Consider the affine $n$-space $\mathbb{A}^n _R= \operatorname{Spec}(R)(R[X_1,, ..., X_n])$ over arbitrary nontrivial ring $R$. (assume furthermore $n \ge 1$)
I want to show that the related structure morphism $s:\mathbb{A}^n _R \to \operatorname{Spec}(R)$ is not proper.
My considerations:
proper means that $s$ is separated, of finite type and universally closed (so closeness of $s$ is stable under base change).
Since $s$ is a morphism between affine schemes it is separated and obviously of finite type.
So the problem is to find a base change $b: X \to \operatorname{Spec}(R)$ such that the induced $s': \mathbb{A}^n _R \times_{ \operatorname{Spec}(R)} X \to X$ from diagram
$$ \require{AMScd} \begin{CD} \mathbb{A}^n _R \times_{\operatorname{Spec}(R)} X @>{s'} >> X \\ @VVprV @VVbV \\ \mathbb{A}^n _R @>{s}>> \operatorname{Spec}(R) \end{CD} $$
isn't a closed morphism.
Does anybody see how to construct such $X$?
Does there exist a conceptual method to find such $X$ in a sophisticated way or it just a "guessing game"?