Solvable subgroups, the radical, and the Borel intersection

First suppose $k$ algebraically closed and choose a Borel $B$. The solvable group $H$ has a fixed point $gB$ in the complete quotient $G\mathrm{/}B$ by [F1], so $H\subseteq gB{g}^{-1}$. If $H$ is normal, its product $HB$ is a closed smooth connected subgroup by [F2], and is solvable: its normal subgroup $H$ and quotient, a homomorphic image of $B$, are solvable. Borel maximality among smooth connected solvable subgroups therefore gives $HB=B$, hence $H\subseteq B$. This holds for every Borel. Over arbitrary $k$, apply this argument after algebraic closure; if $H$ is normal and $B$ is defined over $k$, the geometric inclusion $H_{k^{\mathrm{a}}}\subseteq B_{k^{\mathrm{a}}}$ descends to $H\subseteq B$. This proves all parts of (a).

Over algebraically closed $k$, if $B\subseteq P$, the quotient map induces a surjective morphism $G\mathrm{/}B\to G\mathrm{/}P$, and this morphism remains surjective after any field or scheme base change: pull back the cover $G\to G\mathrm{/}P$ to see the fppf quotient locally as projection with fibre $P\mathrm{/}B$. Its complete source makes $G\mathrm{/}P$ universally closed over $k$: for any base change and closed subset of $G\mathrm{/}P$, its inverse image in $G\mathrm{/}B$ is closed and has exactly the same image in the base. As $G\mathrm{/}P$ is separated and finite type by [F4], it is complete. Conversely, if $G\mathrm{/}P$ is complete, a Borel fixes a point by [F1], so a conjugate Borel is contained in $P$. Completeness descends and ascends along field extension, giving the equivalence over arbitrary $k$ with geometric Borel containment.

Continue over algebraically closed $k$ with $B\subseteq P$. The smooth subgroup variety $P_{\mathrm{r}\mathrm{e}\mathrm{d}}$ is connected: for $p\in P(k)$, both $B$ and $pB{p}^{-1}$ are Borels of $(P_{\mathrm{r}\mathrm{e}\mathrm{d}}{)}^{\circ }$, because they are connected and already maximal solvable in $G$. They are conjugate there; multiplying $p$ by the corresponding point of $(P_{\mathrm{r}\mathrm{e}\mathrm{d}}{)}^{\circ }$ puts it in $N_G(B)(k)=B(k)$ by [F3]. Thus every point of $P_{\mathrm{r}\mathrm{e}\mathrm{d}}$ lies in its neutral component and $P$ is connected. The same argument for $n\in N_G(P)(k)$ puts $n$ in $P_{\mathrm{r}\mathrm{e}\mathrm{d}}(k)$, since it conjugates Borels inside $P_{\mathrm{r}\mathrm{e}\mathrm{d}}$.

Over algebraically closed $k$, put $I=({\bigcap }_{B}B{)}_{\mathrm{r}\mathrm{e}\mathrm{d}}^{\circ }$. The full schematic intersection is conjugation-stable, so its reduced neutral component is normal in smooth $G$ by [F2]; $I$ is smooth connected and solvable, since it lies in a Borel. Hence $I\subseteq R(G)$. Conversely step 1.1 puts the smooth connected normal solvable subgroup $R(G)$ in every Borel, so it factors through their intersection and, by smoothness and connectedness, through its reduced neutral component $I$. Therefore $I=R(G)$, proving (b).

This point argument alone would not settle the scheme normalizer, so put $X=G\mathrm{/}P$, smooth by [F4], and choose maximal $T\subseteq B$. Every $T$-fixed coset is represented by $N_G(T)(k)$: if $g^{-1}Tg\subseteq P$, conjugate this torus to $T$ inside $(P_{\mathrm{r}\mathrm{e}\mathrm{d}}{)}^{\circ }$ by [F3]. The connected normalizer of $T$ is $C_G(T)$ by multiplicative-type rigidity, and this Cartan lies in $B\subseteq P$ by [F3]. Consequently there are finitely many $T$-fixed cosets. The smooth scheme $X^T$ is therefore finite étale. The group quotient $N_G(P)\mathrm{/}P$ is a closed subscheme of $X^T$: closedness descends from $N_G(P)\subseteq G$ along $G\to X$, and normalizer points transport $T$ into $P$. It is finite étale and, by step 1.3, has only its identity point, so it is the trivial group scheme. Thus $N_G(P)=P$. The argument after algebraic closure also proves geometric connectedness; equality of subgroup schemes descends to $k$, proving all assertions of (c).