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Feb 28, 2012 at 9:01 vote accept Christopher Townsend
Feb 28, 2012 at 9:01
Feb 26, 2012 at 17:28 comment added Anton Fetisov @Yosemite Sam, thank you for the pointer! My answer is more or less a reformulation of @Sasha 's one. It's just a step forward: you don't need to check his equality for all l.r.s. $Y$, it suffices to check only affine ones, because all (pre)sheaves are colimits of representables. And then it becomes exactly the representability condition for $X$.
Feb 26, 2012 at 17:20 comment added Anton Fetisov @Qiaochu Yuan, I didn't really describe what a scheme is, I just pointed the direction. There is Zariski-type Grothendieck topology on the category of affine schemes and the corresponding sheaves are schemes. This allows to associate a topological space to any scheme. The structure sheaf appears as a sheaf of morphisms from an open subset to the affine line (affine scheme represented by $\mathbb{Z}[T]$). So description of affine schemes as representable functors is meaningful, although possibly (like any representability) too difficult to check in practice without more concrete theorems.
Feb 24, 2012 at 1:24 comment added Yosemite Sam @Anton: although I agree that locally ringed spaces are evil, I think the OP was asking a different question. (by the way, another great place to learn about the functor-of-points POV is the master course on stacks to be found on Toen's webpage)
Feb 24, 2012 at 1:14 comment added Qiaochu Yuan Aren't you more or less describing affine schemes as the opposite of the category of commutative rings? I don't see how ringed spaces really enter into this description.
Feb 24, 2012 at 1:10 history answered Anton Fetisov CC BY-SA 3.0