Module and Noetherian/Artinian Rings

Question

I am trying to prove that:

Every finitely generated $F$-module $M$ is both Noetherian and Artinian where $F$ is a field.

For this I am looking at the submodules of $F$ and saying that they are in correspondence with the ideals in $F$. Since $F$ has only two ideals it clearly satisfies the condition to be both Noetherian and Artinian. Is this argument correct? I don't see why we need the finitely generated assumption?

Answer 1

Modules over a Noetherian (resp. Artinian) ring need not be Noetherian (resp. Artinian). We can take vector spaces as an example. Consider the vector space $\mathbb{R}^{\infty}$ consisting of all sequences of real numbers where all but finitely many terms are $0$. This has a basis $e_1,e_2,\ldots$ where $e_i$ represents the $i$th coordinate. The submodules generated by $\{e_1,e_2,\ldots\},\{e_2,e_3,\ldots\},\{e_3,e_4,\ldots\}$ etc. form an infinite descending chain, and the submodules generated by $\{e_1\}$, $\{e_1,e_2\}$, $\{e_1,e_2,e_3\}$ etc. form an infinite ascending chain. The way I'd prove that finite-dimensional (finitely generated) vector spaces are Noetherian and Artinian modules is using the concept of dimension.

Answer 2

Since $\mathrm{F}$ is a field, an $\mathrm{F}$-module is the same as an $\mathrm{F}$-vector space (follows by definition). Thus the $\mathrm{F}$-submodules of $M$ are exactly the $\mathrm{F}$-subvector spaces of $M$. Since $M$ is a finitely generated module, $M$ is thus a finitely generated $\mathrm{F}$-vector space, proving that the dimension $\dim_{\mathrm{F}} M$ is finite. Now, every chain of $\mathrm{F}$-subvector spaces of $M$ $$ M_0 \subset M_2 \subset \cdots \subset M_k $$ satisfies $k \leq \dim_{\mathrm{F}} M$ (here $M_i \neq M_{i+1}$ for all $i$). This proves that $M$ is Noetherian and Artinian.