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Let $G$ be a unipotent connected linear algebraic group over a field $F$. Then $G$ is called split if there is a series of closed subgroup schemes $1 = G_0 \subset G_1 \subset \cdots \subset G_t =G$ with each $G_i$ normal in $G_{i+1}$ and $G_i/G_{i+1} \cong_F \mathbb G_a$.

I have read that for $F$ perfect, every such $G$ is split, provided it is connected. And $F$ of characteristic zero, every such $G$ is moreover automatically connected.

I was trying to verify this in the case of $G = \operatorname{Res}_{E/F} \mathbb G_a$, where $E/F$ is a quadratic extension. Does $G$ really have such a composition series?

I tried considering the diagonal embedding $\Delta$ of $\mathbb G_a$ into $G$ and looking at the quotient $G/\Delta$. It should be the case that $G/\Delta \cong \mathbb G_a$, but I'm not able to verify this. The natural map $G/\Delta \rightarrow \mathbb G_a$ given by $(x,y)\Delta \mapsto x- y $ is not defined over $F$.

D_S
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2 Answers2

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It's not that complicated. I believe you have all the needed knowledge, but just have to unpack the definitions.

By fixing a basis of $E/F$, we know that $E$ is isomorphic to $F^2$ as $F$-vector spaces, hence for any $F$-algebra $R$, one has $\mathbb{G}_a(E\otimes_F R) = E\otimes_F R \simeq R^2 = \mathbb{G}_a^2(F)$, where the middle isomorphism is functorial in $R$.

Hence by the definition of restriction of scalars, we see that $\operatorname{Res}_{E/F}\mathbb{G}_a \simeq \mathbb{G}_a^2$.

