Surjective endomorphisms of finitely generated modules are isomorphisms

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.

Solution 1:

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 $T:\mathbb Z\to\mathbb Z:m\mapsto 2m$ !

Solution 2:

The following proof is based on the paper: "Onto Endomorphisms are Isomorphisms", Amer. Math. Monthly 78 (1971), 357-362 by Morris Orzech, Queen's University.

I was told about this paper by KCd in this thread.

The idea is to reduce the theorem to an easy case where $A$ is a Noetherian ring.

Lemma (a slight generalization of Atiyah-Macdonald's Exercise 6.1) Let $A$ be a not-necessarily commutative ring. Let $M$ be a Noetherian $A$-module. Let $N$ be an $A$-submodule of $M$. Let $f\colon N \rightarrow M$ be a surjective $A$-homomorphism. Then $f$ is injective.

Proof Let $K_n = \ker(f^n)$, $n = 1, 2,\dots$. Since $M$ is Noetherian, there exists $n$ such that $K_n = K_{n+1} = \cdots$. Let $x \in K_1$. Since $f$ is surjective, there exist $x_2, \dots, x_n$ such that

$x = f(x_2)$

$x_2 = f(x_3)$

$\dots$

$x_{n-1} = f(x_n)$

$x_n = f(x_{n+1})$

Since $x_{n+1} \in K_{n+1}$, $x_{n+1} \in K_n$. Hence $f^n(x_{n+1}) = 0$. Hence $x = f(x_2) = f^2(x_3) = \cdots = f^n(x_{n+1}) = 0$. QED

Theorem (a generalization of the theorem of Vasconcelos). Let $A$ be a commutative ring. Let $M$ be a finitely generated $A$-module. Let $N$ be an $A$-submodule of $M$. Let $f\colon N \rightarrow M$ be a surjective $A$-homomorphism. Then $f$ is injective.

Proof Let $0 \neq y_0 \in N$. It suffices to prove $f(y_0) \neq 0$. Let $f(y_0) = x_0$.

Let $x_1, \dots, x_n$ be generators for $M$. Let $f(y_i) = x_i$, $i = 1,\dots, n$.

Suppose $f(x_i) = \sum_{j = 1}^{n} a_{i, j} x_j, i = 0, 1,\dots, n$ and $y_i = \sum_{j = 1}^{n} b_{i, j} x_j, i = 0, 1,\dots, n$.

Let $B = \mathbb{Z}[a_{i, j}, b_{i, j}]$. $B$ is a Noetherian subring of $A$.

Let $P = Bx_1 + \cdots + Bx_n$, $Q = By_0 + By_1 + \cdots + By_n$. Since $y_i \in P, i = 0, 1, ..., n, Q \subset P$. Since $f(y_i) = \sum_{j=1}^{n} b_{i, j} f(x_j) \in P, f(Q) \subset P$. Hence $f$ induces a $B$-homomorphism $g\colon Q \rightarrow P$. Since $f(y_i) = x_i, i = 1,\dots, n$, $g$ is surjective. Hence, by the lemma, $g$ is injective. Hence $f(y_0) = g(y_0) \neq 0$ as desired. QED