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I am new to schemes and I would be very grateful if someone would check my argument below.

Important Note. I am following Ravi Vakil's notes:

https://math.stanford.edu/~vakil/216blog/FOAGfeb0717public.pdf

I am currently on page 136 (exercise 4.3.A). So far we have only defined an isomorphism of schemes as an isomorphism of ringed spaces. We have yet to cover morphisms between schemes and so the in the following solution, I cannot merely say "Spec is a functor from commutative rings to affine schemes and, since functors preserve isomorphisms, we are done!"

Many thanks!

Let $A$ and $B$ be commutative unital rings.

Suppose that $\rho:B \to A$ is a ring isomorphism.

I want to show that there is an induced isomorphism of the affine schemes $$\mathrm{Spec }(A)\to \mathrm{Spec}(B).$$ Since Spec is a contravariant functor from CRing to Top, we see that $\rho$ induces a homeomorphism $$\pi:\mathrm{Spec}(A)\to \mathrm{Spec}(B).$$ I must now construct an isomorphism of sheaves $$\Phi:\mathcal{O}_B\to\pi_*\mathcal{O}_A$$ where I have abbreviated $\mathcal{O}_{\mathrm{Spec}(B)}$ (resp. $\mathcal{O}_{\mathrm{Spec}(A)}$) to $\mathcal{O}_B$ (resp. $\mathcal{O}_A$).

For each $f \in B,$ we define $$D_B(f)=\{P \in \mathrm{Spec}(B) \, : \, f \notin P\}.$$ The collection $\mathcal{B}=\{D_B(f) \, : \, f \in B\}$ is a basis for the Zariski topology on Spec$(B)$.

For each $f \in B$ define $$S_{f,B}=\{g \in B \, : \, D_B(f)\subseteq D_B(g)\}.$$

Observe that

  1. $\pi^{-1}(D_B(f))=D_A(\rho(f))$

  2. $\rho(S_{f,B})=S_{\rho(f),A}$

and so we have

$$\pi_*\mathcal{O}_A(D_B(f))=\mathcal{O}_A(\pi^{-1}(D_B(f)))=S_{\rho(f),A}^{-1}A=(\rho(S_{f,B}))^{-1}A.$$

Hence, for each $f \in B,$ we have an induced isomorphism of rings $$\Phi_{D_B(f)}:S_{f,B}^{-1}B\to (\rho(S_{f,B}))^{-1}A.$$ Moreover, it is clear that, if $D_B(f)\subseteq D_B(g),$ then the following diagram commutes $$\require{AMScd} \begin{CD} S_{g,B}^{-1} B @>{\Phi_{D_B(g)}}>> (\rho(S_{g,B}))^{-1}A\\ @VVV @VVV \\ S_{f,B}^{-1} B @>>{\Phi_{D_B(f)}}> (\rho(S_{f,B}))^{-1}A \end{CD}$$ where the vertical arrows are the ring morphisms arising from the inclusion $S_{g,B} \subseteq S_{f,B}.$

In other words, for each $f\in B,$ we have a ring isomorphism $$\Phi_{D_B(f)}:\mathcal{O}_B(D_B(f))\to\pi_*\mathcal{O}_A(D_B(f))$$ and this collection of ring isomorphisms is compatible with the restriction maps for the base $\mathcal{B}.$

We thus have a morphism of "sheaves on the base $\mathcal{B}$" which thus induces a morphism of sheaves $$\Phi:\mathcal{O}_B\to\pi_*\mathcal{O}_A.$$ Since the induced ring morphisms on stalks $$\Phi_p:\mathcal{O}_{B,p} \to (\pi_*\mathcal{O}_{A})_{p}$$ are all isomorphisms (as each $\Phi_{D_B(f)}$ is an isom), it follows that $\Phi$ is an isomorphism of sheaves.

user350031
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    But $A\mapsto\text{Spec}(A)$ is also a functor (contravariant) from rings to schemes, and functors always preserve isomorphisms. – Angina Seng May 26 '17 at 20:41
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    Absolutely but I haven't proved that yet, so I was trying to do it directly :) – user350031 May 26 '17 at 20:42
  • @user350031 But if you're going to put so much effort into it, why not do it for general ring homomorphisms, not just isomorphisms, and prove you do get a functor. – Angina Seng May 26 '17 at 20:50
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    @LordSharktheUnknown: You are absolutely right! My weird take on it came from an exercise that I was doing. Humouring me, does my approach work? – user350031 May 26 '17 at 20:55
  • @user350031 This general approach should work. Two questions: (i) Did you check that your morphisms between rings of sections commute with the restriction morphisms? (ii) Did you check that your sheaf morphism is a morphism of locally ringed spaces? – Kenny Wong May 26 '17 at 20:59
  • @KennyWong Thanks for your response :) I'm a bit confused by what you mean by "locally ringed spaces''... Since we are in the affine case, we just need a morphism of ``ringed spaces'' which is just a continuous map Spec$(A)\to$ Spec$(B)$ together with a morphism of sheaves $\mathcal{O}B \to \pi*\mathcal{O}_A$ - right?

    In our case (since we have isomorphisms) this is just a homeomorphism Spec$(A)\to$ Spec$(B)$ together with an isomorphism of sheaves $\mathcal{O}B \to \pi*\mathcal{O}_A.$ Unless I'm confused? :)

     – user350031 May 26 '17 at 21:03
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    It's a bit more than that. The stalks of these sheaves are local rings: they each have a unique maximal ideal. At the level of stalks, a morphism of locally ringed spaces must send maximal ideals to maximal ideals. Also, did my first comment about restrictions make sense? – Kenny Wong May 26 '17 at 21:08
  • Your first comment makes perfect sense to me - I should and indeed will check this! I'm still a bit confused about the locally ringed condition though. Maybe it would help if I shared the notes that I'm using:

    https://math.stanford.edu/~vakil/216blog/FOAGfeb0717public.pdf

    I'm looking at page 136: the definition of an affine scheme in 4.3.1.

     – user350031 May 26 '17 at 21:14
  • @user350031 Oh I just realised something - for isomorphisms, we don't need to check this locally ringed condition because it automatically follows from the isomorphisms between stalks. We do, however, need to explicitly check this for general morphisms (like what Lord Shark was talking about): see 4.3.6 and 6.3.1 in the June 11, 2013 notes. So that's why you were focusing on isomorphisms! They are slightly simpler! – Kenny Wong May 26 '17 at 21:23
  • Ah good! I'm glad we've got that sorted out :) Thank you so much for your quick responses and help! – user350031 May 26 '17 at 21:25
  • @user350031 Yeah, the edited version looks good! – Kenny Wong May 27 '17 at 12:30
  • @user350031 You should have proved the fact that $A \mapsto \operatorname{Spec}(A)$ is a contravariant functor in exercise 3.4.H. – user5826 Jul 17 '21 at 21:37
  • @user5826 Exercise 3.4.H shows $A\mapsto\mathrm{Spec}(A)$ is a contravariant functor from rings to topological spaces, not (affine) schemes. Actually, OP uses the conclusion of Exercise 3.4.H right at the beginning of the question. – Kenanski Bowspleefi Sep 17 '23 at 16:34

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