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Let $X$ be an integral scheme and $\eta$ its general point. Then we can identify $\mathcal O_{X,\eta}$ with $FF(A)$ where $\operatorname{Spec} A$ is any open affine of $X$, because $\eta$ lives in all open affines. But how can we then extend any $f \in FF(A)$ to a function on the whole of $X$, i.e. to an element of $\mathcal O_{X,\eta}$?

I am not sure how to realize this algebraically because I don't know in what form I should write $\mathcal O_{X,\eta}$. When I think about it geometrically, at least I know that I am not extending $f$ to something bigger than the closure of $\operatorname{Spec} A$ because $X$ is irreducible. But does it really suffice that I am not extending a function outside the closure of its domain? Can I always extend a function to the closure of its domain of definition in algebraic geometry? (here I mean function in the sense of an element of the sheaf of rings, not as a function that should always have a non "1/0 value")

I worked this out for the $\Bbb P^n$ and at least I got the same answers. For $X=\Bbb P^n$, the general point $\eta$ is just the zero ideal in $k[x_0,...,x_n]$. So we get $\mathcal O_{X,\eta} = k(x_0,...,x_n)_0 \cong k(x_1,x_2,...x_n)$ by the dehomogenizing map $-$quotient out by $(x_0-1)$. On the other hand, if we take some affine open $\operatorname {Spec} A$ such that $A =k[x_0/x_0,x_1/x_0...,x_n/x_0]/(x_0/x_0-1)$, we also get $\eta = (0)$ and $\mathcal O_{\operatorname{Spec} A,\eta} = k(x_1/x_0,x_2/x_0,...,x_n/x_0)$.

Rodrigo
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    I don't think I understand the question. I would guess that by a "function on the whole of $X$," you mean an element of $\mathscr{O}X(X)$, but you write ``i.e. to an element of $\mathscr{O}{X,\eta}$." – Keenan Kidwell Jan 24 '14 at 17:44
  • Also, in the last paragraph did you mean to write $k[x_0,\ldots,x_n]/(x_0-1)$, i.e., the coordinate ring of the affine open $D^+(x_0)$ (which can be thought of as the polynomial ring over $k$ in the $n$ variables $x_1/x_0,\ldots,x_n/x_0$). – Keenan Kidwell Jan 24 '14 at 18:08
  • @KeenanKidwell good question! What I had in mind is the following example. Let $X := \Bbb P^2$ and consider the function $f := \frac 1{x_1x_2} \in \mathcal O_{U_0,\eta}$. As an element of $\mathcal O_{U_0,\eta}$ it is not obvious that this function also makes sense in some open set in the complement $X \setminus U_0$ (where open set here is understood in the subspace topology), but in fact it does! This becomes only apparent when we realize that, as an element of $\mathcal O_{X,\eta}$, $f$ is actually defined on $D(x_1x_2)$, which has non-empty intersection with the complement of $U_0$. – Rodrigo Jan 24 '14 at 19:23
  • I equated "function on the whole of $X$" with "element of $\mathcal O_{X,\eta}$" because, any function that lives in the local ring of the general point is defined almost everywhere in the corresponding closed set. Indeed, it is defined everywhere outside of a subset of codimension $1$. – Rodrigo Jan 24 '14 at 19:27
  • Dear @Rodrigo, I still don't understand what your question is. You start with an an element $f\in\mathrm{FF}(A)$, and ask how to extend it to an element of $\mathscr{O}_{X,\eta}$. But these are the same. – Keenan Kidwell Jan 24 '14 at 19:37
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    Dear @KeenanKidwell I read you answer and it made perfect sense to me. But then I computed again my example with the projective space (cf. edit above) and I realize that I was actually getting different answers. Let me read your answer again to see if I figure it out – Rodrigo Jan 24 '14 at 19:41
  • Dear @Rodrigo, The problem in your edit is that for the $\mathrm{Proj}$ of a graded domain $A$, the local ring at the generic point $\eta$ is not the algebraic localization $A_\eta$, but the degree zero part of that localization. – Keenan Kidwell Jan 24 '14 at 19:43
  • @KeenanKidwell correct! – Rodrigo Jan 24 '14 at 20:05

1 Answers1

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If $X=\mathrm{Spec}(A)$ is the spectrum of a domain $A$, with $\eta=(0)$ the generic point of $X$, then the equality $\mathscr{O}_{X,\eta}=A_{(0)}=\mathrm{Frac}(A)$ is built into the definition of the structure sheaf of $X$.

