Is $k[x,y]/(y^2)$ Noetherian or Artinian as a $k$-vector space? As a $k[x]$-module?

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Let $k$ be a field, $R$ be the quotient ring $k[x,y]/(y^2)$. Is $R$ Noetherian or Artinian as a:

(a) $k$-vector space

(b) $k[x,y]/(y^2)$-module

(c) $k[x]$-module

(d) $k[y]/(y^2)$-module

I understand the definitions of Noetherian and Artinian modules but guess I don't quite fully understand the concept as I am unsure of how to approach this question. For part (a), I think that the basis of $R$ as a $k$-vector space is infinite and therefore $R$ cannot be Noetherian or Artinian as a $k$-vector space. For the further parts of the question however I am unsure on how to even begin. Any help would be appreciated.

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$R$ is not Noetherian as a $k$-vector space since it is infinite dimensional. It is also not Artinian for this reason (since we are over a field; generally one may have non finitely generated modules which are Artinian but not Noetherian).

$k[x, y]/(y^2)$ is Noetherian as a module over itself since it is Noetherian as a ring. To see this last fact, use Hilbert basis theorem as well as the fact that ring quotients preserve Noetherian-ness (ideals in the quotient are a subset of ideals in the original). It is not Artinian as a ring since it does not have dimension $0$ (it has the chain $(y) \subsetneq (x, y)$), thus it is not Artinian as a module.

This implies that, for any subring $R'$ of $R$, that $R$ is not an Artinian $R'$-module since every submodule of $R$ as an $R$-module is a submodule of $R$ as an $R'$-module by restriction. In other words, when one has a map $R \to End_R(M)$, one gets a $R'$-module structure as the composition $R' \to R \to End_R(M) \to End_{R'}(M)$.

We claim that $R$ is not a Noetherian $R' = k[y]/(y^2)$ module. To see this, one has the chain of submodules $R' \subsetneq R' \oplus x R' \subsetneq R' \oplus xR' \oplus x^2 R' \subsetneq \dots$.

Finally we are left to check Noetherian-ness for c and d. A $k[x]$-submodule is a subgroup which is closed under the multiplication of $k[x]$. Thus they are classified by the smallest power of $x$ they contain and whether or not they contain $y$. I leave it to you to check that it is Noetherian (one idea: a strict ascending chain of $k[x]$-submodules corresponds to a strict ascending chain of $R$-submodules of at least half the size.)