# Points on algebraic stacks

I'm a bit confused concerning a definition in Laumon--Moret-Bailly. Perhaps someone could shed some light on the following.

It concerns the definition of (closed) point in Chapter 5. More precisely, in 5.5 they define generization and specialization of points. But what are they really saying there? I mean if both x and y are closed points, then how can one be the generization of the other? What am I missing here?

/Daniel

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Perhaps you could include the definition in your question, for those that don't have a copy of Laumon and Moret-Bailly's book close at hand. – Alberto García-Raboso Dec 11 '09 at 17:55
Ahhh, but what I need to define in that case is a bit overwhelming :) I'm putting my hope to the existence of someone: being knowledgeable in algebraic stacks, having LMB in their close proximity, and finally, reading this. – Daniel Larsson Dec 11 '09 at 18:12
OK, I've found the book and the definition in it. It seems to me, though, that he defines points in general in (5.2), without specifying that they should be closed. In that case, I don't see any conflict with generization and specialization as defined in (5.5). – Alberto García-Raboso Dec 11 '09 at 18:26

Are there several versions of the book? In my copy 5.5 says "un point x de X est une générisation d'un point y (et y est une spécialisation de x) si y est adhérent à {x} dans |X|" and this means exactly that y is contained in the closure of {x}, see http://fr.wikipedia.org/wiki/Adh%C3%A9rence_%28math%C3%A9matiques%29 Also x and y are arbitrary points here, not necessarily closed.

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No it says so in my copy also. But given the way LMB define points (5.2) it seems to me they define closed points: images of spectra of fields (modulo some equivalence relation coming from the stackyness). In my world that is a closed point. – Daniel Larsson Dec 11 '09 at 20:17
Maybe this helps: If X is a scheme, the set of points of X is in canonical bijection to the "set" of morphisms Spec K -> X, where K runs through all fields, modulo the equivalence relation that we identify two morphisms Spec K -> X and Spec K' -> X if there is a field K'' with morphisms Spec K'' -> Spec K' and Spec K'' -> Spec K such that the two compositions Spec K'' -> X coincide. This has nothing to do with closed points and the definition in LMB is the obvious generalization of this fact. – Philipp Hartwig Dec 11 '09 at 20:28
Images of spectra of field are not necessarily closed points: Spec Q -> Spec Z is the inclusion of the generic point of Spec Z, and Q is a field. – Alberto García-Raboso Dec 11 '09 at 20:31
Ah, ok, maybe I see where I'm thinking incorrectly. So in the case of a point in an affine scheme $Spec(A)$ corresponding to a non-maximal ideal $p$, the natural field appearing is the fraction field of the residue ring, i.e., $Frac(A/p)$. I mean if take the schematic analog of your explaination. – Daniel Larsson Dec 11 '09 at 20:43
Good point, Alberto. Thanks. – Daniel Larsson Dec 11 '09 at 20:48

Edit: I now realize this is pretty much the same thing that Alberto said in the comments above.

Remember that closed points of a stack $X$ correspond to the $Spec(K)$-valued points of $X$ where $K$ is an algebraically closed field. In Laumon-Moret-Bailley, they define the points of the stack to be all of the field-valued points of $X$ (including the non-algebraically closed fields). Hence specialization and generization still make sense in this context.

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Ok, I can buy this explaination, if that is what they mean. My main problem though was, given this, what is the $\{x\}$ and how can $y$ belong to this set (see 5.5). They must surely mean $\overline{\{x\}}$ (the closure)? – Daniel Larsson Dec 11 '09 at 18:45
Although, come to think about it, why should closed points be defined as $Spec(K)$-valued points where $K$ is algebraically closed? This is not the case for schemes. – Daniel Larsson Dec 11 '09 at 19:04
I agree that they must mean the closure. – Mike Skirvin Dec 11 '09 at 19:05
@Mike: Closed points do not necessarily have residue fields that are algebraically closed: e.g., Spec Z/p -> Spec Z is the a closed point, but Z/p is not algebraically closed! – Alberto García-Raboso Dec 11 '09 at 20:58
Thanks for the correction, Alberto. Looks like things were cleared up in the answer above, so I think I'm just going to leave my answer as is. – Mike Skirvin Dec 11 '09 at 21:25