Noetherian vector space is finite-dimensional Given a field $k$, and a $k$-vector space $V$ which is noetherian as $k$-module, I want to show that $V$ is finite-dimensional.
Is it correct that this follows because since $V$ is noetherian, every submodule of $V$ is finitely generated as $k$-module, in particular $V$ itself, and hence $V$ is already a finite dimensional vector space?
I'm asking because I've seen another proof which looks more complicated, and I want to know whether my simple approach above is right.
 A: For a vector space, having finite dimension is the same as being finitely generated.
One direction is obvious: if the vector space has a finite basis it clearly is finitely generated.
Suppose $V$ is finitely generated. From any finite spanning set $S$, one can extract a minimal spanning set $B$, in the sense that $B\subseteq S$ is a spanning set and no proper subset of $B$ is a spanning set. Prove that $B$ is linearly independent, hence a basis.
If $V$ is noetherian as a module, then it is finitely generated (along with all of its submodules).

If you don't know or can't use the characterization that a module is noetherian if and only if every submodule is finitely generated, but you can only use the ascending chain condition, you can do it in the following way.
Suppose $V$ is finite dimensional. Then no infinite chain of subspaces can exist, because if $U_1$ is a proper subspace of $U_2$, then $\dim U_1<\dim U_2$.
Assume $V$ is not finite dimensional, which means that no linearly independent finite set spans $V$. Then you can build inductively a chain of subspaces in the following way: since $V\ne\{0\}$, there is $v_1\in V$, $v_1\ne0$.
Assume you have found a linearly independent set $\{v_1,\dots,v_k\}$; then there is $v_{k+1}\in V\setminus\langle v_1,\dots,v_k\rangle$ and the set $\{v_1,\dots,v_k,v_{k+1}\}$ is again linearly independent. Then you have the infinite ascending chain
$$
\langle v_1\rangle \subsetneq
\langle v_1,v_2\rangle \subsetneq
\dots\subsetneq
\langle v_1,\dots,v_k\rangle \subsetneq
\langle v_1,\dots,v_k,v_{k+1}\rangle \subsetneq\dotsb
$$
and $V$ is not noetherian.
