Exercises:Saul - Algebraic Topology - 2/Exercise 2.5

Exercises

Exercises 2.5

Show that [ilmath]\mathbb{Q} [/ilmath] is not isomorphic to a free abelian group.

Solution

Our solution will comprise of two steps:

1. Showing [ilmath]\mathbb{Q} [/ilmath] does not have rank ("dimension" as cardinality of "basis") [ilmath]\ge 2[/ilmath] - or an infinite basis.
2. Supposing it is of rank [ilmath]1[/ilmath] and reaching a contradiction

Proof of claims:

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

         [ilmath]\implies mp_1q_2+np_2q_1\eq 0[/ilmath]

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

This completes the proof.