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I can construct a scheme by patching that represents a projective space over an arbitrary ring. I can also prove that, if the ring is a Jacobson domain, the only regular functions on it are constants.

If the ring is not Jacobson, (for instance ${\mathbb{Z}}_{(2)}$ --- integers localized at 2 (or for matter, even the 2-adic integers), it appears that all maximal ideals of $\mathbb{Z}_{(2)}[X_1,\dots,X_n]$ contain 2 so that a polynomial $f\in\mathbb{Z}_{(2)}[X_1,\dots,X_n]$ with even coefficients, evaluates to 0 at all closed points of $\mathrm{Spec}\mathbb{Z}_{(2)}[X_1,\dots,X_n]$. It follows that polynomials like $f(X,Y)=3+2X+4Y$ are effectively constant since $f(t\cdot x,t\cdot y)=f(x,y)$ at every closed point of $\mathrm{Spec}\mathbb{Z}_{(2)}[X,Y]$, and induce a function on the projective space. Does this seem like a reasonable argument?

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  • $\begingroup$ I don't know if this is helpful or not, but I think the moral is that, in this situation you really need to think about both the special fiber and generic fiber over $\mathbb{Z}_{(2)}$. That function you write down is indeed constant on the special fiber, but is certainly not constant on the generic fiber, and so shouldn't be thought of as constant even if it happens to have the same value at the closed points in the "total space." $\endgroup$
    – Ramsey
    Commented Jul 18, 2012 at 20:54
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    $\begingroup$ It's not true that every maximal ideal in $\mathbb Z_{(2)}[X]$ contains 2. For example, the ideal $(2X-1)$ is maximal. $\endgroup$
    – Angelo
    Commented Jul 19, 2012 at 4:00
  • $\begingroup$ $(2X-1)$ is not maximal. $(2X-1,Y)$ is larger. I admit that $(2X-1,2)$ appears to generate the whole ring, though. $\endgroup$
    – Justin Smith
    Commented Jul 19, 2012 at 10:32
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    $\begingroup$ To Justin: What I wrote is correct, I am taking $n = 1$, and there is no $Y$ in my ring. $\endgroup$
    – Angelo
    Commented Jul 19, 2012 at 10:38
  • $\begingroup$ Sorry! Thinking about this some more, I'm inclined to regard the question of evaluating functions on points (closed or otherwise) as irrelevant. In $RP^1$, functions from open affines pull back to polynomials in $X/Y\in R(X,Y)$ or $Y/X$. But regular functions on $\mathbb{A}^2$ are functions in $R[X,Y]$. Since such functions do not have $X^{-1}$ or $Y^{-1}$ in them, higher coefficients must vanish. Evaluating over points (closed or otherwise) is implicitly taking the quotient with the nilradical, which defeats the purpose of allowing nilpotent regular functions. $\endgroup$
    – Justin Smith
    Commented Jul 19, 2012 at 14:14

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