# Contents

## Idea

The basis theorem for vector spaces states that every vector space $V$ admits a basis, or in other words is a free module over its ground field of scalars. It is a famous classical consequence of the axiom of choice (and is equivalent to it by a result of Andreas Blass, proved in 1984).

## Statement and proofs

###### Basis theorem

If $V$ is a vector space over any field (or, say, a left vector space over a skewfield) $K$, then $V$ has a basis.

###### Proof

We apply Zorn's lemma as follows: consider the poset consisting of the linearly independent subsets of $V$, ordered by inclusion (so $S \leq S'$ if and only if $S \subseteq S'$). If $(S_\alpha)$ is a chain in the poset, then

$S = \bigcup_\alpha S_\alpha$

is an upper bound, for each $v$ in the span of $S$ belongs to some $S_\alpha$ and is uniquely expressible as a finite linear combination of elements in $S_\alpha$ and in any $S_\beta$ containing $S_\alpha$, hence uniquely expressible as a finite linear combination of elements in $S$. Thus the hypothesis of Zorn’s lemma obtains for this poset; therefore this poset has a maximal element, say $B$.

Let $W$ be the span of $B$. If $W$ were a proper subspace of $V$, then for any $v$ in the set-theoretic complement of $W$, $B \cup \{v\}$ is a linearly independent set (for if

$a_0 v + a_1 w_1 + \ldots + a_n w_n = 0 \qquad (w_1, \ldots, w_n \in B)$

we must have $a_0 = 0$ – else we can multiply by $1/a_0$ and express $v$ as a linear combination of the $w_i$, contradicting $\neg(v \in W)$ – and then the remaining $a_i$ are 0 since the $w_i$ are linearly independent). This contradicts the maximality of $B$. We therefore conclude that $W = V$, and $B$ is a basis for $V$.

I’ll write out a proof of the converse, that the axiom of choice follows from the basis theorem, as soon as I’ve digested it – Todd.

###### Equipotence of vector space bases (Steinitz)

Any two bases of a vector space are of the same cardinality.

## Generalisations

Given any linearly independent set $A$ and spanning set $C$, if $A \subseteq C$, then there is a basis $B$ with $A \subseteq B \subseteq C$; the theorem above is the special case where $A = \empty$ and $C = V$. The proof of this more general theorem is a straightforward generalisation of the proof above.

Revised on March 30, 2016 05:35:35 by Zoran Škoda (161.53.28.4)