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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.

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In your last example, $A[x]$ is not a finitely generated $A$-module. –  Fredrik Meyer Nov 17 '12 at 17:45
    
@FredrikMeyer Thanks, I edited that. I need to find a counter example for an injective endomorphism which is not invertible. –  LinAlgMan Nov 17 '12 at 18:21
    
The following thread may be useful. math.stackexchange.com/questions/153516/… –  Makoto Kato Nov 17 '12 at 19:54
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3 Answers

up vote 8 down vote accepted

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

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Thank you very much, this proof is much more related to what we learned in class. Now I need to find a counter-example for the case when $T$ is injective and not surjective. –  LinAlgMan Nov 17 '12 at 19:24
    
You're welcome. I've added a counter-example of the type you require in an edit called Caveat. –  Georges Elencwajg Nov 17 '12 at 19:32
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The following proof is based on the paper: "Onto Endomorphisms are Isomorphisms", Amer. Math. Monthly 78 (1971), 357-362. 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

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By an application of the snake lemma, you can reduce to the case $M=A^r$, i.e. $M$ is a free $A$-module.

So you have an exact sequence $$ 0 \to \ker T \to A^r \stackrel T \to A^r \to 0 $$

These are free modules, so the sequence splits, i.e. we can write the middle one as $\ker T \oplus \text{im } T$:

$$ 0 \to \ker T \to \ker T \oplus A^r \stackrel T \to A^r \to 0 $$

Now, the only way dimensions can add up here, is if $\ker T = 0$. So $T$ is an isomorphism. (this is perhaps unnecessarily complicated, but it works) (this was wrong, but see the comments for the details on how to reduce to free modules. Perhaps they admit an easier proof.)

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I don't understand this proof: how do you reduce to free modules and to what do you apply the snake lemma? What do you mean by "the only way dimensions can add up..."? Modules don't have dimension in general. –  Georges Elencwajg Nov 17 '12 at 18:18
    
There must be more elementary proof (e.g. one that doesn't use the Snake Lemma which is behind the scope of the course). –  LinAlgMan Nov 17 '12 at 18:39
    
No, I don't think there is an elementary proof of this result. –  Georges Elencwajg Nov 17 '12 at 18:48
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@Georges: To reduce to free modules: by finite generation we have a surjective map $A^r \to M$. Lift the map $T:M \to M$ to a map $\bar{T}:A^r \to A^r$, and draw the appropriate diagram. Applying the snake lemma, one sees that that $\ker \bar{T} = \ker T$. So the kernel is trivial if and only if the kernel of the lifted map is trivial. But I see now that my second argument is nonsense. –  Fredrik Meyer Nov 18 '12 at 14:55
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Dear Frederik, thanks for the clarification and congratulations for your last sentence which shows intelligence and honesty, a not so common combination: +1 –  Georges Elencwajg Nov 18 '12 at 18:10
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