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?



Lie algebra and base change
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. 


