Artinian $k$-vector space $\implies$ finite-dimensional (using dcc)

Question

Let $k$ be a field and $V$ a $k$-vector space. Prove that $V$ is Artinian $\implies$ $V$ is finite dimensional.

I have seen proofs of this before but am wondering if it is possible to prove this using just the descending chain condition? I am thinking along the lines of supposing that $V$ is not finite dimensional and prove that it is therefore not Artinian, but I'm unsure of how to go about this.

Any help would be appreciated!

Answer 1

Sledgehammer method

1. artinian implies finitely cogenerated
2. finitely cogenerated implies the socle is finitely generated

As a consequence, any artinian module over a semisimple ring is finitely generated, because it is its own socle.

Lighter weapons.

Assume $V=V_0$ is not finitely generated. Choose $v_1\in V$, $v_1\ne0$. Then $\langle v_1\rangle$ has a complement $V_1\ne\{0\}$ in $V_0$. In particular, $V=\langle v_1\rangle\oplus V_1$.

Choose $v_2\in V_1$, $v_2\ne0$, and a complement $V_2$ of $\langle v_2\rangle$ in $V_1$. In particular, $V=\langle v_1\rangle\oplus\langle v_2\rangle\oplus V_2$, so $V_2\ne\{0\}$.

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The ellipsis should actually be the following:

Suppose we have chosen $v_1,\dots,v_n$ and $V_1\supset V_2\supset\dots\supset V_n$ where $V_k$ is a complement of $\langle v_k\rangle$ in $V_{k-1}$. In particular $V=\langle v_1\rangle\oplus\dots\oplus\langle v_n\rangle\oplus V_n$. Since $V$ is not finitely generated, $V_n\ne\{0\}$. Choose $v_{n+1}\in V_n$, $v_{n+1}\ne 0$ and a complement $V_{n+1}$ of $\langle v_{n+1}$ in $V_n$.

This procedure doesn't stop and creates a strictly decreasing infinite descending chain in $V$.

Answer 2

I'm not sure what an Artinian vector space is, but it's more or less obvious that a vector space that satisfies a dcc on subspaces is finite dimensional.

If $V$ is not finite dimensional there is an infinite independent set $x_1,x_2,\dots$. If you let $V_n$ be the span of the $x_j$ for $j>n$ then it's clear that the sequence $V_n$ violates the dcc.