I'm working on a problem to ultimately show that an open subset of $\text{Spec}\mathbb{Z}[t]$ is not an affine scheme. The way the problem suggests to do this is to find the neighborhoods $X_1 = D(2)$ and $X_2 = D(t)$, the domains of the two prime ideals $(2), (t)$, and compute the structure sheaf element $\mathcal{O}(X_1\cup X_2)$. My strategy is to compute $\mathcal{O}(X_1\cup X_2)$ as the limit of $\mathcal{O}(X_1)$ and $\mathcal{O}(X_2)$. Since the structure sheaf elements are the localizations around the specified prime ideals, I have
\begin{eqnarray*} \mathcal{O}(X_1) & = & \left (\mathbb{Z}[t]\right )_{(2)} \;\; =\;\; \left \{\left .\frac{p(t)}{q(t)} \right | \; 2 \; \text{doesn't divide} \;q(t) \right \} \\ \mathcal{O}(X_2) & = & \left (\mathbb{Z}[t]\right )_{(t)} \;\; =\;\; \left \{\left .\frac{p(t)}{q(t)} \right | \; t \; \text{doesn't divide} \;q(t) \right \} \\ \end{eqnarray*}
Given that we have restriction morphisms between these two sets, is it possible to conclude
$$ \mathcal{O}(X_1\cup X_2) \;\; =\;\; \left \{\left .\frac{p(t)}{q(t)} \right | \; \text{Neither} \; t \; \text{nor} \;2 \; \text{divide} \;q(t) \right \} $$
If this is true, then how do we know that $X_1\cup X_2$ isn't isomorphic to $\text{Spec}R$ for any ring $R$ and $\mathcal{O}(X_1\cup X_2)$ to corresponding affine scheme elements? Part of my issue here is that I'm not totally sure what the Zariski topology looks like in $\text{Spec}\mathbb{Z}[t]$.