There is a functorial approach to define an affine scheme.

Let $\mathbf{LRS}$ and $\mathbf{SRng}$ be respectively the category of locally ringed spaces and a small category of (commutative unital) rings.

The functor $S\colon\mathbf{LRS}\to\mathbf{Set}^{\mathbf{SRng}}$ which sends the locally ringed space $X$ to the functor $Nat(\operatorname{Spec}-,X)$ admits a left adjoint $Rg$ called geometric realisation which sends a functor $F\colon\mathbf{SRng}\to\mathbf{Set}$ to the locally ringed space $\operatorname{\underset{(A,\rho)\in I^°_F}{colim}Spec(A)}$ where $I_F$ is the category with objects the couples $(A,\rho)$, $A\in Ob(\mathbf{SRng})$, $\rho \in F(A)$ and morphisms between $(A,\rho)$ and $(A',\rho')$ the $f\colon A\to A'\in Mor(\mathbf{SRng})$ such that $Ff(\rho)=\rho'$.

An affine scheme can be defined as the geometric realisation of a representable functor, and this definition make sense (in the classical sense) via the Yoneda lemma.


I'm trying to find some examples, so I wanted to find the geometric realisation $Rg(F)$ of the forgetful functor $F\colon\mathbf{SRng}\to\mathbf{Set}$. I think it's the singleton $\{*\}=\operatorname{Spec(0)}$ since $0$ is the terminal object in $I_F$ but i'm not sure of this agrument. Furthermore, can anybody show some other interesting example?

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    $\begingroup$ Your example is correct. It's nice to define projective spaces this way, as here: math.stackexchange.com/questions/121105/… $\endgroup$ – Kevin Carlson Jan 5 '17 at 20:18
  • $\begingroup$ Another reasonment, which seems more correct to me, shows that Rg(F)=Spec($\mathbb Z[x]$). Via Yoneda's lemma and the adjunction proprety we know that if a functor F is represented by an object A, then Rg(F)=Spec(A) since they represent the same functor. So the question becomes, why the other reasoning is wrong? $\endgroup$ – Dinisaur Jan 6 '17 at 14:54
  • $\begingroup$ There's no terminal object in $\bf SRng$ as long as you want $0\neq 1$ (but there's plenty of ways you can obtain the one-point space as a spectrum, of course) $\endgroup$ – Fosco Jan 7 '17 at 9:47
  • $\begingroup$ @dinisaur You're right, I missed that the domain we're colimiting over is the opposite of $I_F$. This the colimit will be the value at the initial object of $I_F$ when it exists, which is $(\mathbb{Z}[x],x)$ for $F$ the forgetful functor. $\endgroup$ – Kevin Carlson Jan 8 '17 at 6:10

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