This is question 1.10 of chapter VI in Aluffi's Algebra: "Let $R$ be a commutative ring, and let $F = R^{\oplus B}$ be a free module over $R$. Let $\mathfrak{m}$ be a maximal ideal of $R$, and let $k = R/\mathfrak{m}$ be the quotient field. Prove that $F/\mathfrak{m}F \simeq k^{\oplus B} $ as $k$-vector spaces."
I think I have a proof but I'm not sure if the approach is valid, and if it is, I'm unsure about some of the steps:
Since $k$ is a field, then $B$ is a basis for $k^{\oplus B}$, so I thought that showing that $B + \mathfrak{m}F$ is a basis for $F/\mathfrak{m}F$ should be enough to prove the isomorphism of the vector spaces. If this is not correct, then ignore what comes next; assuming it is, what I did was the following:
First show that any distinct $b_1, b_2 \in B$ (seen as basis elements of $F$) belong to different cosets. Assume the opposite then:
$$b_1 = b_2 + m\sum r_b b = b_2 + \sum m_b b$$
where $r_b \in R$, $m \in \mathfrak{m}$, and each $mr_b = m_b$ are also elements of $\mathfrak{m}$ since it's an ideal. Taking out $b_1$ and $b_2$ from the sigma and factorizing:
$$\sum' m_b b + (m_2+1)b_2 + (m_1-1)b_1= 0$$
Since $B$ is a basis for $F$, it is linearly independent, which implies $m_1 = 1$. Contradiction since $\mathfrak{m}$ is a maximal ideal so it can't contain 1.
Now it's just a matter of showing independence and span. Consider:
$$\sum (r_b + \mathfrak{m})(b + \mathfrak{m}F) = \sum r_b b + \mathfrak{m}F = 0 + \mathfrak{m}F$$
We wish to show this can only happen when $r_b + \mathfrak{m} = 0 + \mathfrak{m}$. From the middle expression we get that $\sum r_b b \in \mathfrak{m}F$ and so:
$$ \sum r_b b = \sum m_b b \implies \sum (r_b-m_b)b = 0 $$
As before, since $B$ is a basis for $F$, the elements $b$ are independent and so $m_b = r_b$, implying that $r_b + \mathfrak{m} = 0 + \mathfrak{m}$ and the independence of $B + \mathfrak{m}F$ (or so I hope).
Finally, let $x \in F/\mathfrak{m}F$, then $x = \sum r_b b +\mathfrak{m}F = \sum (r_b + \mathfrak{m})(b + \mathfrak{m}F)$, where the first equality just comes from translating how elemements of $F/\mathfrak{m}F$ can be written as. (I'm not sure if there may be some trouble with the cosets from jumping between $r_b b +\mathfrak{m}F$ and $(r_b + \mathfrak{m})(b + \mathfrak{m}F)$, but I think it's fine since multiplication by elements of $k$ is well-defined for an $R$-module.) In any case, this means any element can be written as a linear combination of elements in $B + \mathfrak{m}F$ as required.
And so, for any element of $F/\mathfrak{m}F$, there is one in $k^{\oplus B}$ that can be obtained by jumping from one base to the other, implying the isomorphism. That's it, is the strategy valid? If it is, are the steps correct? Any other comments are of course welcome!
Question: " Any other comments are of course welcome!"
Answer: If $F:=\oplus_{i\in I} Re_i$ is a free $R$-module, there are canonical maps
$$\phi: R/I\otimes_R F \rightarrow \oplus_{i} R/I e_i $$
and
$$\psi: \oplus_i R/I e_i \rightarrow \oplus_{i} R/I \otimes_R F $$
defined by $\phi(x \otimes \sum_i a_i e_i):= \sum_i xa_i e_i\in \oplus_i R/I e_i$ and
$\phi(\sum_i x_ie_i):= \sum_i x_i \otimes e_i \in R/I \otimes_ R F.$ You may check that $\phi \circ \psi = \psi \circ \phi =Id$ and hence $\phi, \psi$ are isomorphisms of $R/I$-modules.