# Geometrical meaning of spectrum in algebraic geometry

1. Sep 8, 2009

### navigator

The notion of spectrum in algebraic geometry seem to be a bit abstract to me. Is it a set of points? Is it the analogue of spectrum in Fourier transform?

2. Sep 8, 2009

### morphism

Maybe it's easier to first think about MaxSpec of a (commutative) ring R. This is the set whose points are the maximal ideals of R. How does this come up? Well, think back to affine varieties, say over $\mathbb{C}$. The main idea of algebraic geometry is, of course, to translate geometry into algebra and vice versa. The first thing you probably learn in this vein is that to each variety $V \subset \mathbb{C}^n$ we can associate a ring $\mathbb{C}[V]$--the coordinate ring of V--which we can use in our geometry-algebra dictionary. Then you learn that, by the Nullstellensatz, the maximal ideals of $\mathbb{C}[V]$ are in 1-1 correspondence with points in V. In other words, the sets $\text{MaxSpec}\mathbb{C}[V]$ and V are pretty much the same. And we can say a bit more: the Zariski topology (which is just a fancy way of keeping track of sub-geometric objects--or subvarieties--of V) induces a topology on $\text{MaxSpec}\mathbb{C}[V]$. This is done as follows. A subvariety U of V gives us an ideal I(U) containing the ideal I(V), and so U corresponds to the maximal ideals of $\mathbb{C}[V]=\mathbb{C}[x_1,\ldots,x_n]/I(V)$ that contain I(U). So we say that a subset of $\text{MaxSpec}\mathbb{C}[V]$ is (Zariski-)closed if it consists entirely of maximal ideals of $\mathbb{C}[V]$ which contain a given ideal of $\mathbb{C}[V]$, and you can check that this indeed generates a topology.

Now, given an arbitrary ring R, we can look at its MaxSpec and give it this type of Zariski topology. And then we try to see how far MaxSpecR behaves as a MaxSpec of a coordinate ring of a variety. And then one notices that using prime ideals instead of maximal ideals has some "benefits."

Finally, as to your other question about a relation to the spectrum "in" Fourier transform, then this depends on what you mean by "spectrum." Personally I've never seen a definition for the spectrum of the transform itself. However, the transform on say the commutative Banach algebras $L^1(\mathbb{R})$ or $L^1(S^1)$ (or indeed on any L^1 of a locally compact abelian group G) is nothing other than the Gelfand transform of these Banach algebras -- something very closely related to the "spectrum" of the algebras. The spectrum specA of commutative Banach algebra A is defined to be the set of maximal ideals of A--surprise, surprise! (Well, not quite. We actually need maximal "modular" ideals: a (closed) ideal I of A is said to be modular if the algebra A/I is unital. If memory serves me, I think for A=L^1 all the maximal ideals are modular, but I could very well be wrong.) One can put a topology on specA and then one has a theorem that gives a map (called the Gelfand transform) $A \to C_0(\text{spec}A)$, where $C_0$ is the space of continuous functions vanishing at infinity. For the case of A=L^1(R) or L^1(S^1), one can prove that specA=R or Z, respectively, and then one sees that the Gelfand transform in this case is the Fourier transform.

Last edited: Sep 8, 2009
3. Sep 8, 2009

### morphism

I just noticed a https://www.physicsforums.com/showthread.php?t=309849 [Broken] where you asked about the relation between the spectrum in alg geom and the spectrum in functional analysis.

In the second part of my post above I gave a connection between the alg geom spectrum (well, a common one, at least) and one notion of spectrum that comes up in functional analysis. But there are other spectra that show up in functional analysis, and I assume the one you had in mind was the most familiar of these, namely the generalization of the set of eigenvalues of a finite-dimensional operator. That is, if T is a (bounded) operator on a Hilbert space $\mathcal{H}$ then its spectrum is defined to be the set $\sigma(T) = \{\lambda \in \mathbb{C} \colon \lambda - T \text{ is not invertible}\}$. Let me show you how one can tie all this together. Let's assume that T is a normal operator. Then the unital, self-adjoint $\mathbb{C}$-algebra generated by T in the space $\mathcal{B}(\mathcal{H})$ of bounded operators on $\mathcal{H}$ (i.e. $\mathbb{C}[I, T, T^\ast] = \{ \text{polynomials over } \mathbb{C} \text{ in } I, T \text{ and } T^\ast\}$) is commutative and thus its closure in the operator norm is a unital commutative Banach algebra (C*-algebra, in fact) which we denote by C*(T). So as I mentioned in my previous post, we can look at specC*(T)={maximal ideals in C*(T)} (here because C*(T) is unital we do not need modularity). The interesting thing, and my main point, is that specC*(T) is homoeomorphic to $\sigma(T)$.

Last edited by a moderator: May 4, 2017
4. Sep 8, 2009

### navigator

Thank you ,I think I have to spend some time to digest what your said:-)

5. Sep 15, 2009

### navigator

I have read some stuff a bit more, and learned the correspondences:
radical ideal <-> variety
prime ideal <-> generic point/subvariety
maximal ideal <-> point
spectrum <-> variety
I am not quite sure whether I have repeated them correctly,and I still confuse about the questions below:
1.What is the difference between radical ideal and spectrum ,or what is the "benefit" of spectrum ,compared to radical ideal.
2.What is the meaning of generic point? a set of points within a subvariety? Why do we use the term "point" to denote a point set?

6. Sep 15, 2009

### Hurkyl

Staff Emeritus
It is a common and useful thing to study topological spaces by turning them into algebraic objects -- for example, studying the ring of continuous, real-valued functions on a manifold.

The reverse is true too -- it is often useful to study algebraic objects by turning them into topological spaces. The (prime) spectrum is one of the more generally useful methods.

The correspondences are just ways of translating problems back and forth between the different domains of discourse.

It's an actual point. Prime spectra are generally non-Hausdorf; the set consisting of a generic point is not a closed set! The closure of a generic point consists of all of the ordinary points in the corresponding subvariety. (along with the generic points of every subvariety of that subvariety)

Among the many uses of a generic point are that they tend to behave like a "typical" point of the variety without having any idiosyncracies.

e.g. many planar curves have singularities -- such as the point (0,0) on the curve y²=x³ or y²=x³-x² -- but the generic point is not a singularity, reflecting the "generic" appearance of the curve. (except possibly for badly behaved cases)

Last edited: Sep 15, 2009
7. Sep 16, 2009

### navigator

Thanks.
:-)

8. Sep 16, 2009

### Hurkyl

Staff Emeritus
Well, the first part of my post was addressing that. The literal difference is obvious -- a radical ideal is an ideal of a ring, and the prime spectrum is a (locally ringed) topological space -- so I assume you weren't asking about that!

Another benefit of the spectrum that I didn't mention (which also explains some of the benefit over just doing ordinary coordinate algebra) is that spectra can describe more subtle things. For example, in 2-dimensional complex space, we might consider the intersection of the two subvarieties defined by
y = 0​
and
y = x²​

The naïve thing is to say they simply intersect at a point -- but if we look at the ideals, these equations generate the ideal
(y,x²)​

and we get more subtle information by defining the intersection to be the spectrum of the ring
C[x,y] / (y, x²)​
or isomorphically
C[t] / (t²)​

The underlying topological space of Spec C[t]/(t²) does, in fact, consist of a single point as you'd expect. But the coordinate ring attached to it remembers the additional information that this is actually a "double point".

Last edited: Sep 16, 2009