How simple does a $\mathbb{Q}$-simple group remain after base change to $\mathbb{Q}_{\ell}$?

Question

Of course the general answer to the question in the title is: not very simple. I could not think of a better title, so let me explain my question in more detail.

I have a number field $E/\mathbb{Q}$, and a simple group $H/E$. Let $G$ denote the Weil restriction of scalars of $H$ from $E$ to $\mathbb{Q}$. Then $G$ is a $\mathbb{Q}$-simple group. The group $G_{\ell} := G_{\mathbb{Q}_{\ell}}$ is not $\mathbb{Q}_{\ell}$-simple in general.

Short aside: However, if $n = [E : \mathbb{Q}] \le 3$ there is always an $\ell$ such that $G_{\ell}$ is $\mathbb{Q}_{\ell}$-simple. (For a short proof: consider the Galois action of $\mathrm{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ on the $n$ factors of $G_{\bar{\mathbb{Q}}}$. This action is transitive. Since $n \le 3$, there must therefore be an $n$-cycle in the action, and by Chebotarev there will be (infinitely many) primes $\ell$, such that $\mathrm{Gal}(\bar{\mathbb{Q}_{\ell}}/\mathbb{Q}_{\ell})$ acts transitively on the factors of $G_{\bar{\mathbb{Q}_{\ell}}}$. [Proof copied from D. Lombardo, arXiv: 1402.1478])

My question is about what happens if $n > 3$. (In my particular case $n \le 10$.) For example, I would be very happy, if there is always a prime $\ell$, so that all the $\mathbb{Q}_{\ell}$-simple factors of $G_{\ell}$ are not absolutely simple. Is something like that true?

Answer 1

I assume that the question is: Is it true that there is always a prime $\ell$ such that $G_\ell$ has no absolutely simple direct factors? The answer is YES. There are infinitely many such $\ell$.

Claim. Let $\Gamma$ be a finite group acting transitively on a finite set $S$ of cardinality $n>1$. Then $\Gamma$ has a cyclic subgroup $\Delta\subset \Gamma$ acting on $S$ without fixed points.

Proof of the claim (due to Guntram). Assume the contrary. Then any element $\gamma$ of $\Gamma$ is contained in the stabilizer $\Gamma_s$ of some point $s\in S$, hence $\Gamma=\bigcup_s\Gamma_s$. Since $\Gamma$ acts transitively on $S$, each subgroup $\Gamma_s$ is of index $n$ in $\Gamma$. Since $n>1$ and $1\in \Gamma_s$ for all $s\in S$, we obtain that $$ \#\bigcup_s\Gamma_s\le 1+ n\cdot (\#\Gamma_s-1)<\#\Gamma, $$ contradiction.

Now let $S$ denote the set of embeddings of $E$ into $\overline{\mathbb{Q}}$. The Galois group of $\overline{\mathbb{Q}}$ over ${\mathbb{Q}}$ acts on $S$ via a finite quotient group $\Gamma$. Let $\Delta=\langle\gamma\rangle$ be as in the Claim. Then by Chebotarev's density theorem there exist infinitely many primes $\ell$ unramified in $E$ such that the corresponding Frobenius is conjugate to $\gamma$. For such an $\ell$, the tensor product $E\otimes_{\mathbb{Q}} {\mathbb{Q}}_\ell$ does not contain ${\mathbb{Q}}_\ell$ as a direct summand, and therefore, $G_\ell$ has no absolutely simple direct factors.

