## As an F-vector space, an infinite direct sum of F has strictly smaller dimension than an infinite direct power of F over the same index set

Let $F$ be a field. Prove that a vector space $V$ over $F$ having basis $B$ (regardless of the cardinality of $B$) is isomorphic as a vector space to $\bigoplus_B F$. Prove that $\prod_B F$ is also an $F$-vector space which has strictly larger dimension than that of $\bigoplus_B F$.

(So a free module on any set is isomorphic to a direct sum. We’ve never gotten around to proving this in the best possible generality, though, so we’ll just prove it for vector spaces here.)

Note that, by the universal property of free modules, the natural injection $B \rightarrow \bigoplus_B F$ which sends $b$ to the tuple which has 1 in the $b$th component and 0 elsewhere induces a vector space homomorphism $\varphi : V \rightarrow \bigoplus_B F$. This mapping is clearly surjective, and also clearly injective. So $V \cong_F \bigoplus_B F$.

Certainly $\prod_B F$ is an $F$-vector space which contains $\bigoplus_B F$, so that $\mathsf{dim}\ \bigoplus_B F \leq \mathsf{dim}\ \prod_B F$. Suppose these two dimensions are in fact equal. Identify $B$ with the usual basis of $\bigoplus_B F$. By this previous exercise, there is a basis $D$ of $\prod_B F$ which contains $B$, and as argued above, $\prod_B F \cong_F \bigoplus_D F$. By our hypothesis, in fact $B$ and $D$ have the same cardinality, and so there exists a bijection $\theta : B \rightarrow D$. Now $\theta$ induces a vector space isomorphism $\Theta : \bigoplus_B F \rightarrow \prod_B F$.

However, note that $|\prod_B F| = |F|^{|B|}$, while $|\bigoplus_B F| = |\bigcup_{T \subseteq \mathcal{P}(X), T\ \mathrm{finite}} \prod_T F|$ $= \sum_{|B|} |F|^{|T|}$ $= \sum_{|B|} |F|$ $= |B| \cdot |F|$. Since $|B| \cdot |F| < |F|^{|B|}$, we have a contradiction. Thus the dimension of $\bigoplus_B F$ is strictly smaller than that of $\prod_B F$.

• Josh Swanson  On August 31, 2011 at 8:58 pm

How does |F| * |B| < |F|^|B| follow when |B| = |F|, can just use Cantor’s diagonalization argument, since |F| * |B| = max{|F|, |B|}.

That an infinite cartesian product increases the cardinality of a set is intuitive, at least.

• Josh Swanson  On August 31, 2011 at 9:01 pm

My comment above was mangled. I guess the angle brackets were eaten as HTML. What I wrote was…

How does |F| * |B| less than |F|^|B| follow when |B| is smaller than |F|? When |B| is greater than or equal to |F|, we can just use Cantor’s diagonalization argument, since |F| * |B| = max{|F|, |B|}, so

|F| * |B| = |B| is less than 2^|B| is less than or equal to |F|^|B|

That an infinite cartesian product increases the cardinality of a set is intuitive, at least.