Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

share|improve this question
add comment

1 Answer

up vote 10 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.