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Let $R$ be a commutative ring with unit, $I$ an ideal of $R$ and consider the following three constructions.

  1. The localization $R_I$ of $R$ at $I$ (i.e. the localization of $R$ at the multiplicative system $R\setminus I$) gives a morphism $$ f_1:\operatorname{Spec}(R_I)\to \operatorname{Spec}(R)=:X $$
  2. The completion $\widehat{R}_I$ of $R$ at $I$ gives a morphism $$ f_2:\operatorname{Spec}(\widehat{R}_I)\to X $$
  3. For the special case $I=(a)$, the localization $R_a$ (i.e. the localization of $R$ at the multiplicative system $\{1,a,a^2,\ldots\}$) gives a morphism $$ f_3:\operatorname{Spec}(R_a)\to X $$

My question is:

What is the geometric meaning of all three constructions and how are they related?

This is what I ''know'' already or describes at least the style of answer I would appreciate. As remarked in the comments, $I$ has to be a prime ideal.

  • $R\to R_I$ is injective iff $R\setminus I$ contains no zero divisors which is the case if $I$ is a prime ideal. The scheme $\operatorname{Spec}(R_I)$ is the intersection of all the neightbourhoods of $I$ in $X$. This is a little contra-intuitive for me since the last statement sounds like ''$f_1$ is injective'' but the $\operatorname{Spec}$-operator should turn 'injective'' and ''surjective'' around somehow (I know this is literally not true but I only want to get a feeling like ''is contained'', ''is bigger'', ''is smaller'', etc.).
  • $\operatorname{Spec}(R_a)$ is somehow the opposite of $\operatorname{Spec}(R_{(a)})$ (= the intersection of all the neightbourhoods of $(a)$ in $X$) because $\operatorname{Spec}(R_a)$ seems to be something like the union of all the open sets of $X$ not containing the point $(a)$.
  • $R\to\widehat{R}_I$ is injective iff $\cap I^n=(0)$ and this holds very often (e.g. for $R$ noetherian and either an integral domain or a local ring). Hence (this is probably false intuition as remarked before), $f_2$ should be something like a ''projection'' (from something ''big'' into something ''small''). But what is geometrically the difference between localization and completion. I don't have a geometric idea of completion at all.

As remarked by Qiaochu Yuan in the comments below, one should not think of $\operatorname{Spec}$ as injective-surjective switching.

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3  
There are in fact two reasons a functor like Spec might not switch injective and surjective things. One is that contravariant functors just don't switch epis and monos in general (equivalently, covariant functors just don't preserve epis or monos in general), and the other is that epis in the category of rings look weird and in particular do not consist only of surjective maps (e.g. localizations are epis). –  Qiaochu Yuan Apr 3 '13 at 9:58
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Matt's answer to math.stackexchange.com/q/346588 gives a nice explanation of the geometric meaning of completion in the context of plane curves. –  Martin Apr 3 '13 at 10:05
    
Asides: I believe that's only the right formula for $R_I$ in the case that $I$ is prime. As for $R_a$, a better notation is $R[a^{-1}]$ or $R[1/a]$ or $a^{-1} R$. –  Hurkyl Apr 3 '13 at 10:24
    
There are some fundamental problem with what you write. $R\setminus I$ is only a multiplicative set if $I$ is a prime ideal. This makes the first statement in the first bullet point somewhat strange. –  Tobias Kildetoft Apr 3 '13 at 12:40
    
Example: take the spectrum of a local ring, and its residue field's spectrum as a subscheme. –  Loki Clock Apr 3 '13 at 13:58

1 Answer 1

up vote 3 down vote accepted

First of all, localization $R_I$ is only defined if $R\setminus I$ is actually a multiplicative system, i.e. if for $a\notin I$ and $b\notin I$, you have $ab\notin I$. that translates to $ab\in I$ $\Rightarrow$ $a\in I$ or $b\in I$, so $I$ has to be a prime ideal in any case. For intuition, I would advise thinking of $R$ as a finitely generated (and reduced) $k$-algebra, the quotient of the polynomial ring in $n$ variables over an algebraically closed field $k$: Then, $\mathrm{Spec}(R)$ (or at least the maximal ideals among the prime ideals) is just a subvariety of the affine space $k^n$. Think of $R$ as the functions on $X=\mathrm{Spec}(R)$.

  1. The ring $R_I$ is the ring where you are allowed to invert any function which is not in $I$: If $I$ is the ideal corresponding to a closed subvariety $Z=Z(I)$ of $X$, this means you may invert anything that does not vanish on $Z$. The intuition is that we think of $R_I$ as the functions that are defined locally around $Z$: If you have any function $f$ defined in a neighbourhood $U$ of $Z$ and it does not vanish on $Z$ itself, then by removing $Z(f)$ from $U$, you still have a neighbourhood of $Z$, but now $f$ is invertible everywhere on $U$. So, if you shrink the support of a function around $Z$ far enough, it is either a unit or it vanishes on $Z$.

    Recall that the prime ideals of $R_I$ are precisely the prime ideals of $R$ which do not meet $R\setminus I$, i.e. the ones that are contained in $I$. Those again correspond to subvarieties $Z'$ containing $Z$. So I personally think of $\mathrm{Spec}(R_I)$ as the space that parametrizes the $Z'$.
  2. Let me quote Chapter 7 of David Eisenbud's book Commutative Algebra with a view toward algebraic geometry on this one:

    A localization $R_I$ of the affine ring of a variety at the maximal ideal $I$ of a point on the variety represents and reflects the properties of Zariski open neighborhoods of the point; the completion $\hat R_I$ represents the properties of the variety in far smaller neighborhoods. For example, over the complex numbers, the information available from $\hat R_I$ is (roughly speaking) infor- mation about arbitrarily small neighborhoods in the "classical topology" induced by the fact that the variety is a closed subspace of some $\mathbb C$ with its ordinary topology.

    He follows it up with a nice example, I suggest you give it a look.

  3. Now this time, we are allowed to invert everything in $(a)$. In other words, you may divide by a function that before may have had zeros. What has happened geometrically? We have removed the zeros of that function from $X$. We think of $R_a$ as the regular functions on the so-called standard open set $D(a)=X\setminus Z(a)$. Again, $\mathrm{Spec}$ passes us from functions to points and we think of $\mathrm{Spec}(R_a)$ as the open subvariety $X\setminus Z(a)$. That is, in fact, very accurate.
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Thank you for the answer Jesko. –  Ronald Bernard Apr 4 '13 at 7:25

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