$\operatorname{rank}(F) = \operatorname{dim}_{k}(\frac{F}{mF})$

Question

Let $R$ be a commutative ring with unit; $m$ is a maximal ideal; $F$ a free $R$-module. We know that $\frac{F}{mF}$ is a vector space over $\frac{R}{m} = k$ .

I have to prove that $\operatorname{rank}(F) = \operatorname{dim}_{k}(\frac{F}{mF})$.

$\operatorname{rank}(F) \geq \operatorname{dim}_{k}(\frac{F}{mF})$ is obvious. I have problems with the other inequality. Any hint ?

Answer 1

Hint: $F/mF\cong F\otimes_R R/\mathfrak{m}$.