The generic point of the affine line has no relative k-valued coordinate

If $f,g\in k[t]$ are nonzero, the leading coefficient of $fg$ is the nonzero product of their leading coefficients; hence $k[t]$ is a domain and $(0)$ is a point of ${\mathbb{A}}_k^1$. By [F1], its closure is all of $\mathrm{Spec}k[t]$.

Evaluation at $0$ identifies $k[t]\mathrm{/}(t)$ with the domain $k$, so $(t)$ is prime and strictly contains $(0)$. Thus the closure in step 1.1 is not a singleton, and $(0)$ is not closed. quotient-domain criterion

A relative $k$-valued coordinate representing a point $\mathfrak{p}$ is a [F1, step 2.1, algebra] $k$-algebra map $k[t]\to k$ whose kernel is $\mathfrak{p}$. Such a map fixes $k$, hence is surjective and has maximal kernel. By [F1] its closure is the set of primes containing it, which is the singleton consisting of that maximal ideal. Its kernel is therefore closed, whereas $(0)$ is not, so the generic point $(0)$ has no relative $k$-valued coordinate.