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.


5 Answers 5


Yes, a surjective $A$-linear endomorphism $T:M\to M$ of a finitely generated $A$-module $M$ is an isomorphism.

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 keep forgetting what Nakayama says, look here.

Of course an injective endomorphism of a finitely generated module needn't be surjective: take $A=\mathbb Z$ and $T:\mathbb Z\to\mathbb Z:m\mapsto 2m$ !

  • $\begingroup$ 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. $\endgroup$
    – LinAlgMan
    Commented Nov 17, 2012 at 19:24
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    $\begingroup$ You're welcome. I've added a counter-example of the type you require in an edit called Caveat. $\endgroup$ Commented Nov 17, 2012 at 19:32
  • $\begingroup$ I know its an old post, but I have a question. Could you explain why subjectivity of $T$ translates into $M=XM$, I dont see this, and why is $M=IM$. I would really appreciate an answer. Thanks. $\endgroup$
    – user117449
    Commented Oct 4, 2014 at 16:11
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    $\begingroup$ Dear @George, surjectivity of $T$ means $M=T(M)$ and this translates as $M=X\star M$or more concisely $M=X M$. The other point is that since $X\in I=(X)A[X]\subset A[X]$, we have $M=X M\subset IM\subset M$, so that $IM=M$. $\endgroup$ Commented Oct 4, 2014 at 20:29
  • $\begingroup$ @GeorgesElencwajg great answer! Under these conditions (i.e. $M$ is a finitely generated $A$-module and $T$ is a surjective endomorphism), is it true that $T$ is injective?? $\endgroup$ Commented Jun 9, 2015 at 11:57

Here is a quick and easy proof which I have found at the stacks project: If $M$ is cyclic, then $M \cong A/I$ for some ideal $I$. Then replace $A$ by $A/I$ so that wlog $M = A$. But then it's easy. In general, we do induction, but first we do the trick mentioned in Georges Elencwajg's answer: We endow $M$ with an $A[X]$-module structure such that multiplication with $X$ is the given endomorphism $T$. Thereby, we may assume that $T$ is multiplication with an element $X$ of the base ring $A$ - this means that every submodule is stable under this endomorphism(!). So we just pick some generator, look at its generated submodule $M'$ and apply the Five Lemma to $$\begin{array}{c} 0 & \rightarrow & M' & \rightarrow & M & \rightarrow & M/M' & \rightarrow & 0 \\ & & X\downarrow ~~~~&& X\downarrow ~~~~ && X \downarrow ~~~~ & & \\ 0 & \rightarrow & M' & \rightarrow & M & \rightarrow & M/M' & \rightarrow & 0\end{array}$$ and we are done.


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

$x = f(x_2)$

$x_2 = f(x_3)$


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

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

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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    $\begingroup$ I don't understand how the $f(x_i)$ in the proof of the Theorem make sense... Isn't $f$ only defined on $N$ ? $\endgroup$ Commented Dec 13, 2014 at 2:01
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    $\begingroup$ I think the proof of the Lemma has a similar mistake. The iterated function $f^n$ is used. This is not defined when $N \neq M$. That being said, it might even fail for noetherian rings. Cont'd: math.stackexchange.com/questions/1065786 $\endgroup$ Commented Dec 13, 2014 at 11:07

Consider the commutative ring $B\subset\mathrm{End}_A(M)$ generated by $T$ and all multiplication by $a\in A$ maps, i.e. the maps $\mathsf{m}_a:M\to M, m\mapsto am$. Let $m_1,\dots,m_n$ generate the $A$-module $M$. We can find a matrix $(b_{ij})\in M_n(A)$ with $$\forall j\in\{1,\dots,n\},\quad Tm_j=\sum_{i=1}^n b_{ij}m_i$$ The usual "multiplication by the comatrix" trick done in the commutative ring $B$ shows that $T$ satisfies a monic polynomial equation $$T^n+\sum_{i=0}^{n-1}b_i T^i=T^n+\cdots+(-1)^n\mathsf{m}_\beta=0$$ with coefficients $b_i$ in $A$ and where we wrote $\beta=\det(b_{ij})$.

Since $T$ is onto, the $Tm_i$ generate $M$ and there is a matrix $(a_{ij})\in M_n(A)$ with $$\forall j\in\{1,\dots,n\},\quad m_j=\sum_{i=1}^n a_{ij}Tm_i$$ so that $$(a_{ij})(b_{ij})\begin{pmatrix}m_1\\\vdots\\m_n\end{pmatrix}=\begin{pmatrix}m_1\\\vdots\\m_n\end{pmatrix}$$ and so (again using the comatrix trick) $\mathsf{m}_\alpha\circ\mathsf{m}_\beta=\mathrm{id}_M$ where $\alpha=\det(a_{ij})$. Therefore, composing the monic polynomial relation from earlier with $\mathsf{m}_\alpha$, we get $$\mathrm{id}_M=ST=TS$$ with $S=(-1)^{n-1}\alpha\big(T^{n-1}+\sum_{i=1}^{n-1}b_i T^{i-1}\big)$ showing that $T$ is invertible with inverse $S$.


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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    $\begingroup$ 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. $\endgroup$ Commented Nov 17, 2012 at 18:18
  • $\begingroup$ There must be more elementary proof (e.g. one that doesn't use the Snake Lemma which is behind the scope of the course). $\endgroup$
    – LinAlgMan
    Commented Nov 17, 2012 at 18:39
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    $\begingroup$ @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. $\endgroup$ Commented Nov 18, 2012 at 14:55
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    $\begingroup$ Dear Frederik, thanks for the clarification and congratulations for your last sentence which shows intelligence and honesty, a not so common combination: +1 $\endgroup$ Commented Nov 18, 2012 at 18:10
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    $\begingroup$ An elementary (even constructive) proof of the free case is in Richman's little gem Nontrivial uses of trivial rings. $\endgroup$ Commented Feb 8, 2016 at 17:07

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