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My professor made a comment the "Local rings are not really local in the Zariski topology (on say $\mathbb{C}^{n}$). Thus we take completions of local rings at all points of a smooth variety which turn out to be the same. However in the classical topology local rings are local because Zariski topology has very large open sets."

He does not expect us to understand what completions are, but I donot even intuitively understand the first part of the statement. Can someone make the statement precise and tell exactly what he means by 'local rings not 'really local' in Zariski topology, however they are in normal topology' ?

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    That statement appears to be backwards--it makes much more sense to say local rings are local in the Zariski topology but not in the classical topology. – Eric Wofsey May 28 '17 at 04:36

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I will try to share my insight on this topic. To start, I agree with @Eric Wofsey: Local rings are local with respect to the Zariski topology, but they are not local with respect to the usual Euclidean topology (and our intuition).

Zariski topology is very coarse: no ball in the Euclidean topology is contained in the complement of a Zariski open. Thus, the information that a Zariski open carries about the whole variety is very strong. On the other hand, a complex manifold can always be covered by an atlas of polydisks (hence, if we look close enough in the Euclidean topology, all manifolds look the same). That is why in algebraic geometry very often we look at birational classes of varieties (i.e. varieties $X$ and $Y$ having isomorphic Zariski open sets $U_X$ and $U_Y$). For instance, if we look at an open subset of $\mathbb{P}^n$, we see there are many lines isomorphic to $\mathbb{A}^1$ in there. On the other hand, if we look at the open set of an abelian variety (i.e. a complex torus that is defined by algebraic equations in some projective space), we will never find a curve isomorphic to $\mathbb{A}^1$ in there. So, no matter how close we look in the Zariski topology, we can always tell apart the projective space from an abelian variety.

The local ring $\mathcal{O}_{X,p}$ at a point $p \in X$ is local in the Zariski topology by construction: It is the projective limit of the information carried by each open set containing $p$. For instance, let $X=Bl_q \mathbb{P}^2$, the blow-up of the projective plane at the point $q$. Assume that $p \neq q$ is another point in $\mathbb{P}^2$. Then, we can regard $p$ as a point in $X$, just identifying the open sets $\mathbb{P}^2 \setminus \lbrace q \rbrace$ and the open set $X \setminus E$, where $E$ is the exceptional curve of the blow-up. Thus, we have $\mathcal{O}_{X,p}=\mathcal{O}_{\mathbb{P}^2,p}$, i.e. the local ring at $p$ is local enough to forget about birational modification that happen away from it.

On the other hand, a local ring is pretty much global from the point of view of Euclidean topology. As mentioned before, birational varieties are the same up to close subsets that in the Euclidean topology have Lebesgue measure 0. Now, one can show that two varieties are birational if and only if the field of their rational functions are isomorphic.

Now, fix a variety $X$, and a local ring $( \mathcal{O}_{X,p},\mathfrak{m}_p )$ at some point $p \in X$. By construction (look at it in the affine case), the function field $k(X)$ of the variety $X$ is the localization at the maximal ideal $\mathfrak{m}_p$ of $\mathcal{O}_{X,p}$. Thus, we have that a local ring detects the birational class of $X$, and therefore all the geometry of $X$ off a set of Lebesgue measure 0.

Stefano
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