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The following is the problem 206 from Golan's book Linear Algebra a Beginning Graduate Student Ought to Know. I've been unable to make any progress.

Definition: A Hamel basis is a (necessarily infinite dimensional) basis for $\mathbb{R}$ as a vector space over $\mathbb{Q}$.

Problem: Let $B$ be a Hamel basis for $\mathbb{R}$ as a vector space over $\mathbb{Q}$ and fix some element $a\in\mathbb{R}$ with $a\neq 0,1$. Show there exists some $y\in B$ with $ay\notin B$.

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  • $\begingroup$ Just stating two relatively obvious facts: If we suppose that $b\in B$ $\Rightarrow$ $ab\in B$ for some $a\in\mathbb R$, then this $a$ must be trancendental and we also have $b\in B$ $\Rightarrow$ $a^{-1}b\in B$. $\endgroup$
    – Martin Sleziak
    Jun 2, 2012 at 4:55
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    $\begingroup$ According to Jonathan Golan's answer, this question can be generalized. If $F$ is a proper field extension of a field $K$ with an extension basis $B$, then for any $a\in K\setminus\{0,1\}$, $ab\notin B$ for some $b\in B$. $\endgroup$
    – Batominovski
    Aug 12, 2016 at 21:44

2 Answers 2

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Here is the proof of the exercise:

Let $B$ be a Hamel basis. Then any real number $r$ can we written uniquely as $\Sigma_{x \in B} {r_x}x$ where the $r_x$ are rational numbers only finitely many of which are nonzero. The function $\alpha : r \to \Sigma_{x \in B} r_x$ is a linear transformation of vector spaces over the rational numbers. Now suppose that $a \ne 1$ and $ax \in B$ for all $x \in B$. Then $\alpha(ar)=\alpha(r)$ for all real numbers $r$. In particular, if $x \in B$ and if $r = x(a-1)^{-1}$ then $1 = \alpha(x) = \alpha([a-1]r) = \alpha(ar) - \alpha(r) = 0$. Contradiction!

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  • $\begingroup$ I am glad you like it. In case you don't know, a third (expanded) edition was published by Springer a few months ago. $\endgroup$
    – Jonathan Golan
    Jun 3, 2012 at 1:56
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Completely Revised: Let $f:\Bbb R\to\Bbb R:x\mapsto ax$, and suppose that $f[B]\subseteq B$. Since $f[B]$ is a basis for $\Bbb R$, we must have $f[B]=B$. In particular, $a^nb\in B$ for each $b\in B$ and $n\in\Bbb Z$, and it follows that $a$ must be transcendental.

Define a relation $\sim$ on $B$ by $b_0\sim b_1$ iff $b_1=a^nb_0$ for some $n\in\Bbb Z$; $\sim$ is easily seen to be an equivalence relation. Let $T\subseteq B$ contain exactly one representative of each $\sim$-equivalence class. Fix $t\in T$; there are $m\in\Bbb Z^+$ and for $k=1,\dots,m$ distinct $t_k\in T$ and Laurent polynomials $p_k$ with non-zero rational coefficients such that

$$\frac{t}{a+1}=\sum_{k=1}^mp_k(a)t_k\;,$$

and hence $$t-\sum_{k=1}^m(a+1)p_k(a)t_k=0\;.$$

But this implies that $m=1$, $t_1=t$, and $(a+1)p_1(a)=1$, making $a$ algebraic, which is impossible. Thus, $f[B]\nsubseteq B$.

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    $\begingroup$ I'm a bit confused as to why $B_{\eta + 1}$ is independent. For ease, let's just focus on $B_1$. A linear combination of elements in $B_1$ looks like $p(e) + q(e)x_1$ where $p$ and $q$ are Laurent series with only finitely many positive and negative powers of $e$ appearing. Suppose this combination is $0$. If $q(e) = 0$, we're done by independence of $B_0$. Else, we have $x_1 = -\frac{p(e)}{q(e)}$. Why is this a problem? More specifically, it seems as though, for example $x_1$ could be $\frac{e}{e+1}$, which, if I'm computed correctly is not in the span of $B_0$. $\endgroup$
    – Jason DeVito - on hiatus
    Jun 2, 2012 at 2:10
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    $\begingroup$ @Jason: You’re right. I was definitely having a bad day. (In a way I’m relieved, since I didn’t expect the problem to be in error.) I’ll have another look. $\endgroup$
    – Brian M. Scott
    Jun 2, 2012 at 3:48

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