any invertible matrix to make its <determinant> one. Homogeneity then makes the polynomial vanish on every invertible matrix, a dense open subset, so it is zero. Denote the image by $R_n$. These spaces increase because multiplication by $\Delta$ realizes $R_{n-1}\subset R_n$, and their union is $k[G]^T$. The finite-dimensional image $\Phi(V)$ therefore lies in one $R_n$, and in any sufficiently larger filtration piece.
Choose $q=p^s$ with $q-1$ at least that bound, and set $n=q-1$. The identity $(1+z)^q=1+z^q$ in characteristic $p$ gives
$$
\binom{q-1}{j}=(-1)^j\ne0\quad\text{in }k,\qquad0\le j\le q-1.
$$
The <invariant tensor>
$$
D=\Delta^n=\sum_{j=0}^n(-1)^j\binom nj\,a_0^{n-j}a_1^j b_0^j b_1^{n-j}\in P_n\otimes P_n
$$
therefore defines an invertible equivariant <linear map> $D:P_n^*\to P_n$: its matrix is anti-diagonal with all entries nonzero. Interpret any tensor $B\in P_n\otimes P_n$ as a map $P_n^*\to P_n$ and send it to $BD^{-1}\in\operatorname{End}(P_n)$. Equivariance of $D$ makes this identification intertwine the diagonal tensor action with conjugation on endomorphisms. Crucially, $D$ restricts to the constant function $1$ on $G$, and is sent to the identity endomorphism.
Under the resulting equivariant <linear map> $V\to\operatorname{End}(P_n)$, the vector $v$ is sent to $I$. Since $\dim P_n=n+1=q$, the endomorphism <determinant> is a <homogeneous polynomial> of positive degree $q$, invariant under conjugation. Pull it back to $V$. The resulting polynomial satisfies
$$
\boxed{F\in k[V]^{SL_2(k)},\qquad\deg F=q=p^s>0,\qquad F(v)=\det I=1.}
$$
This constructs the required invariant for every fixed nonzero vector and proves \b[$SL_2(k)$ is geometrically reductive], including <characteristic two>. No averaging by a group order or division by $q$ is used; those operations would fail in the field's characteristic.
Back to article page
