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For any ring $A$, define a functor $\text{Spec}(A)$ from rings to sets by $$\text{Spec}(A)(R) = Hom_{\text{Rings}}(A,R)$$

Call a functor $X$ an affine scheme if it is isomorphic to a functor of the form $\text{Spec}(A)$.

The Yoneda lemma says that for any such functor $X$, and any ring $A$, the set of natural transformations from $\text{Spec}(A)$ to $X$ bijects with $X(A)$.

Let $\mathscr{O}_X$ be the class of natural transformations from $X$ to $\mathbb{A}^1$, where $\mathbb{A}^1$ is the forgetful functor from Rings to Sets.

The question is given these definitions/theorems, show that $$\text{Spec}(A \otimes B) = \text{Spec}(A) \times \text{Spec}(B)$$

(I guess equivalently, $\mathscr{O}_{A \times B} = \mathscr{O}_A \otimes \mathscr{O}_B$)

I guess the obvious way is to show that this definition of affine schemes is the same as the usual algebraic geometry one, which in turn is equivalent to the opposite category of rings, and then it follows because the coproduct of rings is just the tensor products.

Can one do it directly from the definitions as above?

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up vote 3 down vote accepted

Start with the map $\text{Hom}(A,R) \times \text{Hom}(B,R) \rightarrow \text{Bilin}(A \times B, R)$, given by sending $(f,g)$ to $(f \times g)(a,b) = f(a)g(b)$. Then identify $\text{Bilin}(A \times B, R)$ with $\text{Hom}(A \otimes B, R)$ in the usual way. Then show that this does indeed define an isomorphism of functors $$\text{Spec}(A) \times \text{Spec}(B) \overset{\cong}\rightarrow \text{Spec}(A \otimes B).$$

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