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Let $X=Spec(R)$ be an irreducible noetherian scheme and $\eta$ the unique minimal prime ideal of $R$. Let $U$ and $V$ be open sets in $X$ and $p$ a point with $p\in V\subseteq U\subseteq X$.

If $X$ is additionally integer (i.e. additionally reuced, i.e. $R$ is a domain and $\eta=(0)$) then one has a diagram

$$ \begin{array}{ccccccl} \mathcal{O}_X(U)& \hookrightarrow &\mathcal{O}_X(V)& \hookrightarrow&{\mathcal{O}}_{X,p}&\twoheadrightarrow& k(p)=Frac(R/p)&\\ &&&\searrow&\downarrow&&&\\ &&&&{\mathcal{O}}_{X,\eta}&\xrightarrow{=}& k(\eta)=Frac(R/\eta)&(=Frac(R)) \end{array} $$

where the indicated maps ($\hookrightarrow$), the diagonal map ($\searrow$) and ${\mathcal{O}}_{X,p}\hookrightarrow {\mathcal{O}}_{X,\eta}$ are inclusions. I have some questions relating to this situation.

$\mbox{1.}$ Is the map $k(p)\to k(\eta)$ an inclusion?

One may identify all the images of the inclusions in ${\mathcal{O}}_{X,\eta}=Frac(R)$ with their domains and has $$\mathcal{O}_X(U)=\bigcap_{p\in U}{\mathcal{O}}_{X,p}.~~(*)$$

This identification is very helpful for me since one can really "work" then inside the big ring ${\mathcal{O}}_{X,\eta}$.

$\mbox{2.}$ I would like to understand function fields and stalks in the non-reduced case (but $X$ still irreducible). Then one can write down the same diagram as above (instead of the equality $Frac(R/\eta)=Frac(R)$). Which of the arrows remain inclusions ? Can I write down something like $(*)$ in this case, too?

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1 Answer 1

up vote 4 down vote accepted

Concerning 1.: Why do you think that there should be a map $k(p)\to k(\eta)$?

Consider for example the case $R=\mathbb{Z}$. Then for any two different points $p$ and $q$ of $\mathrm{Spec}\,\mathbb{Z}$ one has $\mathrm{Hom}(k(p),k(q))=\emptyset$ because these are two fields of different characteristic.

Concerning 2., the ring $R=k[x,y]/(x^2,xy)$ is enlightening. Topologically it is identical to $k[x,y]/(x)=k[y]$, so its spectrum is just a line, but with the origin infinitesimally thickened in the $x$-direction. The ideal $(x)$ is its unique minimal prime ideal. Whenever $U$ is an open subset not containing the point $(x,y)$ the restriction map $\mathcal{O}_X(X)\to \mathcal{O}_X(U)$ is not injective, because $x$ maps to zero in the latter ring. In particular $\mathcal{O}_X(X)\to \mathcal{O}_{X,\eta}$ is not injective.

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Well said, Philipp, very lucid (+1, of course). –  Georges Elencwajg Jun 27 '11 at 20:24
    
Yes, thank you, Philipp! Can I still write $\mathcal{O}_X(U)$ as something having to do with the stalks $\mathcal{O}_{X,p}$ in the non-reduced case, like in (*)? This would help me much to understand what sections on $U$ should be. –  geometrystudent Jun 28 '11 at 12:33

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