TITLE: Lie algebra and base change
QUESTION [5 upvotes]: 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?

REPLY [10 votes]: 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.