I used this "fact" when I hand in my take home exam. To my surprise this was returned with a remark "$M$ is not noetherian!". But I remember the criterion for noetherian is $M$'s submodules are all finitely generated. Then assume $M$ has submodule $M_{1}$, the map $$R^{n}\rightarrow M\rightarrow M_{1}$$ must be surjective since $M$ is finitely generated. So I do not know where I got wrong.

  • 1
    $\begingroup$ What is the map $M\to M_1$? I don't think there is a canonical choice for an arbitrary submodule. $\endgroup$
    – Andrew
    Commented Nov 28, 2012 at 1:19
  • $\begingroup$ Can I use the projection? $\endgroup$ Commented Nov 28, 2012 at 1:24
  • $\begingroup$ For example, if $M=\Bbb Z$ and $M_1=n\Bbb Z$ for some $n\ge 2,$ then what is the map $\Bbb Z\to n\Bbb Z$? $\endgroup$
    – Andrew
    Commented Nov 28, 2012 at 1:53
  • $\begingroup$ @Andrew: I would think the map $m\mapsto nm$ would work fine there.... In fact in any PID $R$ and ideal $I=(a)$ of $R$ there will be a surjective map $R\mapsto I$ given by $r\mapsto ra$. $\endgroup$
    – froggie
    Commented Nov 28, 2012 at 1:56
  • $\begingroup$ @froggie, I agree with you, my contention was mainly with the definite article, i.e. "the" map. Some choices would not work, even in this example. Also, is it obvious that $m\mapsto nm$ deserves to be called a "projection." $\endgroup$
    – Andrew
    Commented Nov 28, 2012 at 2:00

2 Answers 2


What you claim is false; a finitely generated module need not be Noetherian. As a counterexample, consider the polynomial ring $\mathbb{Q}[x_0,x_1,\dots]$ in infinitely many variables as a module over itself. It is finitely generated (by 1), but its submodule $\langle x_0,x_1,\dots\rangle$ isn't, so the module itself isn' Noetherian.

What is true is that a module over a Noetherian ring is Noetherian iff it is finitely generated (this might require the ring to be commutative with unity).

  • $\begingroup$ Im having difficulty understanding why it i generated by <1> - only. What about the $x_i$'s? Does the $x_i$'s and the a/b come from the ring? $\endgroup$
    – Endre Moen
    Commented Dec 3, 2020 at 9:48
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    $\begingroup$ @EndreMoen We're looking at the ring as a module over itself. So 1 can be multiplied by any scalar (including $x_1$ or $x_2$, etc.) to give any element of the ring. $\endgroup$ Commented Dec 4, 2020 at 15:03
  • $\begingroup$ and an element in the module? $\endgroup$
    – Endre Moen
    Commented Dec 4, 2020 at 15:44
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    $\begingroup$ @EndreMoen The ring and the module are one and the same here. $\endgroup$ Commented Dec 4, 2020 at 15:51

If $M_1$ is a submodule of $M$, then there is a canonical injective module homomorphism $M_1\to M$, given by the inclusion map. However, there is not necessarily a surjective module homomorphism $M\to M_1$. This is where your argument went wrong.

As an example, consider the case when $M = R$, so that submodules of $M$ are exactly ideals of $R$. Let $I$ be an ideal of $R$, and suppose that there is a surjective $R$-module homomorphism $f\colon R\to I$. Let $a = f(1)$. Then, since $f$ is surjective, $I = f(R) = Rf(1) = Ra$, that is, $I$ is necessarily a principal ideal, generated by $a$. We can conclude from this that if $I$ is an ideal of $R$ that is not principal, then there cannot be a surjective $R$-module homomorphism $R\to I$.

In general it is true that if $R$ is a Noetherian ring and $M$ is a finitely generated module over $R$, then $M$ is Noetherian. Your argument is close to right. Since $M$ is finitely generated, there is a surjective homomorphism $R^n\to M$, so $M$ is a quotient of $R^n$. Because $R$ is Noetherian, $R^n$ is Noetherian. Since quotients of Noetherian modules are Noetherian, $M$ is Noetherian. This argument fails when $R$ is not a Noetherian ring.


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