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Let $R$ be an integral domain and also Noetherian ring. Let $f$ be an element in Quot$R$ (field of fractions of $R$). Let $R[f]$ be the subring of Quot$R$ generated by $R$ and $ \{ f \}$. Let suppose that $R[f] \subseteq M$ where $M$ is an $R$-submodule of Quot$R$ and $M$ is finitely generated as $R$-module. Is it true that $R[f]$ is also finitely generated as $R$-module?

If is it true, can you give me a rigorous and possibly elementary proof?

Minato
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2 Answers2

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The following conditions are equivalent for the commutative ring $R$:

  1. $R$ satisfies the ascending condition on ideals
  2. every ideal of $R$ is finitely generated
  3. every submodule of a finitely generated submodule is finitely generated

If $R$ satisfies one (hence all) of these properties, it is said to be Noetherian.

egreg
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  • I didn't know that the 3. statement is equivalent to 1. and 2. I will try to prove this equivalence. Meanwhile, i was tinking:
    1. https://math.stackexchange.com/questions/533873/finitely-generated-modules-over-a-noetherian-ring-are-noetherian
    2. So $M$ is Noetherian
    3. Noetherian is closed by submodules, so $R[f]$ is Noetherian as $R$-module
    4. Noetherian modules are always finitely generated https://math.stackexchange.com/questions/2331871/are-noetherian-modules-over-noetherian-ring-and-artinian-modules-over-artinian-r

    Are they true?

     – Minato Nov 11 '17 at 19:00
  • In this case i have proved the "strong" fact that $R [f]$ is noetherian as $R $ - module, hence also finitelty generated. But actually sems that if R is Noetherian, then for an R module it is the same to be noetherian or finitely generated, so it isn't a really strong fact. – Minato Nov 11 '17 at 19:17
  • @Minato Yes: when you have proved that 2 implies 3, you have that $R[f]$ is also noetherian. The hint is to prove the fact first for $R^n$. – egreg Nov 11 '17 at 20:15
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Here is a quick outline at a pretty basic level. First, a finitely generated module over $R$ is also Noetherian (acc on modules). You can do that by induction on the number of generators and the fact (pretty easy to verify) that if $M$ is an $R$-module, $M'$ a submodule of $M$, then $M$ is noetherian if and only if $M'$ and $M''$ are and take $M'$ to be the submodule generated by one element of your generating set for $M$). Now use this same fact applied to $R[f]\subseteq M$.

John Brevik
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