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The following problem is from Golan's linear algebra book. 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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The existence of Hamel bases to all vector spaces is equivalent to the axiom of choice. The way this is written suggests that a basis for $\mathbb R$ over $\mathbb Q$ may be equivalent to the axiom of choice which is very untrue. – Asaf Karagila Jun 1 '12 at 21:31
You are right. I will fix it. – Potato Jun 1 '12 at 21:31
Thanks. I removed [axiom-of-choice] because it is completely irrelevant to the question (even before the edit). – Asaf Karagila Jun 1 '12 at 21:32
Is it $a\neq 0$? If $a=0$ you will never get it into $B$. – rschwieb Jun 1 '12 at 21:34
Yes. Thanks for catching my typo. – Potato Jun 1 '12 at 21:35
up vote 8 down vote accepted

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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Please visit more often, Dr. Golan :) – rschwieb Jun 2 '12 at 19:13
Thanks! Your book is fantastic, by the way. – Potato Jun 2 '12 at 19:53
I am glad you like it. In case you don't know, a third (expanded) edition was published by Springer a few months ago. – Jonathan Golan Jun 3 '12 at 1:56

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


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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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 $B_0$. – Jason DeVito Jun 2 '12 at 2:10
@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. – Brian M. Scott Jun 2 '12 at 3:48

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