Let $k$ be a field of positive characteristic and let $G$ be a connected semisimple algebraic group over $k$ with fundamental group $\mu$. Note that $\mu$ can be nonsmooth. It is stated in Sansuc's 1981 Crelle paper Groupe de Brauer et arithmétique des groupes algébriques..., Lemma 6.9(iii), p.41, that ${\rm Pic}\, G=\mu^{*}(k)$ is the group of $k$characters of $\mu$. Sansuc refers the reader to the FossumIversen paper On Picard groups of algebraic fiber spaces, Proposition 4.6, for this. Now, it is not clear to me that the latter paper really works at the required level of generality. To confuse matters even more (for me), Raynaud, in his thesis (Lecture Notes in Math. 119), discusses in particular the canonical map $\mu^{*}(k)\to {\rm Pic}\, G$ (Proposition VII.1.5, p.106). By examining Raynaud's proof, all I can see is that the latter map is an injection whose cokernel is a torsion group annihilated by some power of the characteristic of $k$. So, my question is: has it really been proved in the literature that Sansuc's statement ${\rm Pic}\, G=\mu^{*}(k)$ is true in positive characteristic even when $\mu$ is not smooth?

1$\begingroup$ Minor comment: what you are calling "fundamental group" is not generally (or should not generally) go by that name; it is the finite etale Cartier dual $\mu^*$ that is called the fundamental group (with $\mu$ the kernel of the central isogeny from the simply connected central cover $\widetilde{G} \rightarrow G$). For example ${\rm{PGL}}_n$ has fundamental group $\mathbf{Z}/n\mathbf{Z}$ (not $\mu_n$). $\endgroup$– nfdc23Jul 5 '17 at 21:27

1$\begingroup$ I was under the impression that the kernel of the central isogeny you allude to is what is usually called the fundamental group. Anyway, we can change names but the problem remains... $\endgroup$– Cristian D. GonzalezAvilesJul 5 '17 at 22:17

2$\begingroup$ For covariance reasons you're right about this terminology (and anyway, as you note, this has nothing to do with the real content in your question). Also, you are correct that section 4 of FossumIversen does not operate at the correct level of generality (they are tacitly assuming in their proofs that the ground field is algebraically closed). For example, their Corollary 4.4 that Pic($G$) is finite for smooth connected affine $k$groups $G$ is false in the 1dimensional unipotent case over every imperfect field. I'll check with someone I know who has done much work near this recently. $\endgroup$– nfdc23Jul 6 '17 at 1:10

