This question is motivated by the question link text, which compares the infinite direct sum and the infinite direct product of a ring.

It is well-known that an infinite dimensional vector space is never isomorphic to its dual. More precisely, let $k$ be a field and $I$ be an infinite set. Let $E=k^{(I)}=\oplus_{i \in I} k$ be the $k$-vector space with basis $I$, so that $E^{*}$ can be identified with $k^I = \prod_{i \in I} k$. Then a stronger result asserts that the dimension of $E^{*}$ over $k$ is equal to the cardinality of $k^I$. This is proved in Jacobson, Lectures in Abstract Algebra, Vol. 2, Chap. 9, $\S$ 5 (Jacobson deduces it from a lemma which he attributes to Erdös and Kaplansky). Summarizing, we have

\begin{equation} \operatorname{dim}_k (k^I) = \operatorname{card} k^I. \end{equation}

Now, if $V$ is any $k$-vector space, we can ask for the dimension of $V^I$. Does the Erdös-Kaplansky theorem extend to this setting ?

Is it true that for any vector space $V$ and any infinite set $I$, we have $\operatorname{dim} V^I = \operatorname{card} V^I$ ? More generally, given a family of nonzero vector spaces $(V_i)$ indexed by $I$, is it true that $\operatorname{dim} \prod_{i \in I} V_i = \prod_{i \in I} \operatorname{card} V_i$ ?

If $V$ is isomorphic to $k^J$ for some set $J$, then the result holds as a consequence of Erdös-Kaplansky. In the general case, we have $V \cong k^{(J)}$, and we can assume that $J$ is infinite. In this case I run into difficulties in computing the dimension of $V^I$. I can only prove that $\operatorname{dim} V^I \geq \operatorname{card} k^I \cdot \operatorname{card} J$.

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    $\begingroup$ Dear François: Here is a tiny bit of nitpicking. I find your question very nice, but I think it would be more correct to write in your blockquote "a family of nonzero vector spaces" instead of "a family of vector spaces". $\endgroup$ – Pierre-Yves Gaillard May 30 '14 at 14:26
  • $\begingroup$ @Pierre-YvesGaillard Thanks, corrected. $\endgroup$ – François Brunault Oct 30 '15 at 9:10

The answer to both questions is yes.

As a preliminary, let's prove that for any infinite-dimensional vector space $V$, that

  • Lemma: $card(V) = card(k) \cdot \dim V$

Proof: Since $card(k) \leq card(V)$ and $\dim V \leq card(V)$, the inequality

$$card(k) \cdot \dim V \leq card(V)^2 = card(V)$$

is obvious. On the other hand, any element of $V$ is uniquely of the form $\sum_{j \in J} a_j e_j$ for some finite subset $J$ of (an indexing set of) a basis $B$ and all $a_j$ nonzero. So an upper bound of $card(V)$ is $card(P_{fin}(B)) \sup_{j \in P_{fin}(B)} card(k)^j$. If $B$ is infinite, then $card(P_{fin}(B)) = card(B) = \dim(V)$, and for all finite $j$ we have $card(k^j) \leq card(k)$ if $k$ is infinite, and $card(k^j) \leq \aleph_0$ if $k$ is finite, and either way we have

$$card(V) \leq \dim V \cdot \max\{card(k), \aleph_0\} \leq \dim V \cdot card(k)$$

as desired. $\Box$

The rest is now easy. Suppose $I$ is an infinite set, and suppose without loss of generality that $V_i$ is nontrivial for all $i \in I$. Put $V = \prod_{i \in I} V_i$. We have

$$\dim V \geq \dim k^I = card(k)^I \geq card(k)$$

where the equality is due to Erdos and Kaplansky. Therefore

$$\dim(V) = \dim(V)^2 \geq \dim V \cdot card(k) = card(V) = \prod_i card(V_i)$$

by the lemma above.

  • $\begingroup$ Dear Todd: This is just to tell you that I posted a minor complement to your great answer as a community wiki answer. $\endgroup$ – Pierre-Yves Gaillard May 30 '14 at 15:19

Here is a self-contained version of Todd Trimble's wonderful answer.

Let $K$ be a field. "Vector space" shall mean "$K$-vector space", "linear" shall mean "$K$-linear", $\dim$ shall mean $\dim_K$, $\operatorname{Hom}$ shall mean $\operatorname{Hom}_K$, and $|X|$ shall denote the cardinal of $X$ for any set $X$.

Let $V$ be the product of a family of nonzero vector spaces $(V_i)_{i\in I}$: $$ V=\prod_{i\in I}V_i. $$

As we have $$ \dim V=\sum_{i\in I}\dim V_i $$ if $I$ is finite, we can (and will) assume from now on that $I$ is infinite.

Main Theorem. We have, in the above notation, $\dim V=|V|$. In words: the dimension of the product of an infinite family of nonzero vector spaces is equal to its cardinal.

As a corollary, let us express explicitly the dimension of the product $V$ of the $V_i$ in terms of the $d_i:=\dim V_i$. Setting $$ \mu:=\max(\aleph_0,|K|),\quad\alpha:=|\{i\in I\ |\ d_i < \mu\}|, $$ we get $$ \dim\prod_{i\in I}V_i=|K|^\alpha\prod_{d_i\ge\mu}d_i. $$

Let us prove the Main Theorem.

Lemma. If $V$ is a vector space which is infinite as a set, then we have $$ |V|=|K|\cdot\dim V. $$

Proof. It is easy to see that, if $S$ is an infinite generating subset of a group $G$, then $S$ and $G$ are equipotent. Putting $$ G:=V,\qquad S:=\{\lambda b\ |\ (\lambda,b)\in K\times B\}, $$ where $B$ is a basis of $V$, we get the conclusion. QED

Let $V$ be an infinite dimensional vector space.

Say that $V$ is large if $\dim V\ge \max (|K|, \aleph_0)$.

By the lemma, $V$ is large if and only if $\dim V=|V|$.

Erdős-Kaplansky Theorem. The vector space $K^{\mathbb N}$ is large.

It is clear that the Erdős-Kaplansky Theorem implies the Main Theorem. So we are left with proving the Erdős-Kaplansky Theorem.

Proof of the Erdős-Kaplansky Theorem. Let $B$ be a $K$-basis of $K^{\mathbb N}$, and suppose by contradiction $|B|<|K|$. Let $K_0$ be the prime subfield of $K$, and put $$ K_1:=K_0(\{b_j\ |\ b\in B,\ j\in\mathbb N\}). $$ As $|K_1|<|K|$ and $K$ is infinite, there is an $x$ in $K^{\mathbb N}$ whose coordinates are $K_1$-linearly independent. There are $c_1,\dots,c_n$ in $B$ such that $x$ is a $K$-linear combination of the $c_j$. Since $c_{ij}$ is in $K_1$ for all $i,j$, there is a nonzero $\lambda$ in $K_1^{n+1}$ such that $$ \sum_{j=0}^n\lambda_j\,c_{ij}=0 $$ for $1\le i\le n$, and we have $$ \sum_{j=0}^n\lambda_j\,x_j=0, $$ in contradiction with the choice of $x$. QED

  • $\begingroup$ Dear Pierre -- Yes, I noticed, and I like it. +1. $\endgroup$ – Todd Trimble May 30 '14 at 17:11

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