Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $ay\notin B$.

share|improve this question
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
show 3 more comments

2 Answers

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!

share|improve this answer
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
add comment

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, $f[B]\nsubseteq B$.

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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