Does the category of locally ringed spaces have products?

The category of schemes has all fibered products, but the proof uses affine schemes in a crucial way. I want to understand whether this is true for the category of locally ringed spaces. The standard sources for categorical properties (nLab and Stacks project) do not say or contradict this fact explicitly. Though they say that the fiber product of schemes is automatically fiber product in the category of locally ringed spaces (remark 16.2 in the Schemes chapter of Stacks project).

The simplest locally ringed space which is not a scheme is a single point with a local ring (which is not a field) as a structure sheaf. The tensor product of local rings need not be a local ring, but it seems possible that this tensor product is always the direct product of local rings. For fields this is true, and I was unable to verify it for the general case. Even if this holds (which I highly doubt), it's still possible that more complicated fibered product do not exist. Maybe there is an obvious counterexample which I missed?

• I don't know the immediate answer, but this mathoverflow.net/questions/13616/… overflow post has a comment by @MartinBrandenburg which indicates that it is true. Nov 23 '14 at 6:28
• I've seen that comment, but unfortunately the link to the file (probably containing proofs) doesn't work. Nov 23 '14 at 9:29

I don't know whether the following is available in published form somewhere, I learned it from Jens Franke. I have written notes in german, and Martin has them in english, I think?

Suppose $f: X\to Z$ and $g: Y\to Z$ are morphisms of locally ringed spaces. The fiber product $X\times_Z Y$ of $f$ and $g$ in the category $\textbf{LRS}$ can be described as follows:

• Underlying set: The set underlying of $X\times_Z Y$ is given by $$X\times_Z Y := \{(x,y,{\mathfrak p})\ |\ x\in X, y\in Y, f(x)=g(y)=:z,\\\quad\quad\quad\quad\quad\quad{\mathfrak p}\in\text{Spec}({\mathcal O}_{X,x}\otimes_{\mathcal O_{Z,z}}{\mathcal O}_{Y,y}),\\ \quad\quad\quad\quad\quad\quad\quad\quad\ \iota_{x,y,X}^{-1}({\mathfrak p})={\mathfrak m}_{X,x}, \iota^{-1}_{x,y,Y}({\mathfrak p}) = {\mathfrak m}_{Y,y}\}$$ Here, $\iota_{x,y,X}: {\mathcal O}_{X,x}\to{\mathcal O}_{X,x}\otimes_{{\mathcal O}_{Z,z}}{\mathcal O}_{Y,y}$ and $\iota_{x,y,Y}: {\mathcal O}_{Y,y}\to{\mathcal O}_{X,x}\otimes_{{\mathcal O}_{Z,z}}{\mathcal O}_{Y,y}$ are the canonical maps.

• Topology: For $U\subset X$ and $V\subset Y$ open and $f\in{\mathcal O}_X(U)\otimes_{{\mathcal O}_Z(Z)}{\mathcal O}_Y(V)$ put $${\mathcal U}(U,V,f)\ :=\ \{(x,y,{\mathfrak p})\in X\times_Z Y\ |\ x\in U, y\in V, (\text{im. of } f\text{ in } {\mathcal O}_{X,x}\otimes_{\mathcal O_{Z,z}}{\mathcal O}_{Y,y})\notin {\mathfrak p}\}.$$ This defines the base for a topology on $X\times_Z Y$.

• Structure sheaf: For $(x,y,{\mathfrak p})$ denote ${\mathcal O}_{X\times_ ZY,(x,y,{\mathfrak p})} := ({\mathcal O}_{X,x}\otimes_{{\mathcal O}_{Z,z}}{\mathcal O}_{Y,y})_{\mathfrak p}$. For $W\subset X\times_Z Y$ put $${\mathcal O}_{X\times_Z Y}(W) := \{(\lambda_{x,y,{\mathfrak p}})\in\prod\limits_{(x,y,{\mathfrak p})\in W} {\mathcal O}_{X\times_Z Y,(x,y,{\mathfrak p})}\ |\ \text{for every } (x,y,{\mathfrak p})\in W\text{ there ex. } \\ \text{std. open }{\mathcal U}(U,V,f)\subset W\text{ cont. } (x,y,{\mathfrak p})\text{ and }\mu\in({\mathcal O}_X(U)\otimes_{{\mathcal O}_Z(Z)}{\mathcal O}_{Y}(V))_f\\ \text{s.t. for all }(x^{\prime},y^{\prime},{\mathfrak p}^{\prime})\in{\mathcal U}(U,V,f)\text{ we have } \lambda_{(x^{\prime},y^{\prime},{\mathfrak p}^{\prime})}=\mu_{(x^{\prime},y^{\prime},{\mathfrak p}^{\prime})}\}$$ (The stalk of ${\mathcal O}_{X\times_Z Y}$ at $(x,y,{\mathfrak p})$ it then indeed $({\mathcal O}_{X,x}\otimes_{{\mathcal O}_{Z,z}}{\mathcal O}_{Y,y})_{\mathfrak p})$

• Structure morphisms: One has canonical projections $X\leftarrow X\times_Z Y\to Y$, details ommitted for now.

There are many things to be checked, but maybe you want to think about them yourself to familiarize with the definitions?

• That's.....gross. +1 Nov 23 '14 at 10:02
• The topology is not quite correct, you also need open subsets of $Z$ to enter there. Jan 17 '16 at 16:01
• @MartinBrandenburg Are you sure? What to change? Jan 17 '16 at 19:25
• @Hanno: It's been a few years, but I feel like I'm gonna re-read this answer myself. I think what Martin Brandenburg meant is that you need an arbitrary open subset $W$ of $Z$ and consider only those pairs $(U,V)$ where $U \subseteq f^{-1}(W)$ and $V \subseteq g^{-1}(W)$. Then you have morphisms $\mathcal O_Z(W) \to \mathcal O_X(f^{-1}(W)) \to \mathcal O_X(U)$ and $\mathcal O_Z(W) \to \mathcal O_Y(g^{-1}(W)) \to \mathcal O_Y(V)$ that you can use to tensor both rings over $\mathcal O_Z(W)$. Feb 5 '21 at 6:19
• Hey @PatrickDaSilva: I may well be missing and misremembering stuff - it is a very long time ago - but I don't think it matters whether we consider $f$ in the 'refined' tensor product you suggest, or the one used in the answer, because it only enters the definition through its images in the local tensor products. Feb 5 '21 at 6:40