most recent 30 from http://mathoverflow.net 2013-05-18T14:59:08Z http://mathoverflow.net/feeds/question/44516 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44516/non-representable-functor-representable-on-locally-noetherian-schemes jlk 2010-11-02T04:20:33Z 2010-11-02T04:51:24Z

<p>What is an example of a functor $F : \mathbb{C}\text{-Sch.} \to \text{Sets}$ with the property that the restriction of $F$ to locally Noetherian $\mathbb{C}$-schemes can be represented by a locally Noetherian $\mathbb{C}$-scheme, but that scheme does not represent $F$.</p> <p>I'd be particularly nice to see a "real-world" example (though this might be stretching the notion of "real-world").</p>

http://mathoverflow.net/questions/44516/non-representable-functor-representable-on-locally-noetherian-schemes/44518#44518 BCnrd 2010-11-02T04:51:24Z 2010-11-02T04:51:24Z

<p>Define $F(X) = {\rm{Hom}}_{\mathbf{C}}(X,{\rm{Spec}}(R/tR))$ where $R$ is the valuation ring of an algebraic closure of $\mathbf{C}((t))$. Note that every element of the maximal ideal of $R/tR$ is nilpotent yet also an $N$th power for arbitrarily large $N$. For any noetherian $\mathbf{C}$-algebra $A$, every $\mathbf{C}$-algebra map $R/tR \rightarrow A$ carries the maximal ideal into the nilradical of $A$. But the nilradical of $A$ has all elements with vanishing $n$th power for some uniform $n$ (depending on $A$) since $A$ is noetherian, so in fact the maximal ideal of $R/tR$ is killed by any such map. In other words, the restriction of $F$ to the full subcategory of locally noetherian objects is represented by ${\rm{Spec}}(\mathbf{C})$. </p> <p>This is a perfectly "real world" example, since valuation rings of algebraic closures of complete discretely-valued fields come up all the time in number theory and rigid-analytic geometry. </p>