Question: Why they are all assume it trivial, but I cannot make it commute the diagram in the first axiom of being, linear algebraic group (for $A=k[s]$).
Definition 3.48.(ref.Mukai An Introduction to Invariants and Moduli (Cambridge Studies in Advanced Mathematics) Let $A$ be a finitely generated $k$-algebra. Then $G=\mathrm{Spm} A$ is called an affine algebraic group if there exist $k$-algebra homomorphisms $$ \begin{aligned} & \mu: A \rightarrow A \otimes_k A \text { (coproduct) } \\ & \epsilon: A \rightarrow k \quad \text { (coidentity) } \\ & \iota: A \rightarrow A \quad \text { (coinverse) } \\ & \end{aligned} $$ satisfying the following three conditions. (i) The following diagram commutes:
$\require{AMScd}$ \begin{CD} A @>{\mu}>> A \otimes_k A\\ @V\mu VV @VV1_A\otimes\mu V\\ A \otimes_k A @>{\mu\otimes 1_A}>> A \otimes_k A \otimes_k A \end{CD}
we have also second and third axioms to be satisfied (I have checked there are no problem)
Problem Example 3.50. The spectrum of a polynomial ring in one variable $A=k[s]$ can be given the structure of an affine algebraic group by defining $\mu(s)=$ $s \otimes 1+1 \otimes s, \epsilon(s)=0$ and $\iota(s)=-s$. This group is denoted by $\mathbb{G}_{\mathrm{a}}$. As a functor, it assigns to a $k$-algebra $R$ the additive group $R$ itself. As a variety, of course, it is isomorphic to $\mathbb{A}^1$.
The attempt Since our $k-$ Algebra is a polynomial ring $k[s]$ let us try with only $s\in A=k[s]$
$s$ is sent to $s \otimes 1+1 \otimes s$ then, by $1\otimes\mu $, sent to $s\otimes 1 \otimes 1+ s\otimes1\otimes1+ 1\otimes s \otimes 1 +1\otimes 1\otimes s$
in other way
$s$ is sent to $s \otimes 1+1 \otimes s$ then, by $\mu \otimes1 $, sent to $s\otimes 1 \otimes 1+ 1\otimes s \otimes1+ 1\otimes 1 \otimes s +1\otimes 1\otimes s$
So in this calculations $3$ terms in each sides are cancelled but $s\otimes 1\otimes 1$ and $1\otimes 1\otimes s$ clearly since $s$ is not in the field $k$ we cannot take it to in front of the equation.
So what am I missing, can you point it out?
