This is a reality check for the following computations that I did:
Consider the map $(\operatorname{id}, \iota): \Bbb A_\Bbb Z^1 \rightarrow \Bbb A_\Bbb Z^1\times \Bbb A_\Bbb Z^1$ from the definition of a group scheme. In the coordinate rings it is given by $$\Bbb Z[t] \otimes \Bbb Z[t] \cong \Bbb Z [x,y]\rightarrow \Bbb Z[t]\\ x \mapsto t\\y\mapsto -t.$$
I'm trying to find the preimage of the prime ideal $(t^2+1) \subset \Bbb Z[t]$. It contains $(x^2+1, y^2+1, -xy+1)$. Is that all?
Aside: I complexified everything to have a more geometric picture and I found that the image of the closed set $I=V(t^2+1)$ under this map is $\{(\pm i, \mp i)\}$, which is also closed. This brought me the side question: when should I expect the image of closed sets to be closed?
In $\mathbb{Z}[x,y]/(x^2+1,y^2+1,xy-1)$ we see that $x=x(xy)=x^2 y = -y$. It follows immediately that $\mathbb{Z}[x,y]/(x^2+1,y^2+1,xy-1) \to \mathbb{Z}[t]/(t^2+1)$, $x \mapsto t$, $y \mapsto -t$ is an isomorphism.