Does a finite field ${\bf F}_q$, viewed as a vector space over another finite field, have a basis of squares?

Solution 1:

Yes, this is because the $\Bbb{F}_{q'}$ vector space generated by the elements of $\Bbb{F}_q^{\times 2}$ has at least $1+\frac{q-1}2$ elements.

And its cardinality is $(q')^m$ where $m$ is the $\Bbb{F}_{q'}$-dimension.

Solution 2:

Even better: If $R$ is a commutative ring where $2$ is a unit, then every unital $R$-algebra is spanned (as an $R$-module) by squares. Namely, for every $\alpha$, $$ \alpha = \tfrac12(\alpha+1)^2-\tfrac12\alpha^2-\tfrac12 1^2 $$

This holds in particular whenever $R$ is a field of characteristic $\ne 2$ and the algebra is a field extension.

This uses too many different squares to immediately give you a basis, but when you have a spanning set of a vector space, you can always select a subset that is a basis.

In particular, if you already know some basis $\{1,\alpha_2,\alpha_3,\ldots,\alpha_n\}$ for the extension field, then all you have to do is discard those elements of $\{1,\alpha_2^2,(\alpha_2+1)^2, \alpha_3^2, (\alpha_3+1)^2, \ldots, (\alpha_n+1)^2\}$ that are linear combinations of earlier elements in the list.

In characteristic 2, the argument in Just a user's answer takes care of the question when the fields are finite, but that's as far as we can go: If $K$ is any field of characteristic 2, then the field of fractions of $K[X]$ is an example of an extension that does not have a basis (over $K$) consisting of squares.