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Suppose I have a morphism of affine schemes $(f, f^\sharp) : \operatorname{Spec} A \to \operatorname{Spec} B$, then my question is - why does the diagonal morphism $(\Delta_f, \Delta_f^\sharp) : \operatorname{Spec} A \to \operatorname{Spec} A \times_{\operatorname{Spec} B} \operatorname{Spec} A = \operatorname{Spec} (A \otimes_B A)$ correspond the following morphism of rings $\phi : A \otimes_B A \to A$ given by $\phi(a \otimes a') = a\cdot a'$?

This was my attempt to try and asnwer that. I know that we have the result that $$\operatorname{Hom}_{\textsf{Sch}}\left(\operatorname{Spec} A, \operatorname{Spec} (A \otimes_B A)\right) \cong \operatorname{Hom}_{\textsf{Ring}}(A \otimes_B A, A)$$ with the bijection given by $(g, g^\sharp) \mapsto g^\sharp_{\operatorname{Spec} A}$ so one possible way to show that the diagonal morphism does correspond to the claimed morphism of rings would be to show that ${\Delta_f}_{_{\operatorname{Spec} A}}^\sharp = \phi$ but I was unable to actually verify this.

Another thing I was thinking of was the following. If I let $(p_1, p_1^\sharp) : \operatorname{Spec} (A \otimes_B A) \to \operatorname{Spec} A$ denote the projection map that comes with the fibered product, then we know that $p_1 \circ \Delta_f = 1_{\operatorname{Spec} A}$ the identity map on $\operatorname{Spec} A$. Now I guess, (but I cannot show), that the morphism $(p_1, p_1^\sharp)$ corresponds to the morphism of rings $\iota : A \to A \otimes_B A$ given by $\iota(a) = a \otimes_B 1_A$, because then we see that $\phi \circ \iota = 1_A$ the identity map on $A$.

If it were the case that the morphism $(p_1, p_1^\sharp)$ corresponds to $\iota$, then since taking global sections is a contravariant functor (and really what that bijection above is) I would see that ${\Delta_f}_{_{\operatorname{Spec} A}}^\sharp\circ \iota = 1_A$ as well, but this isn't enough to prove that ${\Delta_f}_{_{\operatorname{Spec} A}}^\sharp = \phi$.

So I have in addition two further follow up questions in addition to the original one:

  • How does one prove that the morphism $(p_1, p_1^\sharp)$ corresponds to $\iota$?
  • Is there a more efficient way to prove that a morphism of schemes corresponds to a morphism of rings?

If it is relevant, you can assume that the construction of the fibered product I am working with is the same one given in Theorem 3.3 in Hartshorne's book.

Qiaochu Yuan
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Perturbative
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2 Answers2

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Note that $\Delta : \operatorname{Spec}(A) \to \operatorname{Spec}(A \otimes_B A)$ corresponds to a map $\Delta^*: A \otimes_B A \to A$ as you say. The map $\Delta$ is defined via the universal property of the fibre product with $(\mathrm{id}_{\operatorname{Spec}(A)}, \mathrm{id}_{\operatorname{Spec}(A)})$, so $\Delta^*$ is given (recall that the correspondence is natural, it is an equivalence of categories) by $(\mathrm{id}_A, \mathrm{id}_A)$ via the universal property of pushouts. In particular, it is the unique $B$-algebra homomorphism such that $$ a \otimes 1 \mapsto a \quad\text{and} \quad 1 \otimes a \mapsto a$$ for every $a \in A$. In other words, $\Delta^*: a \otimes a' \mapsto aa'$.

The important part here is that the correspondence $$\operatorname{Hom}_{\mathbf{AffSch}}(\operatorname{Spec}(A), \operatorname{Spec}(B)) \cong \operatorname{Hom}_{\mathbf{Ring}}(B,A)$$ is not only a bijection but rather a natural bijection, it comes from an equivalence of categories $\mathbf{AffSch} \simeq \mathbf{Ring}^{\mathrm{op}}$. It really preserves diagrams! That should answer your first additional question. Unfortunately, this is often not stressed enough. In most textbooks I only see a proof that there is a bijection whereas the naturality is extremely important. It was definitely one part that wasn't entirely clear to me until I realized that the functoriality is an essential part of this assertion and is used over and over again.

I do not know of a more efficient way to prove it than to define the mutually inverse maps and checking that they define an equivalence of categories by hand. I haven't thought about it but you can probably also check that there is a fully faithful, essentially surjective functor $\mathbf{Ring}^{\mathrm{op}} \to \mathbf{AffSch}$, but the first approach should still be the preferred approach to understand both maps in the correspondence.

Qi Zhu
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    You could view the fact as a specialization of the fact that $\operatorname{Spec}$ is adjoint to the global sections functor $X \mapsto \Gamma(X, \mathscr{O}X)$. (Whether it's a right or left adjoint is a bit muddled since $\operatorname{Spec}$ and the global sections functors are contravariant - but the point is you have a natural isomorphism $\operatorname{Hom}{\mathbf{Sch}}(X, \operatorname{Spec} A) \simeq \operatorname{Hom}_{\mathbf{CommRing}}(A, \Gamma(X, \mathscr{O}_X))$ more generally.) – Daniel Schepler Jan 06 '21 at 09:21
  • Which also leads to an argument which sounds different at first but amounts to the same thing in the end: as an adjoint functor, $\operatorname{Spec}$ must transform limits into colimits, including that the projection morphisms must transform into the coprojection morphisms. – Daniel Schepler Jan 06 '21 at 09:26
  • Thank you very much for your answer! So when you say that "$\Delta^{}$ is given by the universal property of pushouts", do you mean by the universal property of pushouts in the category of Rings here? And furthermore, when you say it is the unique $A$-algebra homomorphism, did you get that just by showing that the given $A$-algebra homomorphism satisfied the universal property of pushouts in the category of Rings and so it would by $\Delta^$ by uniqueness? If you could expand on this slightly, it would be very helpful! – Perturbative Jan 07 '21 at 20:46
  • @Perturbative Yes, that's the universal property in $\mathbf{Ring}$. It being the unique $A$-algebra homomorphism is a reformulation of the fact that it is the universal morphism given by the pushout. (The universal property for $A_1 \otimes_B A_2$ as a pushout in $\mathbf{Ring}$ corresponds to the universal property of the coproduct in $\mathbf{Alg}_B$, the category of $B$-algebras.) – Qi Zhu Jan 08 '21 at 06:25
  • I also see and have fixed a typo, it should be a $B$-algebra homomorphism. – Qi Zhu Jan 08 '21 at 06:51
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Question: "How does one prove that the morphism (p1,p♯1) corresponds to ι?"

Answer: By the universal property of the fiber product, it follows the diagonal map $\Delta: X\rightarrow X\times_S X$ at the level of schemes, is the unique map commuting with the projection maps $p,q:X\times_S X \rightarrow X$. The multiplication map has the same property: The map $m:A\otimes_k A\rightarrow A$ defined by $m(a\otimes b):=ab$ is the unique map commuting with the maps $p,q:A \rightarrow A\otimes_k A$ defined by $p(a):=a\otimes 1, q(a):=1\otimes a$. Hence it follows the maps $\Delta$ and $m$ determine each other when $X:=Spec(A)$.

hm2020
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