Extension of Scalars is well-defined The reason I'm asking this, is because as an exercise, I'm asked to prove the following:
Let $A$, $B$ be rings, $f:A\to B$ a ring homomorphism inducing $A$-module structure on $B$, and $M$ a flat $A$-module. Then $M_B:=B\otimes_A M$ is a flat $B$-module.
However, I'm struggling to show that $M_B$ is well-defined as a B-module, that is to say, if $0=\sum_{i=1}^n b_i \otimes_A m_i\in B\otimes_A M$ and $b_0\in B$, that $0=\sum_{i=1}^n b_0 b_i \otimes_A m_i$. If I can prove this, I think I can do the rest of the exercise, but unfortunately, I'm stuck here.
 A: The tensor $\sum_{i=1}^n b_i \otimes m_i$ is zero exactly when, whenever $G$ is an abelian group and $\phi : B \times M \to G$ is a bilinear map such that
$$\forall a \in A, \forall b \in B, \forall m \in M, \phi(ba,m) = \phi(b,am), \tag{*}$$
then $\sum_{i=1}^n \phi(b_i, m_i) = 0$.
Now fix some $b_0$ and some bilinear map $\phi$ as above, and define $\psi : B \times M \to \mathbb{Z}$ by $\psi(b,m) = \phi(b_0 b, m)$. This is still a bilinear map satisfying the condition $(\text{*})$, hence the sum $\sum_i \psi(b_i, m_i)$ vanishes. In other words, for all the possible $\phi$, the sum $\sum_i \phi(b_0 b_i, m_i)$ vanishes. But this exactly means that $\sum b_0 b_i \otimes m_i$ vanishes in $B \otimes_A M$.

The characterization in the first paragraph comes from the definition of the tensor product as satisfying a universal property, see e.g. Wikipedia. 
Suppose that $\sum b_i \otimes m_i = 0$ and take some balanced map $\phi : B \times M \to G$. Using the universal property, this corresponds to some linear map $\phi' : B \otimes_A M \to G$ such that $\phi'(b \otimes m) = \phi(b,m)$ for all $b \in B$ and $m \in M$. But now since $\phi'$ is linear, $\phi'(0) = 0$, hence
$$0 = \phi'\bigl(\sum b_i \otimes m_i \bigr) = \sum \phi'(b_i \otimes m_i) = \sum \phi(b_i, m_i).$$
Conversely, suppose that for all balanced maps $\phi : B \times M \to G$, $\sum \phi(b_i, m_i) = 0$. Then in particular for $G = B \otimes_A M$, consider the canonical balancer map $\otimes : B \times M \to B \otimes_A M$. Then $\sum \otimes(b_i, m_i) = 0$, i.e. $\sum b_i \otimes m_i = 0$.
