Lie ideals and normal connected subgroups in characteristic zero

For every $k$-algebra $R$ and $x\in G(R)$, $\mathrm{Lie}(c_x)=\mathrm{Ad}(x)$ on ${\mathfrak{g}}_R$, and for $Y\in \mathfrak{g}$ the dual-number point $e^{\epsilon Y}\in G(k[\epsilon ])$ acts by $\mathrm{Ad}(e^{\epsilon Y})=\mathrm{id}+\epsilon \mathrm{ad}(Y)$ on $\mathfrak{g}{\otimes }_{k}k[\epsilon ]$.

The identity component. $H^{\circ }$ is a smooth closed connected subgroup scheme of $G$ with $\mathrm{Lie}(H^{\circ })=\mathfrak{h}$: it is smooth by Cartier's theorem, and the identity component of the smooth group scheme $H$ has the same Lie algebra as $H$.

Part (b). Assume that $H$ is normal in $G$. For $Y\in \mathfrak{g}$ the point $e^{\epsilon Y}\in G(k[\epsilon ])$ normalizes $H_{k[\epsilon ]}$, so by step 1.1 the automorphism $\mathrm{Lie}(c_{e^{\epsilon Y}})=\mathrm{id}+\epsilon \mathrm{ad}(Y)$ of $\mathfrak{g}\otimes k[\epsilon ]$ preserves $\mathfrak{h}\otimes k[\epsilon ]$. Writing $X\in \mathfrak{h}$ as $X$, this says $X+\epsilon [Y,X]\in \mathfrak{h}\otimes k[\epsilon ]$, hence $[Y,X]\in \mathfrak{h}$; as $Y\in \mathfrak{g}$ was arbitrary, $[\mathfrak{g},\mathfrak{h}]\subseteq \mathfrak{h}$.

Part (a), reverse. Assume $[\mathfrak{g},\mathfrak{h}]\subseteq \mathfrak{h}$. By [F2] applied to $(\mathfrak{g},\mathrm{Ad})$, the subspace $\mathfrak{h}$ is $G$-stable. Put $N=N_G(H^{\circ })$, a closed affine subgroup scheme of $G$. The normalizer formula in The Lie functor: exactness, fixed points and generation, part (c), identifies $\mathrm{Lie}(N)$ with the inverse image of $(\mathfrak{g}\mathrm{/}\mathfrak{h}{)}^{H^{\circ }}$. Since $\mathfrak{h}$ is a Lie ideal, the differential action of $\mathrm{Lie}(H^{\circ })=\mathfrak{h}$ on $Q=\mathfrak{g}\mathrm{/}\mathfrak{h}$ is zero. Equip $Q\oplus k$ with the quotient representation and a trivial last summand. For each $v\in Q$, the line $k(v,1)$ is Lie-stable, hence $H^{\circ }$-stable by [F2] applied to the smooth connected characteristic-zero group $H^{\circ }$. For every $R$ and $h\in H^{\circ }(R)$, its last coordinate forces the scalar by which $h$ preserves this line to be $1$, so $h(v,1)=(v,1)$. Thus every $v$ is fixed and the quotient representation is trivial. Consequently $(\mathfrak{g}\mathrm{/}\mathfrak{h}{)}^{H^{\circ }}=\mathfrak{g}\mathrm{/}\mathfrak{h}$, and $\mathrm{Lie}(N)=\mathfrak{g}$. Cartier [F4] makes $N$ smooth; [F3] for the nested inclusion $N\subseteq G$ now gives $N=G$. Therefore $H^{\circ }$ is normal in $G$.

Part (a), forward. If $H^{\circ }$ is normal in $G$, then step 2.1 applied to the normal smooth closed subgroup $H^{\circ }$ with Lie algebra $\mathfrak{h}$ gives $[\mathfrak{g},\mathfrak{h}]\subseteq \mathfrak{h}$.

If $H$ is connected then $H=H^{\circ }$ and steps 3.1 and 2.2 give the equivalence of normality with $\mathfrak{h}$ being an ideal of $\mathfrak{g}$; steps 2.1, 3.1 and 2.2 together prove all three assertions.