Assume $k$ is a division ring and $M$ an infinite set. There are two vector spaces over $k$ to consider: $k^M$ and $k^{(M)}$ (the latter is the subset of $k^M$ containing all infinite sequence with only a finite number of nonzero components).

Obviously $|k^{(M)}|\leq |k^M|$. I know that $k^{(M)}$ has a canonical basis $(e_\mu)_{\mu\in M}$. This means that $(e_\mu)_{\mu\in M}$ is a family of linearly independent elements of $k^M$. How can I show that the $e_\mu$'s don't generate $k^M$? I have to find an element of $k^M$ that isn't a linear combination of the $e_\mu$. What is a good choice for this supposed element? (Or is a proof by contradiction more appropriate?)

  • 1
    $\begingroup$ Hint: take a nonzero constant sequence. $\endgroup$ Jan 15, 2021 at 18:33
  • $\begingroup$ One way to do is to note that your first observation is sharp: $|k^{(M})|<|k^M|$. Another way is to show that $k^{(M)}$ doesn't have the universal property of being the product of $|M|$ many $k$'s. $\endgroup$
    – Berci
    Jan 15, 2021 at 18:34
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    $\begingroup$ @Berci The inequality need not be strict, for example if $k$ has the cardinality of the continuum and $M$ is countably infinite. $\endgroup$ Jan 15, 2021 at 18:38
  • $\begingroup$ @AndreasBlass: Indeed, that's right. $\endgroup$
    – Berci
    Jan 15, 2021 at 18:40

1 Answer 1


Finding such an element would not be enough, as it would simply show that the inclusion $k^{(M)}\to k^M$ isn't an isomorphism, but that doesn't mean that there is no isomorphism.

For the record, the constant family with value $1$ is not in the span of the $e_\mu$.

The usual way to prove that they are not isomorphic is to prove that any $M$-indexed family does not span $k^M$ - in fact that would be equivalent because one can extract a basis from any such family, and it must then have cardinality $\geq |M|$ and $\leq |M|$ so $=|M|$. But for some reason I always forget the standard argument for this - it's a sort of diagonal argument where you build a new element not in the span of any $M$-indexed family of elements.

So another way to proceed is to prove that $k^{(M)}$ has too many linearly independent linear forms. Indeed, $(k^{(M)})^* \cong k^M$ (canonically), and so these linearly independent linear forms contradict the existence of an isomorphism.

For this I will use the following well-known set-theoretic lemma:

If $M$ is an infinite set, there is a bijection between the set of (nonempty) finite subsets of $M$, and $M$.

The proof then goes as follows (hidden, to leave some suspense):

It follows from this lemma - actually, a variation on that lemma that I'll let you figure out- that $k^{(M)}$ is isomorphic, as a $k$-vector space, to $k[(X_\mu)_{\mu\in M}]$, the polynomial algebra in $M$-many indeterminates. Now for any $z\in k^M$, $ev_z$,"evaluation at $z$", defined by sending a polynomial $P$ to $P(z)$, defines a linear form on $k[(X_{\mu})_{\mu\in M}]$.

The claim is that these are linearly independent. But this is classical polynomial algebra: if you have a finite family $(z_i)_i \in k^M$, then I can find a finite subset $J\subset M$ such that the restrictions of these families to $J$ are distinct - and now in a finite number of variables, I can of course design a polynomial that takes any value I like on a finite list of points, in particular I can make the values leave any hyperplane. This is left as an exercise.

The coup-de-grâce is in claiming that these are too many. But Cantor's theorem ensures that $|k^M|> |M|$, as $k$ is a field and so $|k|\geq 2$, so we are done.


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