In general, for $X$ any scheme and $U\subseteq X$ an open subscheme, for any $x\in U$, the map $\mathscr{O}_{X,x}\rightarrow\mathscr{O}_{U,x}$ induced by the open immersion $U\hookrightarrow X$ is an isomorphism, so stalks can be ``computed" in any open subscheme. Now if $X$ is integral with generic point $\eta$ and $U=\mathrm{Spec}(A)$ is an affine open, then $\eta\in U$ and we have $\mathscr{O}_{X,\eta}=\mathscr{O}_{U,\eta}=\mathrm{Frac}(A)$. Elements of $\mathscr{O}_{X,\eta}$ are equivalence classes of pairs $(f,V)$, where $V\subseteq X$ is open and $f\in\mathscr{O}_X(V)$. In particular, any element $f\in A=\mathscr{O}_X(U)$ gives rise to the equivalence class of $(f,U)$ in $\mathscr{O}_{X,\eta}$, which is identified with the image of $f$ under the canonical injection $A\hookrightarrow\mathrm{Frac}(A)$ under the identification above. But there is no reason that an element of $\mathscr{O}_{X,\eta}$ should necessarily come from a global section of $X$.

Maybe you mean that an element of $\mathscr{O}_{X,\eta}$ should come from a global section of $U$, i.e., from an element of $A$? But you shouldn't expect this either, because the map $A\hookrightarrow\mathrm{Frac}(A)$ is not surjective unless $A$ is already a field.

Keenan Kidwell
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  • I meant that every element of $\mathcal O_{X,\eta}$ should come from a global section of $V$, where $V$ is any non-empty open set of $X$ (which therefore automatically contains the general point of $X$). The problem that I had is that the intersection $W := V \cap (X \setminus U)$ might be non-empty and it is not clear how to evaluate elements of $\mathcal O_{U,\eta}$ at the points of $W$. In other words, I get how we can start with $f\in \mathcal O_{X,\eta}$, map it to $\tilde f \in \mathcal O_{U,\eta}$ by the isomorphism you gave, but how do we take/evaluate now – Rodrigo Jan 24 '14 at 20:03
  • $\tilde f(\mathfrak w) = \tilde f /\mathfrak w$ for $\mathfrak w \in W$? – Rodrigo Jan 24 '14 at 20:03
  • This isn't true though. An element of $\mathscr{O}{X,\eta}$ comes from some open subset $V$ of $X$, but not necessarily from any open subset of $X$. Consider the case of $X=\mathrm{Spec}(\mathbf{Z})$. The section $1/2\in\mathscr{O}{X,\eta}=\mathbf{Q}$ comes from $D(2)=\mathrm{Spec}(\mathbf{Z}[1/2])$, but not from $X$ itself. – Keenan Kidwell Jan 24 '14 at 20:25
  • Yes. You are right. Forgive me, I did not use English quantifiers in the right order. You said exactly what I meant. For every element of $\mathcal O_{X,\eta}$ you can find some open set V from which it comes, but of course in general circumstances you'll not find an open set for all elements of $\mathcal O_{X,\eta}$. But yeah, with exactly this interpretation you gave, do you now understand my problem? Again, if we pick some chart $U =\Bbb A^2$ of $X =\Bbb P^2$, then 1) $U\ni \eta$ and 2) it is not clear that the elements of $\mathcal O_{U,\eta}$ take values on the line at infinity. – Rodrigo Jan 24 '14 at 20:55
  • It is not clear because the line at infinity is not in their domain of definition (because they are in $\mathcal O_{U,\eta}$, their domain of definition is, a priori, an open set inside $U$. However, these functions do take values at infinity (at least in some open set there (in the subset topology). – Rodrigo Jan 24 '14 at 20:59