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Why were algebraic geometers in the 19th Century thinking of m-Spec as the set of points of an affine variety associated to the ring whereas, sometime in the middle of the 20 Century, people started to think Spec was more appropriate as the "set of points".

What are advantages of the Spec approach? Specific theorems?

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I don't think that in the 19th century geometers were really thinking of m-Spec. –  Emerton Dec 17 '10 at 2:06
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The basic reason in my mind for using Spec is because it makes the category of affine schemes equivalent to the category of commutative rings. This means that if you get confused about what's going on geometrically (which you will), you can fall back to working with the algebra. And if you have some awesome results in commutative algebra, they automagically become results in geometry.

There's another reason that Spec is more natural. First, I need to convince you that any kind of geometry should be done in LRS, the category of locally-ringed spaces. A locally-ringed space is a topological space with a sheaf of rings ("the sheaf of (admissible) functions on the space") such that the stalks are local rings. Why should the stalks be local rings? Because even if you generalize (or specialize) your notion of a function, you want to have the notion of a function vanishing at a point, and those functions that vanish at a point should be a very special (read: unique maximal) ideal in the stalk. Alternatively, the values of functions at points should be elements of fields; if the value is an element of some other kind of ring, then you're not really looking at a point.

Suppose you believe that geometry should be done in LRS. Then there is a very natural functor LRS→Ring given by (X,OX)→OX(X). It turns out that this functor has an adjoint: our hero Spec. For any locally ringed space X and any ring A, we have HomLRS(X,Spec(A))=HomRing(A,OX(X)) ... it may look a little funny because you're not used to contravariant functors being adjoints. This is another reason that spaces of the form Spec(A) (rather than mSpec(A)) are very special.

Exercise: what if you just worked in RS, the category of ringed spaces? What would your special collection of spaces be? Hint: it's really boring.

Edit: Since there doesn't seem to be much interest in my exercise, I'll just post the solution. The adjoint to the functor RSRing which takes a ringed space to global sections of the structure sheaf is the functor which takes a ring to the one point topological space, with structure sheaf equal to the ring.

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This wasn't my question, but this answer was uncommonly informative. Thanks! –  Jeremy West Dec 16 '10 at 21:11
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I'm impressed by the intrinsicality of this definition of affine schemes. It seems to me that it raises the following naive question. Consider the category of locally "Lie algebra"ed spaces. Does its global section functor have an adjoint ? –  Cédric Bounya Apr 14 '11 at 11:02
    
@Cédric: to get an interesting result, you have to answer the question, "what is a 'local Lie algebra'?" If you don't impose some condition on the stalks of the sheaf of Lie algebras, then the adjoint will simply send a Lie algebra to the one point space with that Lie algebra on it. In the case of rings, we could force the adjoint to be interesting by imposing the condition that the stalks are local rings. –  Anton Geraschenko Apr 14 '11 at 15:34
    
A Lie algebra is local if it has a unique maximal ideal seems a reasonable definition. –  Cédric Bounya Apr 14 '11 at 18:46
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Atiyah-MacDonald, exercise 1.26, mentions one advantage of spec over max-spec: Given a map of rings A -> B, you get a map spec B -> spec A, but not necessarily a map max-spec B -> max-spec A, since the inverse image of a maximal ideal need not be maximal.

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For example, consider the inclusion of k[x] into its fraction field k(x). –  Anton Geraschenko Oct 16 '09 at 16:05
    
@Anton Hartshorne Exercise 2.3.2 works this out in detail. –  David Zureick-Brown Oct 16 '09 at 16:28
    
In general, any domain R (not a field) injects into its field of fractions, F wherein (0) is a maximal ideal but isn't so in R. –  Abhishek Parab Feb 9 '10 at 3:33
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There was some discussion about this (and other things) at the secret blogging seminar fairly recently: http://sbseminar.wordpress.com/2009/08/06/algebraic-geometry-without-prime-ideals/

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For starters is worth noting that in the case of Jacobson rings (and more generally Jacobson schemes) (http://en.wikipedia.org/wiki/Jacobson_ring for instance has a definition) that the spectrum of maximal points is equivalent to the full spectrum.

However, more generally this is not the case and working with non-closed points allows one more flexibility by using arguments relying on generic points for example. Another example is the common technique of reducing arguments to local statements; in general local rings cannot be Jacobson (in other words one should not view a non-artinian local ring as just a closed point).

An example of what can go wrong with the spectrum of closed points is given by the following http://math.berkeley.edu/~ogus/Math%20_256A--08/bigval.pdf where a quasi-affine scheme with no closed point is constructed.

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The reason why $\operatorname{Spec} A$ is an important notion is because it solves the following problem for a commutative ring $A$:

Find a local ring $\mathcal O$ together with a localisation morphism $A \to \mathcal O$ such that every other localisation morphism $A \to B$ to a local ring $B$ factors as a local morphism over $A \to \mathcal O$, i.e. one looks for a kind of universal localisation of $A$.

Stated as above, this problem has no solution at least as long one is not willing to leave the world of rings in the category of sets. However it has a solution in the following more general setting:

There is a topos $X$ endowed with a local ring object $\mathcal O$ and a localisation morphism $A \to \mathcal O$ such that for every other topos $Y$ together with a local ring object $B$ and a localisation morphism $A \to B$ there is a pair of a geometric morphism $f\colon Y \to X$ and a morphism $f^* \mathcal O \to B$ of local rings, which is unique up to a unique natural isomorphism, such that $A \to B$ is given by the composition of $f^* \mathcal O \to B$ and $A \to f^* \mathcal O$.

In fact, the solution to this problem is the topos $X$ of sheaves on $\operatorname{Spec} A$ together with the structure sheaf $\mathcal O_X$ as a local ring object. Now if you replace $\operatorname{Spec} A$ by the max-spectrum, the locally ringed sheaf topos you get will not solve the universal localisation problem in general.

This means that the usual definition of $\operatorname{Spec} A$ with prime ideals is a correct one (as long as one is working in classical logic with the axiom of choice) but it does not mean that it is the only correct definition: You can, for example, replace $\operatorname{Spec}$ by any other topological space or, more generally, by any other site such that the sheaf topos over it is still equivalent to $X$.

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See also here: mathoverflow.net/questions/8204/… –  Peter Arndt Dec 16 '10 at 21:54
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I like this point of view of Spec, but on the other hand, I think it is very far from motivating for, say, an algebraic geometer who works with closed points all the time. It would be great if you add relevance of this universal property. –  Martin Brandenburg Dec 16 '10 at 22:17
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One example: you need generic points for base change to work well. For instance, there is a base change map A^1 / C \to A^1 / Q. The non-algebraic points map to the generic point.

(This is also a problem in rigid analytic geometry where one does use Max Spec and one reason why Berkovich's more general theory of analytic spaces is useful).

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