WhatsUp
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  • I don't think this answers what I want. The isomorphism $\operatorname{Res}_{E/F} \mathbb G_a \cong \mathbb G_a^2$ is over $E$, not over $F$. – D_S Oct 22 '19 at 21:10
  • The isomorphism is over $F$: both sides are algebraic groups over $F$. You might want to clarify your definition of restriction of scalars. – WhatsUp Oct 22 '19 at 21:12
  • Both are algebraic groups over $F$, but $\operatorname{Res}_{E/F} \mathbb G_a$ is definitely not isomorphic to $\mathbb G_a^2$ as groups over $F$. They have the same $\overline{F}$-points but their Galois actions are totally different. – D_S Oct 22 '19 at 21:14
  • The Galois action is not relevant here. The isomorphism $E\otimes_F R\simeq R^2$ is a natural transform from the functor $R\mapsto \operatorname{Res}_{E/F}\mathbb{G}_a(R)$ to the functor $R\mapsto \mathbb{G}_a^2(R)$, hence gives an isomorphism between the two algebraic groups, which represent the two functors respectively. Maybe you should state explicitly your definition of restriction of scalars? – WhatsUp Oct 22 '19 at 21:20
  • It is the unique group scheme over $F$ defined on $\overline{F}$ points as $\overline{F} \times \overline{F}$, and whose Galois action is given by $$\sigma.(x,y) = \begin{cases} (\sigma(x),\sigma(y)) & \textrm{ if } \sigma|_E = 1 \ (\sigma(y),\sigma(x)) & \textrm{ if } \sigma|_E \neq 1 \end{cases}$$Can equivalently be given by a one-cocycle in $H^1(\operatorname{Gal}(\overline{F}/F), \mathbb G_a)$. I see what you're saying with the Yoneda lemma but that would seem to say that that one-cocycle is equivalent to the trivial one which seems weird to me. – D_S Oct 22 '19 at 21:37
  • I should say $H^1(\operatorname{Gal}(\overline{F}/F), \operatorname{Aut}_{\overline{F}}(\mathbb G_a))$ – D_S Oct 22 '19 at 21:45
  • I see. You have a somehow unusual definition of restriction of scalars. The $\operatorname{Aut}(\mathbb{G}_a)$ is, in fact, $\overline{F}^\times$. Both $\overline{F}^\times$ and $\overline{F}$ have trivial $H^1$: it's Hilbert 90. – WhatsUp Oct 22 '19 at 21:50
  • @D_S Maybe this will clear up your confusion. Showing that these two schemes are isomorphic (which is really where your confusion lies, I think--the Galois action confusion is seem at the scheme level) is the same thing as showing that two $E$-algebras with a semilinear action of $\mathrm{Gal}(E/F)={\sigma,\text{id}}$ are isomorphic as $E$-algebras in a $\mathrm{Gal}(E/F)$-equivariant way. What are these two algebras? The first is $E[x,y]$ with $\sigma$ acting by $x\mapsto x$ and $y\mapsto y$. The second is $E[x,y]$ with $x\mapsto y$ and $y\mapsto x$. – Alex Youcis Oct 23 '19 at 00:32
  • These correspond to your idea of what $(\mathbb{G}{a,F}^2)_E$ and $(\mathrm{Res}{E/F}\mathbb{G}_{a,E})_E$ with their Galois actions correspond to. Let $E=F[\alpha]$. Then an isomorphism is given by $x\mapsto x+\alpha y$ and $y\mapsto x-\alpha y$. This is essentially the picking the basis argument that @WhatsUp made. Note that it also shows why this cannot work for something like $\mathbb{G}_m$ whose Weil restriction is not a split torus. Namely there we have $E[x,y,x^{-1},y^{-1}]$ with the two obvious actions and the above map makes no sense since $x+\alpha y$ isn't invertible. – Alex Youcis Oct 23 '19 at 00:35
  • PS, you should have a cocycle in $H^1(\mathrm{Gal}(\ov{F}/F),\mathrm{Aut}{\ov{F}}(\mathbb{G}{a,\ov{F}}^2))$ since the Weil restriction is a twist of $\mathbb{G}_{a,F}^2$. This is Aut group i just $\GL_2(\ov{F})$ which also has trivial cohomology by non-abelian Hilbert's theorem 90. In fact in characteristic $0$ every abelian unipotent group is just a power of $\mathbb{G}_a$ (see the relevant section of Milne's book on algebraic groups). – Alex Youcis Oct 23 '19 at 00:36
  • Makes sense, thanks for your help. – D_S Oct 23 '19 at 03:46
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Tricky business if you were using the definition of $G = \operatorname{Res}_{E/F}(\mathbb G_a)$ as a form of $\mathbb G_a \times \mathbb G_a$ and you didn't recollect Hilbert's Theorem 90...

Worse, I actually did my PhD thesis on the restriction of scalars of a certain group and I didn't realize that $\operatorname{Res}_{E/F}(\mathbb G_a)$ is the same algebraic group as $\mathbb G_a \times \mathbb G_a$. It's obvious from the Yoneda lemma, but less obvious from the way I was thinking about restriction of scalars.

Here I'm taking $G$ to be the group given on $\overline{F}$-points by $G(\overline{F}) = \overline{F} \times \overline{F}$, with the Galois action

$$\sigma.(x,y) = \begin{cases} (\sigma(x),\sigma(y)) & \textrm{ if } \sigma|_E =1 \\ (\sigma(y),\sigma(x)) & \textrm{ if } \sigma|_E \neq 1 \end{cases}$$ for $\sigma \in \operatorname{Gal}(\overline{F}/F)$. Now choose $\beta \in \overline{F}$ such that $E = F(\sqrt{\beta})$, and define

$$\phi: G \rightarrow \mathbb G_a \times \mathbb G_a$$

$$\phi(x,y) = (x+y, \sqrt{\beta}(y-x))$$

This can be checked to be defined over $F$.

D_S
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  • I guess that your map $\phi$ is essentially the one I wrote above in the comments--they are almost inverses (their composition differs by the map $(x,y)\mapsto (2x,2\beta y)$. – Alex Youcis Oct 23 '19 at 10:43