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Let $X,Y,Z$ are affine schemes say $X=\rm{Spec}(A), Y=\rm{Spec}(B), S=\rm{Spec}(R)$ and we have morphism of schemes $f: X\rightarrow S$ and $g:Y\rightarrow S$.

These maps $f: \rm{Spec}(A)\rightarrow \rm{Spec}(R), g: \rm{Spec}(B)\rightarrow \rm{Spec}(R)$ gives maps $m:R\rightarrow A$ and $n:R\rightarrow B$ such that corresponding map $m^{\#},n^{\#}$ on spec equals $f,g$ respectively.

We have inclusion maps $i:A\rightarrow A\otimes_R B$ and $j:B\rightarrow A\otimes_r B$.

This gives $i^{\#}:\rm{Spec}(A\otimes_R B)\rightarrow \rm{Spec}(A)$ and $j^{\#}:\rm{Spec}(A\otimes_R B)\rightarrow \rm{Spec}(B)$ $$\require{AMScd} \begin{CD} \rm{Spec}(A\otimes B) @>i^{\#}>> \rm{Spec}(A)\\ @Vj^{\#}VV @VVf=m^{\#}V\\ \rm{Spec}(B) @>g=n^{\#}>> \rm{Spec}(R) \end{CD}$$

To show that $\rm{Spec}(A\otimes_RB)$ is fibered product, we have to prove atleast that the above diagram commutes i.e., $m^{\#}\circ i^{\#}=n^{\#}\circ j^{\#}$, equivalently $(i\circ m)^{\#}=(j\circ n)^{\#}$, equivalently $i\circ m=j\circ n$ i.e., commutativity of following diagram

$$\begin{CD} R @>m>> A\\ @VnVV @VViV\\ B @>j>> A\otimes_R B \end{CD} $$

and I do not see why should this digram be commutative. For $r\in R$ we have $(j\circ n)(r)=1\otimes n(r)$ and $(i\circ m)(r)=m(r)\otimes 1$ and I do not see why these two should be equal.

Help me to understand this better.

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    Takumi. Thanks.. This looks much better – 2017-02-20
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    No problem! As for your actual question, how are you defining the tensor product? For example, in Atiyah–Macdonald, Prop. 2.12, the tensor product is defined to have the relation you ask for. – 2017-02-20
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    @TakumiMurayama In that theorem, maps are from tensor products to Modules, here maps are from modules to tensor products. How do I relate these two – 2017-02-20
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    @TakumiMurayama : I saw in page 31, there is a result which is exactly the same as this question. But it is left to reader to prove. – 2017-02-20
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    @TakumiMurayama : I was able to prove that.. Thanks.. – 2017-02-20
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    Great! One thing to keep in mind is that the map that Atiyah–Macdonald claim is the structure map $R \to A \otimes_R B$ is not correct, since it is not a ring homomorphism, and that the map in your question is the correct one. Georges Elencwajg [has a MathOverflow post](http://mathoverflow.net/a/42271/33088) explaining this. – 2017-02-20
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    @TakumiMurayama : Thanks for the reference :) :) – 2017-02-26

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