I am confused about a line in Vakil's algebraic geometry notes (November 2017 version, page 136) right after he gives the definition of a scheme.
Suppose we have an affine scheme $(X,\mathcal{O}_X)$. By definition, we have that $(X,\mathcal{O}_X)$ is isomorphic to $(\operatorname{Spec}A,\mathcal{O}_{\operatorname{Spec}A})$ for some ring $A$. Vakil states that we "may recover its ring (i.e. find the ring such that $\operatorname{Spec}A=X$) by taking the ring of global sections, as $X=D(1)$, so $\Gamma(X,\mathcal{O}_X)=\Gamma(D(1),\mathcal{O}_{\operatorname{Spec}A}) = A$ (as $D(1)=\operatorname{Spec}A$)."
My question is: why is the equality above an equality and not an isomorphism? By definition of an isomorphism of ringed spaces, we have a homeomorphism $\pi:\operatorname{Spec}A\rightarrow X$ and an isomorphism of sheaves $\mathcal{O}_X\rightarrow \pi_*\mathcal{O}_{\operatorname{Spec}A}$, so don't we just get that $\Gamma(X,\mathcal{O}_X)\cong\Gamma(\operatorname{Spec}A,\mathcal{O}_{\operatorname{Spec}A})\cong A$? Perhaps the significance lies in the identification of $\operatorname{Spec}A$ with $D(1)$? If so, I am not sure how. What am I missing?
Second, Vakil goes on to say "we get more, and can 'recognize $X$ as the scheme $\operatorname{Spec}A':$ we get an isomorphism $\pi:(\operatorname{Spec}\Gamma(X,\mathcal{O}_X),\mathcal{O}_{\operatorname{Spec}\Gamma(X,\mathcal{O}_X)})\rightarrow (X,\mathcal{O}_X)$." Does this follow from the above somehow?
Best Answer
Ravi is being slightly informal. As in the comments, the correct statement is that there's a natural map coming from the adjunction between schemes and affine schemes and this natural map is an isomorphism.