Is the direct limit of Noetherian rings necessarily Noetherian? And if it is, how to prove this? If it is not, what is a counterexample?

(I was thinking this question: if $A_{m}$ are Noetherian for $m\in \mathbb{N}$, with $A_{m}\subseteq A_{m+1}$ is then $\bigcup_{m \in \mathbb{N}}A_m$ necessarily Noetherian?)

The problem in the bracket is homework (if taken $A_{m}$to be $K_{m}[[x_1,...,x_d]]$, where $K_m \subseteq K_{m+1 }$ as proper field extensions), but the general one is not, so if it is wrong, then I would think of other ways to prove or disprove my homework problem.

And please don't tell me the answer to the specific problem. I don't want to spoil it.

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    $\begingroup$ For your general problem, try to look at $A_m = \mathbb{C}[x_1,\cdots,x_m]$ $\endgroup$ – user27126 Nov 10 '13 at 22:20

In fact, every ring is a direct limit of Noetherian rings (here by "ring" I mean "commutative ring"). For any ring $R$, the set of finitely generated subrings of $R$ is a directed set under inclusion, and $R$ is the direct limit of the directed system they form. But since $\mathbb{Z}$ is Noetherian, any finitely generated ring is Noetherian by the Hilbert basis theorem. This fact is more than just a curiosity--it can sometimes be used to prove statements about general rings by reducing to the Noetherian case. If there's any property that is preserved under direct limits and you can prove it holds for all Noetherian rings, then you get for free that it holds for all rings!

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    $\begingroup$ A (deeper) result with a similar flavor is the Govorov-Lazard Theorem: a module over a commutative ring is flat iff it is a direct limit of free modules (iff it is a direct limit of finitely generated free modules). $\endgroup$ – Pete L. Clark Dec 29 '15 at 7:56

A directed colimit (you don't mean limit) of Noetherian ring is usually not Noetherian. For instance, look at $\mathrm{colim}_{n \in \mathbb{N}} k[x_1,\dotsc,x_n] = k[x_1,x_2,\dotsc]$.

But if $(k_n)_{n \in I}$ is a directed system of fields, then $\mathrm{colim}_n k_n$ is also a field, and it follows that also $\mathrm{colim}_n \, k_n[x_1,\dotsc,x_d] = (\mathrm{colim}_n k_n)[x_1,\dotsc,x_d]$ is Noetherian. The last equation does not hold for formal power series rings. Although $(\mathrm{colim}_n k_n)[[x_1,\dotsc,x_d]]$ is Noetherian, I doubt that the subring $\mathrm{colim}_n \, k_n[[x_1,\dotsc,x_d]]$ is always Noetherian.

  • $\begingroup$ Thanks I believe that it should be. I am thinking on proving it using $(colim_{n}k_n)[[x_1,...,x_d]]$ is the completion of $(colim_{n}k_n)[[x_1,...,x_{d-1}]][x_d]$ with respect to the ideal $(x_d)$. Hope it is right. $\endgroup$ – Alex Nov 10 '13 at 23:37
  • $\begingroup$ How do you want to prove that $\mathrm{colim}_n (k_n[[x]])$ (just one variable) is Noetherian? $\endgroup$ – Martin Brandenburg Nov 11 '13 at 8:20
  • $\begingroup$ because every ideal is generated by a power of x. $\endgroup$ – Alex Nov 11 '13 at 17:32
  • $\begingroup$ Ok how do you prove that? $\endgroup$ – Martin Brandenburg Nov 13 '13 at 10:42

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