The affine line $\mathbb{A}^1$ (over $Spec R$) is, as far as I understand, $Spec R[x]$. My guess is that $0\in \mathbb{A}^1$ is given by the (opposite of the) composite homomorphism $R[x] \to R \to F$ where the first map sends a polynomial to its constant term and the second is the canonical map to the field of fractions $F$ of $R$. What ring homomorphism $R[x] \to F$ gives us $1 \in \mathbb{A}^1$?
Edit:
Ok, closed points of $\mathbb{A}^1$ (over a field k) correspond to ideals $(x-a) \subset k[x]$ or equivalently the quotient maps $k[x] \to k[x]/(x-a)$. So then 0 corresponds to $k[x] \to k$ (consistent with my question as originally stated) and 1 corresponds (I guess) to $k[x] \to k[x]/(x-1)$. This should extend easily to rings with unit. Is this correct? How do we extend this to rings, bearing in mind I may have it wrong that one passes to fields of fractions.