How to proceed in this group scheme? (Explicit calculations) I want to see if the following takes the requirements of a group scheme

The thing is that I am having troubles with the inverse element diagram with the explicit computations of each of the maps, the following is my definition 

Then, How can I proceed in the verification of the commutativity?
Thanks a lot in advance.
 A: What the book in your first image essentially says is that since an affine $Spec(\mathbb{Z})$-group scheme is just a group object in the category of affine $Spec(\mathbb{Z})$-schemes, and since this category is anti-equivalent to the catgory of $\mathbb{Z}$-algebras, then an affine $Spec(\mathbb{Z})$-group scheme is basically the same as a cogroup in the category of $\mathbb{Z}$-algebras.
Thus to verify the group axioms for $G$, you have to "dualize" them in the category of $\mathbb{Z}$-algebras. What are the corresponding maps for everything in your diagram ?


*

*$\pi:G\to S$ corresponds to the structural morphism $\varphi: \mathbb{Z}\to A$;

*$\varepsilon: S\to G$ corresponds to your given counity $\varepsilon^*: A\to \mathbb{Z}$ which is the zero map ;

*$m: G\times_S G\to G$ corresponds to your given comultiplication $m^*:A\to A\otimes_\mathbb{Z} A$ ;

*$1\times inv : G\times_S G\to G\times_S G$ corresponds to $Id\otimes inv^*: A\otimes_\mathbb{Z} A \to A\otimes_\mathbb{Z} A$ where $inv^*$ is your given coinversion, which is actually the identity ; 

*$\Delta: G\to G\times_S G$ corresponds to the multiplication map $\mu: A\otimes_\mathbb{Z} A\to A$ given by $\mu(a\otimes b) = ab$.
Now you have to check that the diagram of $\mathbb{Z}$-algebras you get by replacing all arrows by the ones just above (in particular, all the arrows go in the reverse direction) commutes ; this is just a direct verification going through the definitions of each map.
Since $A= \mathbb{Z}[X]/(X^2+2X)$, and you have to check $\varphi\circ \varepsilon^* = \mu\circ (Id\otimes inv^*)\circ m^*$ where both maps are $A\to A$, you just have to check that the image of $X$ is the same in both cases.
But $\varepsilon^*$ is the zero map, and $inv^*$ is the identity, so it translates to $\mu\circ m^* = 0$.
Now $\mu(m^*(X)) = \mu(1\otimes X + X\otimes 1 + X\otimes X) = 1\cdot X + X\cdot 1 + X\cdot X = X^2 + 2X = 0$ in $A$, precisely by definition of $A$.
