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.
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.