$\begingroup$ Many thanks nfdc23. I think it is important that we try to fill this apparent gap in the literature. $\endgroup$– Cristian D. GonzalezAvilesJul 6 '17 at 14:29
I'm not sure if there is really a literature gap, or maybe it is addressed in some reference that just hasn't come to mind. Anyway, let $G$ be a perfect smooth connected affine group over a field $k$, so $G_K$ has no nontrivial $K$homomorphism to ${\rm{GL}}_1$ for any extension field $K/k$, and hence $K[G]^{\times} = K^{\times}$ by Rosenlicht's Unit Theorem. If $L'$ is a line bundle on ${\rm{Pic}}(G_{k'})$ for a finite Galois extension $k'/k$ such that the isomorphism class of $L'$ is ${\rm{Gal}}(k'/k)$invariant, it therefore follows via Hilbert 90 that the isomorphisms to the Galoistwists can be arranged to satisfy the cocycle condition, so the natural map $${\rm{Pic}}(G) \rightarrow {\rm{Pic}}(G_{k'})^{{\rm{Gal}}(k'/k)}$$ is surjective, and hence by passing to the direct limit on such $k'/k$ we see that the natural map $${\rm{Pic}}(G) \rightarrow {\rm{Pic}}(G_{k_s})^{{\rm{Gal}}(k_s/k)}$$ is surjective.
But this latter map is also injective. Indeed, again by limit considerations it suffices to show if $L$ is a line bundle on $G$ such that $L_{k'}$ admits a nowherevanishing global section $s'$ over $G_{k'}$ for a finite Galois extension $k'/k$ then $L$ admits a nowherevanishing global section. For each $\gamma \in {\rm{Gal}}(k'/k)$, the canonical $\gamma$equivariant action on $L_{k'}(G_{k'}) = k' \otimes_k L(G)$ satisfies $\gamma(s') = c_{\gamma} s'$ for a unique $c_{\gamma} \in k'[G]^{\times} = {k'}^{\times}$, and $\gamma \mapsto c_{\gamma}$ is a 1cocycle. Thus, by Hilbert 90 there exists $u' \in {k'}^{\times}$ such that $c_{\gamma} = \gamma(u')/u'$ for all $\gamma$, so $s'/u'$ is a nowherevanishing global section that is Galoisinvariant and hence corresponds to a trivialization of $L$ over $G$.
Now assume $G$ is a connected semisimple $k$group. Letting $\widetilde{G} \rightarrow G$ be the simply connected central cover over $k$, with central kernel denoted $\mu$, we have ${\rm{Pic}}(\widetilde{G}) = 1$ by the above injectivity applied to $\widetilde{G}$ because ${\rm{Pic}}(\widetilde{G}_{k_s})=1$. To see this latter triviality, note that for any split connected reductive group $H$ over a field $F$, the "open cell" relative to a choice of split maximal $F$torus and Borel $F$subgroup containing it provides a dense open subscheme that is $F$isomorphic to a direct product of copies of $\mathbf{A}^1_F$ and $\mathbf{A}^1_F  \{0\}$. Using such an open subscheme enables one to show that the group ${\rm{Pic}}(H)$ is identified with the group of central extensions $E$ of $H$ by ${\rm{GL}}_1$ as $F$groups by adapting the proof of the WeilBarsotti formula describing the dual abelian variety in terms of extensions by ${\rm{GL}}_1$  see section 4 in the paper "Local Properties of Algebraic Group Actions" by Knop, Kraft, Luna, and Vust for that argument (presented in the setting over algebraically closed fields, but via completely general methods that work for a split connected reductive group over any field). If the $F$split $H$ is connected semisimple and simply connected then there are no such nontrivial extensions $E$ since necessarily $E$ is connected reductive and hence $\mathscr{D}(E) \rightarrow H$ is a central isogeny (so an isomorphism when $H$ is simply connected). This shows ${\rm{Pic}}(H)=1$ in such cases, so ${\rm{Pic}}(\widetilde{G}_{k_s})=1$ as claimed (and from the consequence ${\rm{Pic}}(\widetilde{G})=1$ we obtain the nonobvious fact that $k[\widetilde{G}]$ is a UFD).
The result VII 1.5 that you refer to in Raynaud's thesis provides an injective homomorphism $\mu^*(k_s) \rightarrow {\rm{Pic}}(G_{k_s})$ (this map assigns to any $k_s$homomorphism $\chi:\mu_{k_s} \rightarrow {\rm{GL}}_1$ the $\chi$pushout of the central extension $1 \rightarrow \mu_{k_s} \rightarrow \widetilde{G}_{k_s} \rightarrow G_{k_s} \rightarrow 1$), and it is Galoisequivariant by design. Hence, as long as this is an equality, which is purely a counting question (due to its injectivity), it follows by the preceding arguments that passage to ${\rm{Gal}}(k_s/k)$invariants gives that the natural map $\mu^*(k) \rightarrow {\rm{Pic}}(G)$ is an equality. In this way, we see that the result asserted without proof in Remark VII 1.7(a) of Raynaud's thesis (basically the affirmative answer to the question posed in the split case) gives the desired result over $k_s$ and hence over $k$. This doesn't answer the question of providing a literature reference with a complete proof, but it "confirms" that a complete proof should have been known to experts in Paris since the 1960's.

$\begingroup$ Many thanks @nfdc23! It seems to me that there exists no published proof of the "triviality" that Pic of a connected, semisimple and simplyconnected algebraic group over a separably closed field is zero. All the references I'm aware of address this problem over an algebraically closed field (Iversen, Popov, etc). So I think some version of your answer above ought to appear in print somewhere for ease of reference. Can I include it in one of my papers? (with due credit to you, of course) $\endgroup$ Jul 7 '17 at 14:55

$\begingroup$ For that triviality over a general field, I think it is fine to say "we give a proof due to lack of a convenient reference" and not go give anyone credit for it, since this has been known for a long time and I'm sure the argument has been rediscovered many times (I only tried to prove it after hearing the result from someone else who didn't know a reference). $\endgroup$– nfdc23Jul 7 '17 at 17:04

$\begingroup$ Dear @nfdc23, very good! I'll include your proof in one of my papers and refer my readers to it as often as I need to. BTW, if you look into mathoverflow.net/questions/96987, you'll see that quite a few people "know" the result is true over any field, but nobody can point to an adequate reference. $\endgroup$ Jul 7 '17 at 18:29

$\begingroup$ @CristianD.GonzalezAviles: the discussion in that link looks entirely focused on the case of an alg. closed field aside from one comment by Jim Humphreys (and nobody else commenting there seems to be claiming to "know" that fact). Anyway, this is all a moot issue now. $\endgroup$– nfdc23Jul 7 '17 at 23:37