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?

nothave $X^{-1}$ or $Y^{-1}$ in them, higher coefficients must vanish. Evaluating overpoints(closed or otherwise) is implicitly taking the quotient with the nilradical, which defeats the purpose of allowing nilpotent regular functions. – Justin Smith Jul 19 '12 at 14:14