My Problem: Let $M$ be a finitely generated $A$-module and $T$ an endomorphism. I want to show that if $T$ is surjective then it is invertible.
My attempt:
Let $m_1,…,m_n$ be the generators of $M$ over $A$. For every $b = b_1 m_1 + … + b_n m_n$ with $b_i \in A$ there is $a = a_1 m_1 + … + a_n m_n$ with $a_i \in A$ such that
$$ T(a)=b $$
or in matrix-vector notation
$$ T \vec{a} = \vec{b} $$
where $\vec{x}$ is the column vector of $x_1,…,x_n$ where $x = x_1 m_1 + … + x_n m_n$. I multiply by the adjugate matrix to get
$$ \mathrm{adj}(T) \vec{b} = \mathrm{adj}(T) T \vec{a} = \det(T) I_n \vec{a} = \det(T) \vec{a} \ .
$$
Now take $\vec{b}=0$. Then $\vec{0} = \det(T) \vec{a}$ and thus $T$ is injective if and only if
$\det(T)$ is not a zero divisor.
If I prove that $T$ is injective, then I'll get it is invertible. For that, I think the way is to prove that $\det(T)$ is not a zero divisor.
The importance of finitely generated condition:
Let $M = A^{\aleph_0} =\{ ( a_1 , a_2 , … ) \mid a_i \in A \}$ be a not finitely generated $A$-module. Let $T : M \to M$ defined by
$$ T(a_1, a_2, a_3, … ) = (a_2, a_3, … ) \ . $$
Then clearly $T$ is surjective but not injective ($\ker T = \{ ( a , 0 , 0 , … ) \mid a \in A \}$),
and thus not invertible.
The importance of surjective and not injective condition:
Need to find a counter-example.
Best Answer
Yes, a surjective $A$-linear endomorphism $T:M\to M$ of a finitely generated $A$-module $M$ is an isomorphism.
Proof
Consider $M$ as an $A[X]$-module via the multiplication $P(X)*m=P(T)(m)$ , so that for example $(X^2-7)*m=T(T(m))-7m$.
[this is a classical trick used in advanced linear algebra].
Surjectivity of $T$ translates into $M=XM$ and so a fortiori for the ideal $I=XA[X]$ we have $M=IM$.
Now Nakayama's lemma comes to our rescue : it says that there exists an element $XQ(X)\in I$ such that $(1-XQ(X))*m=0$ for all $m\in M$, which means that $m=TQ(T)m$ and this immediately implies that $T$ is invertible with inverse $T^{-1}=Q(T)$.
This result and its extremely elegant proof are due to Vasconcelos.
And if, like so many of us , you keeep forgetting what Nakayama says, look here.
Caveat
Of course an injective endomorphism of a finitely generated module needn't be surjective: take $A=\mathbb Z$ and $T:\mathbb Z\to\mathbb Z:m\mapsto 2m$ !