Consequence of Nakayama's Lemma Let $R$ be a local ring with maximal ideal $M$. Let $A$ be a finitely generated $R$-module, and $x_1, ..., x_n$ be a minimal generating subset of $A$. I would like to prove that $x_1 + MA, ... , x_n + MA$ is a basis of $A/MA$ over $R/M$.
The book I am using gives a hint to use Nakayama's Lemma, and I used it to show that $x_1 + MA, ... , x_n + MA$ spans $A / MA$. But how do show that $x_1 + MA, ... , x_n + MA$ are linearly independent? 
 A: $R/M$ is a field, if they were dependent, one would be a combination of the others, contradicting the minimality of the generating set.
To clarify the argument we have 
$$x_1=r_2x_2+\cdots +r_nx_n+m_1x_1+\cdots +m_nx_n$$ which gives
$$(1-m_1)x_1=r_2x_2+\cdots +r_nx_n+m_2x_2+\cdots +m_nx_n$$ and $1-m_1$ is invertible as $M$ is maximal.
A: The other answer is missing some steps, so I'm going to elaborate:

Suppose $x_1, \dots, x_n$ is a minimal generating set for $A$. Then the images $x_1 +MA, \dots, x_n+MA$ generate $A/MA$ over the field $R/M$. We want to show that $x_1 +MA, \dots, x_n+MA$ is a basis for $A/MA$ over $R/M$ and so all that is left to show is that they are linearly independent.
Arguing by contraposition, suppose $x_1 +MA, \dots, x_n+MA$ are linearly dependent. Then we can write
$$
\begin{align}
x_1 +MA &= (r_2+M)(x_2 + MA) + \cdots + (r_n+M)(x_n +MA)\\
&=(r_2x_2 +MA) + \cdots + (r_nx_n + MA)\\
&=(r_2x_2 + \cdots + r_nx_n)+MA.
\end{align}$$
Therefore, $x_1-(r_2x_2 + \cdots + r_nx_n) \in MA$ and since $x_1, \dots, x_n$ generates $A$ we can write $$x_1-(r_2x_2 + \cdots + r_nx_n) = m_1x_1 + \cdots +m_nx_n$$ where $m_i \in M$. So,
$$(1-m_1)x_1 = (r_2+m_2)x_2 + \cdots + (r_n+m_n)x_n$$
and since $m_1 \in M$, then $1-m_1 \notin M$ hence $1-m_1$ is a unit. So, we can write
$$x_1 = s_2 x_2 + \cdots s_nx_n$$
where $s_i = (1-m_1)(r_i+m_i)$. So, $x_1, \dots, x_n$ is not minimal.
By the contrapositive, if $x_1, \dots, x_n$ is a minimal generating set for $M$, then $x_1 +MA, \dots, x_n+MA$ are linearly independent.
