Why does the multiplication map on an algebraic group induce the Lie algebra addition? Let $G$ an group scheme (of finite type) over a field $k$ with identity $e$ and multiplication $\mu$. As usual, we can think of the Zariski tangent space $T_e(G)$ either as $(m_x/m_x^2)^\star$ or as $\operatorname{ker}(G(k[\epsilon]) \to G(k))$.
It is claimed various places that
$$
    \mu(k[\epsilon])|_{T_e(G)} : T_e(G) \times T_e(G) = T_{(e,e)}(G \times G) \to T_e(G)
$$
coincides with the usual vector addition map on the Zariski tangent space (addition of functionals). Can anyone prove this, or provide a reference?
My thoughts:
Of course, the functorial description of the tangent space corresponds exactly to $k$-homomorphisms $v_1, v_2 : \mathcal{O}_{G, e} \to k[\epsilon]$, and we restrict these to the maximal ideal to obtain functionals which we can add pointwise.
I tried to walk through how to add two such homomorphisms using the other process, but run into difficulty. Since $(G \times G)(k[\epsilon]) = G(k[\epsilon]) \times G(k[\epsilon])$, it must be possible to combine the two maps to obtain a map
$$
   \mathcal{O}_{G \times G, (e,e)} \to k[\epsilon].
$$
First, it is not so clear to me how to obtain such a map, since the natural map
$$
   \mathcal{O}_{G, e} \otimes \mathcal{O}_{G,e} \to \mathcal{O}_{G \times G, (e, e)}
$$
is not an isomorphism. It is a localization, I believe, so one might hope to descend $v_1 \otimes v_2$ using the universal property, but the structure of this localization is not so clear to me.
Second, even once we have combined these to obtain a map $\mathcal{O}_{G \times G, (e, e)} \to k[\epsilon]$, we still need to precompose with the black-box map $\mu^\# : \mathcal{O}_{G, e} \to \mathcal{O}_{G \times G, (e, e)}$. This whole process is then, on the level of maximal ideals, supposed to recover simple addition! Maybe the approach of working with the local ring maps is not a useful one, but I don't see how else to pass between the two perspectives.
 A: For a $k$-algebra homomorphism $f : A \to B$ and an element $x$ of $X (A)$, where $X$ is a scheme, I will write $f \cdot x$ for the corresponding element of $X (B)$.
Let $r : k [\epsilon] \to k$ be the $k$-algebra homomorphism given by $r (s + t \epsilon) = s$.
When $X$ is a scheme, given $a$ and $b$ in $X (k [\epsilon])$ such that $r \cdot a = r \cdot b$, there is a unique element $\lambda (a, b)$ of $X (k [\epsilon] \times_k k [\epsilon])$ such that $p_1 \cdot \lambda (a, b) = a$ and $p_2 \cdot \lambda (a, b) = b$, where $p_1, p_2 : k [\epsilon] \times_k k [\epsilon] \to k [\epsilon]$ are the two projections.
(That is, $\operatorname{Spec} (k [\epsilon] \times_k k [\epsilon])$ has the universal property of the pushout of $\operatorname{Spec} r : \operatorname{Spec} k \to \operatorname{Spec} k [\epsilon]$ with itself.
This is basically immediate if $X$ is affine, and the general case follows because $\operatorname{Spec} (k [\epsilon] \times_k k [\epsilon])$ is topologically just a point and so factors through an affine open subscheme of $X$.)
The sum of $a$ and $b$ considered as tangent vectors of $X$ is then $v \cdot \lambda (a, b)$, where $v : k [\epsilon] \times_k k [\epsilon] \to k [\epsilon]$ is the $k$-algebra homomorphism given by $v (s + t \epsilon, s + u \epsilon) = s + (t + u) \epsilon$.
Now, suppose $G$ is a group scheme and $a$ and $b$ are elements of $G (k [\epsilon])$ such that $r \cdot a = r \cdot b = e$ is the unit of $G (k)$.
