Let Sch be the category of schemes, and Ring be the category of commutative rings, then Spec and the global sections functor $\Gamma$ are adjoint between these two categories. Ie, for any scheme $X$ and ring $A$ there is a natural bijection $$Hom_{Sch}(X,{\rm Spec}\,A) = Hom(A,\Gamma(X))$$ If $A = \Gamma(X)$, then we get $$Hom_{Sch}(X,{\rm Spec}\,\Gamma(X)) = Hom(\Gamma(X),\Gamma(X))$$ Thus, the identity on the right side corresponds to some special morphism $X\rightarrow{\rm Spec}\,\Gamma(X)$.

Now, the definition of this morphism is given in rather grueling detail in http://stacks.math.columbia.edu/tag/01HX

My question is - how should I think about this intuitively? I suppose one obstacle to thinking about this is the lack of examples. The only examples I'm comfortable with are either affine or projective curves, and in either case this morphism is trivial.

  • 1
    $\begingroup$ Isn't like a point $x$ goes the ideal of points that vanish at $x$? $\endgroup$ – Alex Youcis Sep 7 '16 at 6:29
  • $\begingroup$ I guess there's some baseline motivation for the unit/count (I can never remember which is which) in general, right? Any morphism from $X$ to an affine scheme factors through this thing. $\operatorname{Spec} \Gamma(\mathscr O)$ is the geometric object you get from $X$ by forgetting about everything besides the global functions. $\endgroup$ – Hoot Sep 7 '16 at 15:44
  • $\begingroup$ I would, by the way, like to see interesting uses of this map other than as an intermediary in the proofs of criteria for being affine. $\endgroup$ – Hoot Sep 7 '16 at 15:45
  • $\begingroup$ Since any morphism from a scheme is given by compatible morphisms from an affine cover (of which cover the scheme is an appropriate pushout), you could define $\Gamma(\mathscr O)$ as the ring corresponding to the affine scheme that is the pushout (in affine schemes) of an affine cover. This allows you to translate a formulation schemes glued abstractly from affine schemes (e.g. defined by their functors of points or the gluing construction on affine schemes) to the usual formulation of schemes as certain locally ringed spaces. $\endgroup$ – Vladimir Sotirov Sep 9 '16 at 23:00

The map $z:Hom_{Sch}(X,{\rm Spec}\,\Gamma(X))$ corresponding to the identity $Id:Hom(\Gamma(X),\Gamma(X))$ is easy to interpret geometrically: it associates to a point $x\in X$ the set of global regular functions on $X$ which are zero at $x$.
That set is a prime ideal $j_x\subset\Gamma(X)$, hence corresponds to a point of the target affine scheme ${\rm Spec}\,\Gamma(X)$.
(Note carefully: saying that $f\in \Gamma(X)$ is zero at $x$ means that the class of its germ at $x$ is zero : $[f_x]=0\in \mathcal O_{X,x}/\mathfrak m_x=\kappa(x)$.)

The beauty of scheme theory is that the above description includes the case where $x$ is not closed: then you get the prime ideal $j_x$ corresponding to the set of regular functions zero on a more classical-looking variety, namely the closure $\overline {\{x\}}$ of the point $x$.

In some cases this map $z$ is quite uninteresting: if $X$ is an integral projective (or even complete) scheme, then all $j_x=(0)$ so that the map $z$ sends all points of $X$ to the generic point of ${\rm Spec}\,\Gamma(X)$, in other words it is useless!
The opposite extreme is the case when $X$ is affine: then the map $z$ is an isomorphism of schemes.

The "intermediate" cases can be quite instructive: for example if $X=\mathbb A^2_k\setminus \{\langle x,y\rangle\}\subset \mathbb A^2_k=\operatorname {Spec} k[x,y]$, the punctured affine plane over the field $k$, the map $z$ is almost surjective: its image is all of $\mathbb A^2_k$ except the deleted point $\langle x,y \rangle$.
This constitutes a proof that $X$ is not affine !


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.