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Let $k$ denote a field and let $\mathbb{A}^n$ denote affine $n$-space. Then if I understand correctly, it's best to view $\mathbb{A}^n$ as a ringed space. I'm a bit unsure as to what the structure sheaf is though, and this seems not to be easily found on the internet. If I had to guess, I'd say that for $U \subseteq \mathbb{A}^n$ open in the Zariski topology, the definition of $\mathcal{O}(U)$ is probably $$\left\{\frac{f}{g} : f,g \in k[n], g \neq 0, V(g) \cap U = \emptyset\right\}.$$

(I write $k[n]$ for the set of polynomials in $n$-variables.)

Is this right? I'd also appreciate a reference that describes classical objects like (the set of closed points of) affine space from the ringed space perspective. This seems to be hard to find, as most texts dive into schemes rather quickly.

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  • $\begingroup$ If you mean for the underlying set of $\mathbb{A}^n$ to be just $k^n$, this isn't going to be a ringed space in a nice way unless $k$ is algebraically closed. $\endgroup$ – Eric Wofsey Aug 5 '18 at 3:22
  • $\begingroup$ In particular, presumably you want $\mathcal{O}(\mathbb{A}^n)$ to be $k[n]$, but it's not by your definition because it would include $1/g$ whenever $g$ has no roots over $k$. As far as I know there is no good way to fix this without changing the underyling set of $\mathbb{A}^n$. $\endgroup$ – Eric Wofsey Aug 5 '18 at 3:37
  • $\begingroup$ Anyways, I'm not sure I have correctly interpreted what you want (that you want to do this over non-algebraically closed fields, and you mean for the underlying set to be $k^n$). If you confirm that this is what you mean I can turn these comments into an answer. $\endgroup$ – Eric Wofsey Aug 5 '18 at 3:41
  • $\begingroup$ @EricWofsey, those are reasonable comments, but I'm not sure what your question is. I'm mainly just interested in whatever definitions are standard, if that helps. $\endgroup$ – goblin Aug 5 '18 at 3:54
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    $\begingroup$ Well, what is standard is to use schemes...you seem to want to do something less standard that avoids schemes, but I'm not sure exactly in what about schemes you want to avoid. $\endgroup$ – Eric Wofsey Aug 5 '18 at 4:20
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If $\mathbb{A}^n$ refers to $\operatorname{MaxSpec} k[n]$ as you seem to have in mind based on the comments, then the structure sheaf on $\mathbb{A}^n$ is just the restriction of the structure sheaf on $\operatorname{Spec} k[n]$. In fact, this is really basically the same sheaf: an open subset of $\operatorname{Spec} k[n]$ is determined by the closed points it contains (this is part of what the Nullstellensatz says), so the inclusion map $\operatorname{MaxSpec} k[n]\to \operatorname{Spec} k[n]$ induces an isomorphism on lattices of open sets. So a sheaf on $\operatorname{MaxSpec} k[n]$ is really exactly the same thing as a sheaf on $\operatorname{Spec} k[n]$; you've just removed all the generic points from your open sets.

Concretely, your description is basically correct. If $U\subseteq\mathbb{A}^n$ is a nonempty open set, then $\mathcal{O}(U)$ is the subring of $k(n)$ consisting of rational functions which can be written with a denominator which vanishes nowhere on $U$. (If $U=\emptyset$, then of course $\mathcal{O}(U)$ is the zero ring.) Note that it is important here that "vanishes nowhere on $U$" includes points of $U$ with residue field besides $k$, so we really aren't thinking of polynomials as just functions on $k^n$ as you seem to be implying.

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  • $\begingroup$ Can you explain what you mean by "includes points of $U$ with residue field besides $k$"? I assume by "residue field" you mean a quotient of a ring by a maximal ideal? $\endgroup$ – goblin Aug 5 '18 at 6:01
  • $\begingroup$ Yes, points of $\operatorname{MaxSpec} k[n]$ are by definition maximal ideals in $k[n]$, and the residue field at a point is just the quotient of $k[n]$ by the maximal ideal. $\endgroup$ – Eric Wofsey Aug 5 '18 at 13:56

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