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I am very confused by the following and would appreciate any help.

Let $\mu_p \subset \mathbb{G}_m$ be the $p$-torsion subgroup scheme of the multiplicative group over $\mathbb{Z}_p$. I would like to compute the Lie algebra of $\mu_p$ (at the identity section) to make sure that I understand Lie algebras well. I have heard that "the formation of the Lie algebra at the identity section of a group scheme commutes with base change" and I know that the Lie algebra of $\mu_p$ has to be killed by $p$, so by looking at the residue field I get that this Lie algebra is $\mathbb{Z}/p\mathbb{Z}$.

On the other hand, from the $\epsilon$-points definition, I get that the Lie algebra of $\mu_p$ should be a subfunctor of the Lie algebra of $\mathbb{G}_m$. The Lie algebra of $\mathbb{G}_m$ is just $\mathbb{Z}_p$ because $\mathbb{G}_m$ is smooth of relative dimension $1$. But $\mathbb{Z}/p\mathbb{Z}$ cannot sit inside $\mathbb{Z}_p$, so where have I made a mistake? Does the $\epsilon$-points definition maybe not apply for some reason?

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  • $\begingroup$ Well, what happens depends on the kind of object you want the Lie algebra to be. Most of your question is written for `naive' object, where you expect the Lie algebra to be a $\mathbb{Z}_p$-module. In this setup, the formation of Lie algebra does not commute with base change (unless the group is smooth). $\endgroup$
    – t3suji
    Commented Oct 12, 2015 at 1:02
  • $\begingroup$ On the other hand, if you want your Lie algebra to be a scheme over $\mathbb Z_p$, it is supposed to commute with the base change, but there is no contradiction: the Lie algebra of $\mathbb{G}m$ is the constant scheme with one-dimensional fiber, and it contains a non-flat subscheme sitting over the closed point. $\endgroup$
    – t3suji
    Commented Oct 12, 2015 at 1:04
  • $\begingroup$ Thank you, t3suji, this is very helpful. Could you comment on what is the relationship between the two: is the $\mathbb{Z}_p$-module just the global sections of the scheme version? I am seeing in SGA 3, Expose II, 3.3 that the scheme version should maybe be $\mathbb{V}(e^*(\Omega^1_{\mu_p/\mathbb{Z}_p}))$, but I guess the $\mathbb{V}$ functor is parametrizing global sections of the dual of the coherent sheaf $e^*(\Omega^1_{\mu_p/\mathbb{Z}_p})$, so this must be the source of the bad behavior of global sections with respect to base change beyond the smooth case? $\endgroup$
    – Lisa S.
    Commented Oct 12, 2015 at 1:53

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The quotation you have heard is false because over an affine base $S = {\rm{Spec}}(k)$ for a commutative ring $k$ and a $k$-group scheme $G$, the Lie algebra is the linear dual ${\rm{Hom}}_k(e^{\ast}(\Omega^1_{G/k}),k)$ and that generally does not commute with non-flat base change when $G$ is not $k$-smooth. You have already noticed yourself that this is false for $\mu_p$ over $\mathbf{Z}_p$.

There is always a "base change morphism" ${\rm{Lie}}(G) \otimes_{k} k' \rightarrow {\rm{Lie}}(G_{k'})$ for a $k$-algebra $k'$, but one cannot say more when $k'$ is not $k$-flat or $G$ is not $k$-smooth. Read section A.7 (through A.7.6) for a discussion of Lie algebras of group schemes locally of finite type over rings in the book "Pseudo-reductive groups" (2nd edition) for a discussion of several of the key points from SGA3 (including that ${\rm{Lie}}(G)$ admits a natural Lie bracket compatibly with the base change morphism, even when $G$ isn't $k$-smooth).

The functor $k' \rightsquigarrow {\rm{Lie}}(G_{k'})$ on $k$-algebras is generally not representable if $e^{\ast}(\Omega^1_{G/k})$ is not a vector bundle over $k$, as usually fails if $G$ is not $k$-smooth and $k$ is not a field.

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  • $\begingroup$ Thank you. I am confused by your last sentence: isn't $\mathbb{V}(e^*(\Omega^1_{G/k}))$ the scheme that is representing the functor $k' \mapsto \mathrm{Lie}_{G_{k'}} = \mathrm{Hom}_{k'}((e')^*(\Omega^1_{G/k'}), k')$ (by SGA 3, Expose I, 4.6.3.1)? $\endgroup$
    – Lisa S.
    Commented Oct 12, 2015 at 2:43
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    $\begingroup$ Oops, you are correct: it is indeed represented by Spec of the symmetric algebra of $e^{\ast}\Omega^1_{G/k}$. What fails when the latter isn't a vector bundle is that this representing module scheme is determined by its $k$-module of $k$-points (usually it is not). $\endgroup$
    – nfdc23
    Commented Oct 12, 2015 at 3:05

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