Dimension of free module tensor residue field

Question

Let $A$ be a ring, and $\mathfrak{m}$ be a maximal ideal in $A$. Then why does $(A/\mathfrak{m}) \otimes_A A^n$ have dimension $n$ as a vector space over $A/\mathfrak{m}$? I can show that the elements $\bar{1} \otimes e_i$ generate this vector space, where $e_i$'s are the generators of $A^n$, but how do I show their linear independence?

Answer 1

As Parthiv commented:

$A/\mathfrak{m} \otimes_A A^n = A/\mathfrak{m} \otimes_A (\oplus_{i=1}^nA) \simeq \oplus_{i=1}^n (A/\mathfrak{m} \otimes_A A) \simeq \oplus_{i=1}^n A/\mathfrak{m} = (A/\mathfrak{m})^n$.