Group law of $\operatorname{Spec} \mathbb{Z}[x,x^{-1}]$ I'm trying to understand the group scheme $\mathbb{G}_m= \operatorname{Spec} \mathbb{Z}[x, x^{-1}]$ but I don't have much knowledge of algebraic geometry, so I thought that I should try to write everything down concretely to see what's going on. I started by computing spectrum of $\mathbb{Z}[x, x^{-1}],$ and I have found the following:
Denote by $S$ be the multiplicative set $\{ x^{n} \mid  n \in \mathbb{Z} \},$ then
$\mathbb{Z}[x, x^{-1}] = \mathbb{Z}[x]_S,$ and we have a classification of prime ideals of $\mathbb{Z}[X]$ here:
Classification of prime ideals of $\mathbb{Z}[X]$.
Now let $\mathcal{p}$ be a prime ideal of $\mathbb{Z}[x]$ such that $\mathcal{p} \cap S $ is empty,   then $\mathcal{p} \mathbb{Z}[x, x^{-1}]$ is a prime ideal of $\mathbb{Z}[x, x^{-1}].$  If I'm not missing something, I think this should give us all prime ideals of $\mathbb{Z}[x,x^{-1}].$
Now my question is the following:
Is it possible to write down explicitly what the group law on $\operatorname{Spec}\mathbb{Z}[x,x^{-1}]$ is?
For example: what is the product of the two prime ideals $(x-2)$ and $(2x-1)?$
Thanks in advance.
 A: I think definition group scheme is not “a scheme where underlying set has a group structure”, but “a group object in a category of schemes”. Here’s similar confusion that might happen: in a category of schemes, underlying set of the product $X_1, X_2$ of two schemes is not the same as product of underlying sets. (Even $X_1=X_2=\operatorname{Spec}\mathbb{Z}$ gives a counterexample.) For affine scheme, product of $\operatorname{Spec}(A)$ and $\operatorname{Spec}(B)$ is $\operatorname{Spec}(A\otimes B)$.
In our case, “group operation” is not a function that takes to prime ideals and give another prime ideal, rather a morphism of scheme $\operatorname{Spec}(\mathbb{Z}[x, x^{-1}]) \times \operatorname{Spec}(\mathbb{Z}[x, x^{-1}]) \to \operatorname{Spec}(\mathbb{Z}[x, x^{-1}])$, which should corresponds to a ring map $\mathbb{Z}[x, x^{-1}] \to \mathbb{Z}[x, x^{-1}] \otimes \mathbb{Z}[x, x^{-1}] = \mathbb{Z}[x, x^{-1}, y, y^{-1}]$ that satisfies so-called coassociativity. The map is actually $x\mapsto x\otimes y$, and via this map, you can take a prime ideal of $\mathbb{Z}[x, x^{-1}, y, y^{-1}]$ and get one of $\mathbb{Z}[x, x^{-1}])$, but not in the way you mentioned.
