# Given the multiplication morphism on the scheme functor, how to get it as scheme morphism?

I am studying group schemes. I would say I understand the definition of an $$S$$-group scheme as to give an $$S$$-scheme together with morphisms for multiplication law, identity, and inverse and also the definition as representable functor from the category $$(\text{Sch/S})$$ to the category ($$\text{Gr})$$ of groups by Yoneda lemma. However, I do not know how to go from one to another in Examples. Here is one:

The additive group $$\mathbb{G}_{a,R}$$:

For a ring $$R$$, let $$S = \text{Spec}(R)$$, and $$G= \mathbb{A}_k^1$$.

For any $$k$$-scheme T, give $$G(T) := h_G(T) = \Gamma(T,O_T)$$ its additive group. This gives a group structure on $$G$$ with multiplication law

$$m : h_G\times h_G \to h_G.$$

I want to write the multiplication law as a morphism of schemes $$m_G: G\times_SG \to G.$$

I know there is construction involved Hopf algebra that gives the answer, but is there a way to obtain it using only Yoneda lemma? Below is my try. Is it right? is there a better way to do it? Is there something else to say?

Yoneda lemma says it should be the map $$m(id_{G \times_S G})$$. We should have $$id_{G \times_S G} \in G\times _sG(G\times _s G) = R[X]\otimes R[X].$$ Then from the fiber product universal property the scheme morphism $$id_{G \times_S G}$$ corresponds to the $$R$$-algebra homomorphism $$(X \mapsto X \otimes 1) \otimes (X \mapsto 1 \otimes X).$$ Then the multiplication law is the scheme morphism that corresponds to the $$R$$-algebra homomorphism \begin{align} \tilde{m} : R[X] & \to R[X] \otimes R[X], \\ X & \mapsto X \otimes 1 + 1 \otimes X. \end{align}