33
$\begingroup$

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

$\endgroup$
0

2 Answers 2

32
$\begingroup$

The answer to both questions is yes.

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

  • Lemma: $|V| = |k| \cdot \dim V$

where $|X|$ denotes the cardinality of a set $X$.

Proof: Since $|k| \leq |V|$ and $\dim V \leq |V|$, the inequality

$$|k| \cdot \dim V \leq |V|^2 = |V|$$

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 $|V|$ is $|P_{fin}(B)| \cdot \sup_{j \in P_{fin}(B)} |k|^j$. If $B$ is infinite, then $|P_{fin}(B)| = |B| = \dim(V)$, and for all finite $j$ we have $|k^j| \leq |k|$ if $k$ is infinite, and $|k^j| \leq \aleph_0$ if $k$ is finite, and either way we have

$$|V| \leq \dim V \cdot \max\{|k|, \aleph_0\} \leq \dim V \cdot |k|$$

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

$$\dim V \geq \dim k^I = |k|^I \geq |k|$$

where the equality is due to Erdos and Kaplansky. Therefore

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

by the lemma above.

$\endgroup$
1
  • $\begingroup$ Dear Todd: This is just to tell you that I posted a minor complement to your great answer as a community wiki answer. $\endgroup$ May 30, 2014 at 15:19
21
$\begingroup$

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

$\endgroup$
2

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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