# “Vector spaces” over a skew-field are free?

Are modules over a skew field free?

That is, if $F$ is a skewfield then can any module $M$ be written as $\underset{i \in I}{\bigoplus} F$ for some indexing set $I$?

• The answer is yes. The proof is not hard, do you need a hint? – Nick Feb 22 '14 at 20:36
• Actually... a category of modules over a ring contains only free objects iff the base ring is a division ring :) – AIM_BLB Feb 22 '14 at 20:52

Yes, the same proof as usual works. If $V$ is a vector space over a skew field $F$, then $V$ has a maximal linearly independent subset $B$ (Zorn's Lemma). It is a generating set: If $v \in V$, then either $v \in B$ and we are done, or $B \cup \{v\}$ is linearly dependent (by maximality). Since $B$ is linearly independent, it follows $\lambda v \in \langle B \rangle$ for some $\lambda \in F \setminus \{0\}$, hence (since $\lambda$ is invertible!) $v \in \langle B \rangle$.
Only in the very last step we use that $F$ is a skew field. For other rings the proof breaks down. For example, the empty set is a maximal linearly independent subset of the $\mathbb{Z}$-module $\mathbb{Z}/2$, but of course it is not a generating set.