Theorem 3

Two bases of the same vector space $E$ over a field $K$ are equipotent.

We note first that if $E$ admits an infinite basis $B$, it follows from § 1, no. 12, Corollary 2 to Proposition 23 that every other basis of $E$ is equipotent to $B$. We may therefore confine our attention to the case where $E$ has a finite basis of $n$ elements. We note that every monogenous vector space over $K$, not reduced to 0, is a simple $K$-module (I, § 4, no. 4, Definition 7), for it is generated by each of its elements $\neq 0$, by virtue of the relation $\mu a = (\mu \lambda)(\lambda^{-1}a)$ for $\mu \in K$, $\lambda \in K$ and $\lambda \neq 0$. Hence if $(a_i)_{1 \leq i \leq n}$ is a basis of $E$, then $E = \bigoplus_{i=1}^n Ka_i$ up to isomorphism and the subspaces $E_k = \bigoplus_{i=1}^\kappa Ka_i$ for $0 \leq k \leq n$ form a Jordan-Hölder series of $E$, $E_k/E_{k-1}$ being isomorphic to $Ka_k$. Theorem 3 then follows in this case from the Jordan-Hölder Theorem (I, § 4, no. 7, Theorem 6).

A proof can be given independent of the Jordan-Hölder Theorem, by showing by induction on $n$ that, if $E$ admits a basis of $n$ elements, every other basis $B'$ has at most $n$ elements. The proposition is obvious for $n = 0$. If $n \geq 1$, $B'$ is non-empty; then let $a \in B'$. By Theorem 2 (no. 1) there exists a subset $C$ of $B$ such that $\{a\} \cup C$ is a basis of $E$ and $a \notin C$, since $\{a\} \cup B$ is obviously a generating system of $E$. As $B$ is a basis of $E$, $C = B$ is impossible (no. 1, Corollary to Theorem 2) and hence $C$ has at most $n - 1$ elements. Let $V$ be the subspace generated by $C$ and $V'$ the subspace generated by $B' - \{a\}$; $V$ and $V'$ are both supplementary to the subspace $Ka$ of $E$ and hence are isomorphic (§ 1, no. 10, Proposition 13). As $V$ admits a basis with at most $n - 1$ elements, $B' - \{a\}$ has at most $n - 1$ elements by the induction hypothesis and hence $B'$ has at most $n$ elements.