We want to show that $\mu (a, b) = v \cdot \lambda (a, b)$.
Let $c_1, c_2 : k [\epsilon] \to k [\epsilon] \times_k k [\epsilon]$ be the $k$-algebra homomorphisms given by $c_1 (s + t \epsilon) = (s + t \epsilon, s)$ and $c_2 (s + t \epsilon) = (s, s + t \epsilon)$.
Let $i : k \to k [\epsilon]$ be the unique $k$-algebra homomorphism.
We have:
\begin{align*}
v \circ c_1 & = \textrm{id}_{k [\epsilon]} &
p_1 \circ c_1 & = \textrm{id}_{k [\epsilon]} &
p_2 \circ c_1 & = i \circ r
\\
v \circ c_2 & = \textrm{id}_{k [\epsilon]} &
p_1 \circ c_2 & = i \circ r &
p_2 \circ c_2 & = \textrm{id}_{k [\epsilon]}
\end{align*}
Hence:
\begin{align*}
p_1 \cdot \mu (c_1 \cdot a, c_2 \cdot b) = \mu (p_1 \cdot c_1 \cdot a, p_1 \cdot c_2 \cdot b) & = \mu (a, i \cdot e) = a \\
p_2 \cdot \mu (c_1 \cdot a, c_2 \cdot b) = \mu (p_2 \cdot c_1 \cdot a, p_2 \cdot c_2 \cdot b) & = \mu (i \cdot e, b) = b
\end{align*}
That is, $\mu (c_1 \cdot a, c_2 \cdot b) = \lambda (a, b)$.
Therefore:
$$v \cdot \lambda (a, b) = v  \cdot \mu (c_1 \cdot a, c_2 \cdot b) = \mu (v \cdot c_1 \cdot a, v \cdot c_2 \cdot b) = \mu (a, b)
\tag*{$\blacksquare$}$$
I find it easier to understand the above argument when it is recast geometrically.
Write $D$ for $\operatorname{Spec} (k [\epsilon])$ and $D \vee D$ for $\operatorname{Spec} (k [\epsilon] \times_k k [\epsilon])$.
We can think of $D$ as a point with one tangent vector and $D \vee D$ as a point with two tangent vectors.
(Not to be confused with $D \times D$, which is a point with a tangent plane.)
Write $1$ for $\operatorname{Spec} k$.
We are given
$$\require{AMScd}
\begin{CD}
1 @>>> D \\
@VVV @VV{b}V \\
D @>>{a}> G
\end{CD}$$
with the diagonal $1 \to G$ being $e$.
The universal property of $D \vee D$ gives us a morphism $\lambda (a, b) : D \vee D \to G$.
The argument then comes down to contemplating this diagram:
$$\begin{CD}
D @>{\operatorname{Spec} v}>> D \vee D @>{\lambda (a, b)}>> G \\
@| @V{(c_1 \cdot a, c_2 \cdot b)}VV @| \\
D @>>{(a, b)}> G \times G @>>{\mu}> G
\end{CD}$$
The goal is to show that the diagram commutes.
The left half commutes by definition of $c_1, c_2$ and $v$.
For the commutativity of the right half, by the universal property of $D \vee D$, it suffices to show that these two diagrams commute:
$$\begin{CD}
D @>{a}>> G \\
@V{(a, i \cdot e)}VV @| \\
G \times G @>>{\mu}> G
\end{CD}
\hspace{4em}
\begin{CD}
D @>{a}>> G \\
@V{(a, i \cdot e)}VV @| \\
G \times G @>>{\mu}> G
\end{CD}$$
But this is just the unitality of $\mu$, so we are done.
Even more abstractly, we can see this is as an instance of the Eckmann–Hilton argument: $D$ is a cogroup object in the category of pointed $k$-schemes, and $G$ remains a group object in this category, so the set of pointed morphisms $D \to G$ has two mutually compatible group structures: one induced by $D$ and one induced by $G$; but mutual compatibility implies they coincide, which is what we wanted to show.
