Timeline for Projective spaces with nonconstant regular functions
Current License: CC BY-SA 3.0
7 events
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| Aug 26, 2013 at 17:52 | history | edited | user9072 |
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| Jul 19, 2012 at 14:14 | comment | added | Justin Smith | 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. | |
| Jul 19, 2012 at 10:38 | comment | added | Angelo | To Justin: What I wrote is correct, I am taking $n = 1$, and there is no $Y$ in my ring. | |
| Jul 19, 2012 at 10:32 | comment | added | Justin Smith | $(2X-1)$ is not maximal. $(2X-1,Y)$ is larger. I admit that $(2X-1,2)$ appears to generate the whole ring, though. | |
| Jul 19, 2012 at 4:00 | comment | added | Angelo | It's not true that every maximal ideal in $\mathbb Z_{(2)}[X]$ contains 2. For example, the ideal $(2X-1)$ is maximal. | |
| Jul 18, 2012 at 20:54 | comment | added | Ramsey | 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." | |
| Jul 18, 2012 at 20:20 | history | asked | Justin Smith | CC BY-SA 3.0 |