I have a question concerning a construction introduced in Bosch's "Algebraic Geometry and Commutative Algebra" ( pages 428/ 429). Here the relevant excerpt:
We start with a scheme $\mathbb{A}^1 _U = \mathbb{A}^1 _{\mathbb{Z}} \times_{\mathbb{Z}} U$ over a base scheme $U$.
Then we define a morphism $\psi: \mathbb{A}^1 _U \to \mathbb{A}^1 _U $ via
$$(t,x) \mapsto (\eta(x) \cdot t, x)$$
for a section $\eta \in \mathcal{O}_U(U)$, $t \in \mathbb{A}^1 _{\mathbb{Z}}, x \in U$.
What I don't understand what is concretely $\eta(x) \cdot t$, how the multiplication "$\cdot$" on $t$ is defined and why set theoretically $\eta(x) \cdot t \in \mathbb{A}^1 _{\mathbb{Z}}$?
Below it is explained how $\psi$ is defined on level of rings namely as a map of ringed spaces we have by definition for an open affine subset $Spec(R)= V \subset U$ the description as ring map
$$\psi^{\#} _{\mathbb{A}^1 _{\mathbb{Z}} \times V}: R[\zeta]=\mathbb{A}^1 _U(\mathbb{A}^1 _{\mathbb{Z}} \times V) \to \mathbb{A}^1 _U(\mathbb{A}^1 _{\mathbb{Z}} \times V)$$
$$\zeta \mapsto (\eta \vert _{Spec(R)}) \cdot \zeta$$
Take into account that $\mathbb{A}^1 _U(\mathbb{A}^1 _{\mathbb{Z}} \times V)= \mathbb{A}^1 _{\mathbb{Z}}(\mathbb{A}^1 _{\mathbb{Z}} ) \otimes _{\mathbb{Z}} R = R[\zeta]$.
So on level of local sections the definition of $\psi$ is clear.
What I don't understand how to interpret $(t,x) \mapsto (\eta(x) \cdot t, x)$ on level of Specs. Indeed, $t \in \mathbb{A}^1 _{\mathbb{Z}}, x \in U$. What is $\eta(x) \cdot t$?
Set theoretically since $\mathbb{A}^1 _U = \mathbb{A}^1 _{\mathbb{Z}} \times_{\mathbb{Z}} U$ the $\eta(x) \cdot t$ lives in first factor $\mathbb{A}^1 _{\mathbb{Z}}$.
But how $\eta(x) \cdot$ "acts" on $t$? Can $\eta(x)$ be interpreted as an element in residue field $\mathcal{O}_{U, x}/m_x= \kappa(x)$? If yes, then the prime ideal $\eta(x) \cdot t$ doesn't live in $\mathbb{A}^1 _{\mathbb{Z}}$.
So what is strictly speaking the object $\eta(x) \cdot t$?
A regular function is the same thing as a map to $\mathbb A^1_{\mathbb Z}$, so $\eta(x)$ really does live on $\mathbb A^1_{\mathbb Z}$. It's good to think about why this should be.