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.
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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– Andrew2012-11-28 01:19:02 +00:00Commented Nov 28, 2012 at 1:19
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$\begingroup$ Can I use the projection? $\endgroup$Bombyx mori– Bombyx mori2012-11-28 01:24:53 +00:00Commented Nov 28, 2012 at 1:24
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$\begingroup$ 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– Andrew2012-11-28 01:53:08 +00:00Commented Nov 28, 2012 at 1:53
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$\begingroup$ @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– froggie2012-11-28 01:56:32 +00:00Commented Nov 28, 2012 at 1:56
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$\begingroup$ @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– Andrew2012-11-28 02:00:11 +00:00Commented Nov 28, 2012 at 2:00
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2 Answers
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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).
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$\begingroup$ 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– Endre Moen2020-12-03 09:48:47 +00:00Commented Dec 3, 2020 at 9:48
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3$\begingroup$ @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. $\endgroup$Miha Habič– Miha Habič2020-12-04 15:03:12 +00:00Commented Dec 4, 2020 at 15:03
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$\begingroup$ and an element in the module? $\endgroup$Endre Moen– Endre Moen2020-12-04 15:44:54 +00:00Commented Dec 4, 2020 at 15:44
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4$\begingroup$ @EndreMoen The ring and the module are one and the same here. $\endgroup$Miha Habič– Miha Habič2020-12-04 15:51:15 +00:00Commented Dec 4, 2020 at 15:51
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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.