Answer 2

For any finite separable extension of fields $k'/k$ and geometrically connected smooth affine $k'$-scheme $X'$, the affine finite type Weil restriction $X := {\rm{R}}_{k'/k}(X')$ over $k$ is smooth and geometrically connected since $$X_{k_s} = {\rm{R}}_{(k' \otimes_k k_s)/k_s}(X' \otimes_{k'} (k' \otimes_k k_s)) = \prod_{\sigma} (X' \otimes_{k',\sigma} k_s)$$ where $\sigma$ varies through the set of $k$-embeddings of $k'$ into $k_s$ (with each $X' \otimes_{k',\sigma} k_s$ geometrically connected over $k_s$, so their direct product is also geometrically connected). Here we used (i) the product decomposition of $k' \otimes_k k_s$ into a direct product of copies of $k_s$ (so ${\rm{Spec}}(k' \otimes_k k_s) = \coprod_{\sigma} {\rm{Spec}}(k_s)$, (ii) the compatibility of Weil restriction with any base change, (iii) the natural identification ${\rm{R}}_{(\coprod S'_i)/S}(\coprod X'_i) = \prod {\rm{R}}_{S'_i/S}(X'_i)$ for any finite collection of finite locally free maps $S'_i \rightarrow S$ and affine $\coprod S'_i$-scheme $X'$ of finite type.

Now consider when $G' = X'$ is a connected semisimple $k'$-group. Note that $G'$ is geometrically connected over $k'$ since every connected $k'$-scheme of finite type with a rational point is geometrically connected. Thus, $G := {\rm{R}}_{k'/k}(G')$ is a connected semisimple $k$-group since $G_{k_s} = \prod_{\sigma} (G' \otimes_{k',\sigma} k_s)$ is clearly connected semisimple.

Suppose $k$ is a global field and $v$ is a place of $k$, so $$G_{k_v} = {\rm{R}}_{(k' \otimes_k k_v)/k_v}(G' \otimes_{k'} (k' \otimes_k k_v)) = \prod_{v'|v} {\rm{R}}_{k'_{v'}/k_v}(G'_{k_{v'}})$$ where $v'$ varies through the set of places of $k'$ over $v$ (as $k' \otimes_k k_v = \prod_{v'|v} k'_{v'}$). To understand if the direct factor for each $v$ is $k_v$-simple (and then whether those are absolutely $k_v$-simple or not), there is no harm in replacing $G'$ with its simply connected central cover $\widetilde{G'}$ because Weil restriction through a finite etale cover carries isogenies to isogenies (easy, and false if "etale" is relaxed to "flat").

Hence, now we may and do assume $G'$ is simply connected. Then $G' = {\rm{R}}_{E'/k'}(H')$ for a finite separable extension $E'/k'$ and a connected semisimple $E'$-group $H'$ that is simply connected and absolutely simple over $E'$ (see 6.21(ii) in the Borel-Tits paper on reductive groups, or Prop. A.5.14 in the book "Pseudo-reductive groups"), so $$G = \prod_{w'} {\rm{R}}_{E'_{w'}/k_{v}}(H'_{E'_{w'}})$$ where $w'$ varies through the places of $E'$ over $v$. Moreover, for each $w'$ the corresponding direct factor is $k_v$-simple (see the end of Prop. A.5.14 mentioned above).

To summarize, your question is when ${\rm{R}}_{E'_{w'}/k_v}(H'_{E'_{w'}})$ is absolutely $k_v$-simple (or not). The scalar extension of this $k_v$-simple group to a separable closure $k_{v,s}$ of $k_v$ is equal to the direct product of the scalar extensions of $H'$ to the factor fields of $E'_{w'} \otimes_{k_v} k_{v,s}$ (each such factor field naturally isomorphic to $k_{v,s}$), so absolute $k_v$-simplicity is precisely the condition that there is only one such factor field, which is to say that $E'_{w'} = k_v$.

In other words, the $k_v$-simple factors of $G_{k_v}$ are absolutely simple over $k_v$ if and only if $v$ is completely split in $E'$. So these all fail to be absolutely simple (is that what you really want? Or just one to fail?) precisely when there is no place $w'$ of $E'$ over $v$ that is split over $v$ (i.e., $E_{w'} = k_v$). For example, if $k'/k$ is a nontrivial Galois extension then any $v$ not totally split in $k'$ does the job. You can now deal with the various other cases, depending on the arithmetic of $E'/k'/k$, on your own, presumably guided by whatever is the actual motivation (which was not given; it is puzzling why you are specifically interested in the case $n \le 10$, about which there doesn't seem to be anything special).