Let $X=\operatorname{Spec} A,Y=\operatorname{Spec}B$ and $Z=\operatorname{Spec}C$ be affine schemes, with $A,B,C$ commutative rings. According to Wikipedia, the following holds:

$X \times_Y Z\cong \operatorname{Spec}\left( A\otimes_B C \right)$.

Question: What is the tensor products of rings? Do we view $A$ and $C$ as $B$-algebras in some kind of way?


$A\otimes_B C$ does have a structure of $B-$algebra. The ring structure on $A\otimes_B C$ is defined by $(a\otimes c)\cdot(a'\otimes c')=(aa'\otimes cc').$ This has the structure of a $B-$algebra by $$ b(a\otimes c)=(ba\otimes c)=(a\otimes bc)$$ where we used the definition of the tensor product. In particular, there is a structure map $\operatorname{Spec}(A\otimes_B C)\to \operatorname{Spec}(B)$ fitting into the diagram $\require{AMScd}$ \begin{CD} \operatorname{Spec}(A\otimes_B C) @>{}>> \operatorname{Spec}(C)\\ @VVV @VVV\\ \operatorname{Spec}(A) @>{}>> \operatorname{Spec}(B). \end{CD}

Edit: Following the request below, I will just add the following comment: you should notice that the morphism $\operatorname{Spec}(A)\to \operatorname{Spec}(B)$ corresponds to a morphism of rings $f:B\to A$ (using the antiequivalence of categories between commutative unital rings and affine schemes), so that $A$ has the structure of a $B-$algebra. This lets us define $b\cdot a$ for $a\in A$ by $f(b)a=b\cdot a$. The same thing applies for $\operatorname{Spec}(C)\to \operatorname{Spec}(B)$ so that $C$ has a $B-$algebra structure.

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  • $\begingroup$ Thanks but how are $ba$ and $bc$ defined? This would assume that we are treating $A$ and $C$ as $B$-algebras, wouldn't it? $\endgroup$ – test123 Aug 12 at 14:52
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    $\begingroup$ Yes, but in the context of your question, if you have morphisms $X\to Y$ and $Z\to Y$, e.g. morphisms $\operatorname{Spec}(A)\to \operatorname{Spec}(B)$, then there is a ring map $B\to A$ in the category of rings, which is equivalent to $A$ having a $B-$algebra structure. So, I am using this $B-$algebra structure to define $ba$ for $b\in B$ and $a\in A$, and similarly for $bc$. $\endgroup$ – Alekos Robotis Aug 12 at 14:54
  • $\begingroup$ @AlekosRobotis I think your comment about how you define the $B$-algebra structure on $A$ and $C$ via the maps $B\to A$ and $B\to C$ would be a good addition to your answer. $\endgroup$ – KReiser Aug 12 at 19:16
  • $\begingroup$ Okay, I added it for clarity. $\endgroup$ – Alekos Robotis Aug 12 at 19:35

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