Exercises:Saul - Algebraic Topology - 2/Exercise 2.5

Exercises

Exercises 2.5

Show that $\mathbb{Q}$ is not isomorphic to a free abelian group.

Solution

Our solution will comprise of two steps:

1. Showing $\mathbb{Q}$ does not have rank ("dimension" as cardinality of "basis") $\ge 2$
2. Supposing it is of rank $1$ and reaching a contradiction

Proof of claims:

1. To show that $\mathbb{Q}$ does not have rank $\ge 2$, we will show that for any $a,b\in\mathbb{Q}$ with $a\neq 0$ and $b\neq 0$ that $a$ and $b$ are "linearly dependent" (with respect to $\mathbb{Z}$) and thus cannot constitute (part of) a "basis".
   • Let $a,b\in\mathbb{Q}$ be given, with $a\neq 0$ and $b\neq 0$. Also let $a\eq\frac{p_1}{q_1}$ and $b\eq\frac{p_2}{q_2}$ be their representations as a quotient, for $p_1,q_1,p_2,q_2\in\mathbb{Z}$ with neither $q_i\eq 0$ of course.
     • to be "linearly independent" the only solution to $ma+nb\eq 0$ should be $m\eq n\eq 0$. We must exhibit another $m,n\in\mathbb{Z}$ such that $ma+nb\eq 0$ to show they're "linearly dependent".
       • Well: $m\frac{p_1}{q_1}+n\frac{p_2}{q_2}\eq 0$

         $\implies mp_1q_2+np_2q_1\eq 0$

         • Let $m:\eq t$ for some free variable we introduce to make things easier. We solve for $n$:
       • $np_2q_1\eq -tp_1q_2$ which gives $$n\eq -t\frac{p_1q_2}{q_1p_2}$$
       • We must ask now, what choice of $t\neq 0$ ensures $n$ is an integer, where $t$ is also an integer.
         • It's simple, if we let $t\eq q_1p_2$ - which is an integer - then it cancels with the denominator, the absence of a division and all terms being integer means we have an integer.
       • $$n\eq-q_1p_2\frac{p_1q_2}{q_1p_2}\eq-p_1q_2$$ , so:
         • $m\eq t\eq q_1p_2$ and $n\eq-p_1q_2$ is a non-trivial solution.
       • let us quickly check this. $q_1p_2\frac{p_1}{q_1}-p_1q_2\frac{p_2}{q_2}\eq p_1p_2-p_1p_2\eq 0$ - as required.
     • Thus $a$ and $b$ are "linearly dependent"
   • Since we showed that any non-zero $a,b\in\mathbb{Q}$ are "linearly dependent (WRT $\mathbb{Z}$)" and thus do not form part of a "basis" we see that $\mathbb{Q}$ cannot be of rank 2 or higher.
   • Now suppose $\mathbb{Q}$ has an infinite generator $A\in\mathcal{P}(\mathbb{Q})$. A "Hamal basis" but for a free Abelian group if you will. Then any $B\in\mathcal{P}(A)$ where $B$ is finite must be a linearly independent set (and obviously doesn't contain $0\in\mathbb{Q}$)
     • We have shown that any such $B$ cannot be linearly independent. Thus $\mathbb{Q}$ does not have an infinite basis either.
2. We are left with the case where $\mathbb{Q}$'s rank might be 1. If it's of rank 1 then it's just "cyclic" and $\mathbb{Z}\cong_f\mathbb{Q}$ where $f:\mathbb{Z} \rightarrow\mathbb{Q}$ is an isomorphism.
   • $\mathbb{Z}$ has a generator, $1$. That is $\langle 1\rangle\eq\mathbb{Z}$
   • As $\mathbb{Z}$ and $\mathbb{Q}$ are isomorphic (with $f$ as an isomorphism) we see that $\langle f(1)\rangle\eq \mathbb{Q}$ - that is $\mathbb{Q}$ is generated by the image of $1$ under $f$.
     • Note $\langle f(1)\rangle:\eq\{nf(1)\ \vert\ n\in\mathbb{Z} \}$
   • Note that $\frac{1}{2}f(1)\in\mathbb{Q}$, and as $f(1)$ generates $\mathbb{Q}$ there should be an $n\in\mathbb{Z}$ such that $\frac{1}{2}f(1)\eq nf(1)$
     • As $\mathbb{Q}$ is a field we see the only solution to this is $n\eq\frac{1}{2}$ but $\frac{1}{2}\notin\mathbb{Z}$
       • So $n$ cannot be $\frac{1}{2}$
         • This contradicts that $f(1)$ generates $\mathbb{Q}$
   • Thus $\mathbb{Q}$ cannot be isomorphic to $\mathbb{Z}$

This completes the proof.