Definition of extension of scalars I'm beginning to study commutative algebra following Atiyah\MacDonald's book, and I came across the definition of extension of scalars. I have some problems in grasping why it is defined this way.
Given $A\subseteq B$ two rings and an $A$-module $M$, I can think of $B$ as an $A$-module, therefore I can do the tensor product between $B$ and $M$, obtaining $B\otimes_AM$. Then I can endow $B\otimes_AM$ with a $B$-module structure by saying $$b'(b\otimes x) = b'b\otimes x$$
But what are the pros of defining an extension in this way? I can think of different ways of defining a reasonable action of $B$ over $M$. Intuitivelly I'd say that this kind of construction allows to have nice extensions of $A$-linear maps defined over $M$ into $B$-linear maps defined over $B\otimes_AM$, but it is a (maybe wrong or misleading) intuition I cannot formalize. 
Moreover, once we define the $B$-module structure over $B\otimes_AM$, what are the consequences on the (defining) properties of the tensor product $B\otimes_AM$? Thanks
 A: It might become clearer with an example in mind. Consider $A=\mathbb{R}, B=\mathbb{C}$ and $V$ a real, two dimensional vector space. The dimension is irrelevant, but low dimensions are easier to handle. Now we have to define $(x+iy)v$ where we only have given $xv$ and $yv$. The crucial points are the associative law which requires a definition of $iv$, and the distributive law, i.e. the requirement of bilinearity.
With this example in mind, you should check your statement "I can think of different ways of defining a reasonable action".
The formal proof probably uses the universal property of tensor products. Intuitively it is simply the fact, that a tensor product automatically represents a distributive product, which doesn't affect the original products $xv,yv$. The tensor product is a formalism to write $iv$, namely as $iv=(0+1\cdot i)v=i \otimes v$.
A: We start by looking at some of the other algebraic construction that is often referred to as the "best alternative" in the relevant categories:


*

*Free abelian: If $X$ is a set, the free abelian group $F(X)=\mathbb{Z}^{\oplus X}$ on $X$, together with the canonical injection $\iota:X\to F(X)$, has the following property: For every set map $\varphi:X\to G$ into a group $G$, there exists a unique homomorphism  $\tilde{f}:F(X)\to G$  of abelian groups  such that $\tilde f\circ \iota=f$.

*Field of fractions: If $R$ is an integral domain, the inclusion $\iota:R to K$ of $R$ into its field of fractions $K$ has the following property: for every ring homomorphism $\varphi:R\to F$ into a field, there exists a unique field homomorphism $\tilde{\varphi}:K\to F$ such that $\tilde{\varphi}\circ\iota=\varphi$

*Abelianization: If $G$ is a group, the projection $\pi:G\to G_{ab}=G/[G,G]$ from $G$ to its abelianization has the following property: for every group homomorphism $\varphi:G\to H$ with $H$ abelian, there exists a unique homomorphism $\overline{\varphi}:G_{ab}\to H$ of abelian groups such that $\overline{\varphi}\circ\pi=\varphi$.
Returning to your question, observe the following:

Let $\iota:M\to B\otimes_A M$ denote the $A$-module homomorphism
  $m\mapsto1\otimes m$. For every $B$-module $N$ and
  an $A$-module homomorphism $\varphi: M\to N$, there is a unique
  $B$-module homomorphism $\overline{\varphi}:B\otimes_A\to N$ such that
  $\overline{\varphi}\circ\iota=\varphi$.

Comparing this with the above lists, I am convinced that $B\otimes_A M$ is the best alternative of $M$ in the category of $B$-module. (More formally, one can show that the functor $B\otimes_A -:A\mathsf{-Mod}\to B\mathsf{-Mod}$ is left-adjoint to the forgetful functor $B\mathsf{-Mod}\to A\mathsf{-Mod}$. More information can be found  here and there.)
