# proof in English of Proposition on morphisms into affine schemes

Does anybody know where to find an English proof of the following proposition

Let $$(X, \mathcal{O}_X)$$ be a locally ringed space, $$Y = \operatorname{Spec} A$$ an affine scheme. Then the natural map \begin{align}\operatorname{Hom}(X, Y ) &\to \operatorname{Hom}(A, Γ(X, \mathcal{O}_X)),\\ (f, f^\flat) &\mapsto f^\flat_Y ,\end{align} is a bijection.

This is Prop. 3.4 in Görtz's and Wedhorn's Algebraic Geometry I (there the proof is only given for the case where $$X$$ is a scheme) and Proposition 1.6.3 in the 1971 edition of Grothendieck's Eléments de Géométrie Algébrique (full proof given there, but it's in French and AFAIK there is no translation).

• I imagine you ask for a proof in english, because the proofs have no nationally (as far as I know). – xarles Dec 26 '18 at 18:42
• You can find one in the answer to mathoverflow.net/questions/228137/… – xarles Dec 26 '18 at 18:52
• – xarles Dec 26 '18 at 18:54
• @jgon The problem is most textbooks only seem to prove it in the case where $X$ is a scheme – Carlos Esparza Dec 26 '18 at 20:08
• I explain this in great detail in my answer here: math.stackexchange.com/questions/56854/… – Keenan Kidwell Dec 26 '18 at 21:31

Here's my effort at a translation. The original text uses $$S = Y$$, as I will.


It remains to prove that $$\rho_{\text{loc}}$$ and $$\sigma$$ are mutually inverse. However, the definition given above for $$\varphitilde$$ shows immediately that $$\Gamma(\varphitilde) = \varphi$$, and hence $$\rho_{\text{loc}} \circ \sigma$$ is the identity. To see that $$\sigma \circ \rho_\text{loc}$$ is the identity, we begin with a morphism $$(\psi, \theta): X \to S$$ of locally ringed spaces, and set $$\varphi: \Gamma(\theta)$$; the hypothesis that $$\theta_x^\#$$ is local allows us to deduce that this homomorphism, by passing to quotients, is a monomorphism of fields $$\theta^x : \kappa(\theta(\psi(x)) \to \kappa(x)$$ such that, for each section $$f \in A = \Gamma(S, \O_S)$$, we have $$\theta^x(f(\psi(x)) = \varphi(f)(x)$$; the relation $$f(\psi(x)) = 0$$ is thus equivalent to $$\varphi(f)(x) = 0$$, which shows that $$\varphia = \psi$$ by virtue of the definition of $$\varphia$$. On the other hand, the definitions imply that the diagram [see diagram in text] is commutative, and the same is true of the analogous diagram where $$\theta_x^\#$$ is replaced by $$\varphitilde_x^\#$$, hence $$\varphitilde_x^\# = \theta_x^\#$$ (Bourbaki, Alg. comm., chap. II, $$\S 2$$, no. 1, prop. 1), which implies that $$\varphitilde = \theta$$.