Products of Locally Ringed Spaces

1. Jun 11, 2014

Mandelbroth

I'm relatively new to locally ringed spaces and sheaves. I was aware of them before, but I lacked the mathematical maturity to understand them.

Let $(X,\mathcal{O}_X)$ and $(Y,\mathcal{O}_Y)$ be locally ringed spaces. If I were to take the product of two, what would the corresponding product space look like? I'd imagine that the new underlying topological space would be $X\times Y$. However, I don't know what to do about the structure sheaves.

Any nudges in the right direction would be helpful. Thank you!

2. Jun 11, 2014

micromass

This is not at all trivial! Products do exist, but are difficult to describe. Let $(X,\mathcal{O}_X)$ and $(Y,\mathcal{O}_Y)$ be locally rined spaces. The product is a locally ringed space $(S,\mathcal{O}_S)$ given by

$$S=\{((x,y),\mathfrak{P})~\vert~(x,y)\in X\times Y,~\mathfrak{P}~\text{is a prime ideal of }~\mathcal{O}_{X,x}\otimes \mathcal{O}_{Y,y},~\mathfrak{P}\cap \mathcal{O}_{X,x} = \mathfrak{M}_x,~\mathfrak{P}\cap \mathcal{O}_{Y,y} = \mathfrak{M}_y\}$$

We can endow this set with a topology by choosing a basis as follows:

$$U(U_1,U_2,g) = \{((x,y),\mathfrak{P})\in S~\vert~(x,y)\in U_1\times U_2,~g_{(x,y)}\notin \mathfrak{P}\}$$

with $U_1\subseteq X$ open, $U_2\subseteq Y$ open and $g\in \mathcal{O}_X(U_1)\otimes \mathcal{O}_Y(U_2)$.

The sheaf $\mathcal{O}_S$ is defined by defining the stalks as
$$\mathcal{O}_{S,((x,y),\mathfrak{P})} = (\mathcal{O}_{X,x}\otimes \mathcal{O}_{Y,y})_\mathfrak{P}$$

For more information, see the article "Vollständigkeit von geometrischen Kategorien" by Martin Brandenburg.

3. Jun 11, 2014

Mandelbroth

Wouldn't $\mathfrak{P}$ be unique for each pair $(x,y)$?

Are specifying a basis element $U$, each of which comes from $U_1$, $U_2$, and $g$?

This is the localization, correct?

Where can I find this (preferably in English)? I'm having difficulty finding it with Google.

Last edited: Jun 11, 2014
4. Jun 11, 2014

micromass

Why? Local rings only have a unique maximal ideal, nothing is said about the prime ideals.

Yes, for each $U_1$, $U_2$ and $g$, we have a basis element.

Yes.

5. Jun 14, 2014

Mandelbroth

At the time, I thought, for some reason, that this would collapse to simply the product of topological spaces with a nice sheaf. Clearly, this is not necessarily the case.

I think I'm just lacking intuition for why this is the product. The prime ideal is a little foreign looking to me. Could you help me through an example?

For the sake of familiarity, let's consider the real numbers with its sheaf of smooth functions, $(\mathbb{R}, C^\infty)$. We wish to find the product $(\mathbb{R}, C^\infty)\times (\mathbb{R}, C^\infty)$. The underlying set will be $\mathbb{R}^2$ along with elements of $\operatorname{Spec}(C^\infty_x\otimes C^\infty_y)$ associated to $(x,y)\in\mathbb{R}^2$. That is, a point will look like $((x,y),\mathfrak{P})$. Here's where I hit a roadblock. What are the prime ideals of $C^\infty_x\otimes C^\infty_y$? I cannot think of any.

My wishful thinking about unique prime ideals comes from this example. Intuitively, I'd like the underlying space to just be $\mathbb{R}^2$.

Additionally, is there a simple example, particularly one that is not a manifold, that would be beneficial to work with? I'm slowly making my way through the paper you linked to, but I think an example that does not use my locally Euclidean crutch will go a long way to deeper understanding.

Once again, thank you!

Last edited: Jun 14, 2014
6. Jun 14, 2014

micromass

I think you are now witnessing the fact that the category of locally ringed spaces is not very useful to work with. The products and other constructions do not coincide with the usual constructions on manifolds and algebraic varieties. The complicated form of the product is a striking example of this.

The fact that prime ideals and the spectrum are involved is not unnatural if you're familiar with algebraic geometry. It is however not very natural from the point of view of differential geometry. Locally ringed spaces are useless in differential geometry anyway.

Some simple examples would consist of some basic affine schemes. These are the typical locally ringed spaces. They are still quite complicated though unless your underlying ring is a field or something with very little amount of prime ideals.

As for the the smooth function thing. You should use the theorem that there is a bijective correspondence between prime ideals of a ring $A_\mathfrak{p}$ and the prime ideals of $A$ which are contained in $\mathfrak{p}$.

So you are only wanting to find the prime ideals of $C^{\infty}[\mathbb{R}$ contained in $\{f~\vert~f(x) = 0\}$. I don't know many results on this.

For $C^\infty(\mathbb{R}^2)$, you could consider (connected) embedded manifolds going through the origin like the graph of $y=x^2$ and consider smooth functions vanishing on the embedded manifold. The germs of that should generate a prime ideal.

Things are (like always) easier in the complex case. With complex "manifolds" we have results like the ring germs of holomorphic functions at $0$ in $\mathbb{C}^n$ is isomorphic to the ring of formal power series vanish in $0$. For more information, I recommend the excellent book by Taylor "Several Complex Variables with Connections to Algebraic Geometry and Lie Groups".

In any case, the example of a real manifold (even in 1 dimension) seems complicated. Why not deal with some simple affine schemes first?

Also, if you're interested in ringed spaces (not necessarily locally ringed spaces), then the answer for the product is much nicer. Indeed, given two ringed spaces $(X,\mathcal{O}_X)$ and $(Y,\mathcal{O}_Y)$, the product is given by the product topological space $X\times Y$ with a sheaf whose germs are given by $\mathcal{O}_{X,x}\otimes \mathcal{O}_{Y,y}$.

The forgetful functor $U:LRS/X\rightarrow RS/U(X)$ (where LRS/X and RS/U(X) are the slice categories associated with the category of ringed spaces and locally ringed spaces) has a right adjoint. The right adjoint is not easy to describe and is the reason of products in locally ringed spaces being so complicated. In any case, the right adjoint preserves products, so we can find the products in locally ringed spaces by taking the products in the ringed spaces and applying the right adjoint functor.