# Dimension of infinite product of vector spaces

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 $\operatorname{dim} V^I \geq \operatorname{card} k^I \cdot \operatorname{card} J$.

• 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". – Pierre-Yves Gaillard May 30 '14 at 14:26
• @Pierre-YvesGaillard Thanks, corrected. – 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.

• Dear Todd: This is just to tell you that I posted a minor complement to your great answer as a community wiki answer. – 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

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