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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.

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2 Answers

up vote 5 down vote accepted

If $R$ is a ring (not necessarily a field) then $0$ is not a point of Spec $\mathbb A^1$, it is a section. Namely, there is a natural map Spec $\mathbb A^1 \to$ Spec $R$, corresponding to the map $R \to R[x]$, and for any $r \in R$ (whether it be $0$, $1$, or some other element) the map $R[x] \to R$ given by mapping $x$ to $r$ gives a section Spec $R \to \mathbb A^1$ which is a section to the natural map.

Added: As Mariano writes in his answer, such sections are the so-called $R$-valued points of $\mathbb A^1$.

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Thanks, Matt. I knew about R-points, but wasn't sure what the deal was in relation to working over a ring rather than a field. –  David Roberts Jan 6 '11 at 2:27
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If $r\in R$, there is an $R$-point in $\mathbb A^1$ given by the map $R[x]\to R$ given by evaluation of polynomials at $r$.

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I would vote this up but I don't have enough rep yet. –  David Roberts Jan 6 '11 at 2:28
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Now I do. +1 ....... –  David Roberts Jan 6 '11 at 22:28
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