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We covered the basics of module theory in our Algebra lecture and now we are asked to prove/argue that $\mathbb{R}$ and $\mathbb{Q}$ are NOT free as $\mathbb{Z}$-modules.

I already read up a bit on this on stack exchange, but couldn't find something concrete for exactly this (they were not specifically concerned with the freeness of the modules). But still, I wanted to know if someone has a tip on how to approach this task.

I have two proposals to share, and I'd appreciate thoughts and help.

  1. Arguing with linear (in)dependence.

    The idea is, for $\mathbb{Q}$, at least, that for every $ q, p \in \mathbb{Q}$, you can find $a, b \in \mathbb{Z}$ with $ a,b \neq 0$ such that $aq + bp = 0$. From that follows that no two element of $\mathbb{Q}$ are linearily independent, meaning that $\mathbb{Q}$ would have to be generated by 1 element and would thus be isomorphic to $\mathbb{Z}$, a cyclic group, but $\mathbb{Q}$ isn't cyclic and thus it cannot be.

  2. Looking at the base.

    The idea behind this is to assume that it actually has a base and there exists a finite set of $b_i$ $\in$ $\mathbb{Q}$ and $\in$ $\mathbb{Z}$ resepctively, and $z_i$ $\in$ $\mathbb{Z}$, such that any element in $\mathbb{Q}$ and $\mathbb{R}$ other ring can be represented as a finite sum of prodcuts; meaning for $q \in \mathbb{Q}$ it holds that $q = \sum b_i z_i$ and analogous for $\mathbb{R}$. And then to come to a wrong conclusion, meaning our assumption was false and there exist no such basis in these cases.

Is there maybe a different way to approach this or any tips on how to use the ideas I outlined? Or am I completely on the wrong way with them?

Any help and comment is highly appreciated and TIA.

Kartöffelchen
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    Use idea $1$: $\mathbb{Q}$ is not a free module over $\mathbb{Z}$ since any subset with at least two elements is not independent (given $a/b$ and $c/d$, take $bc(a/b)-ad(c/d)$, which equals zero though the separate summands do not) and $\mathbb{Q}$ is not a cyclic $\mathbb{Z}$-module. (I leave it as a comment since it is like a duplicate of this one) – Bowei Tang Apr 24 '25 at 10:13
  • Similar reasoning works for part ii. The subset containing at least two real numbers cannot be linearly independent which implies that $\mathbb R$ is cyclic i.e. $\mathbb Z \simeq \mathbb R$ which is impossible. – nozalp10 Apr 25 '25 at 20:06

2 Answers2

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Your argument (1) for $\mathbb Q$ is fine.

One method which works for $\mathbb Q$ and $\mathbb R$ is to show that there are no non-zero homomorphisms from $\mathbb Q$ or $\mathbb R$ to $\mathbb Z$. For, we can always divide by non-zero integers in $\mathbb Q$ and $\mathbb R$.

On the other hand, every free module admits a non-zero homomorphism to $\mathbb Z$.

Andrew Hubery
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  • I'm not sure that I know that, actually. I skimmed my lecture notes and the only thing I found adjacent to this is that : "A module homomorphis f between two modules M and N is injective iff ker(f)= {0}" and " Remark: Not every module has a basis, contratry to vector spaces. For example, Z/2Z viewed as a Z-module has no basis. The only option for a basis element is 1 and it holts that 0 = 0 * 1 = 2 * 1 so we have no unique choice of scalar elements."

    Do you mean something like that? I'm sorry, but our instruction language is not English, so there might be a bit of a language barrier.

     – Kartöffelchen Apr 24 '25 at 11:37
  • Sorry for the long comment, but I just wanted to ask.

    How would I go about your advice exactly?

     – Kartöffelchen Apr 24 '25 at 11:38
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    Assume $f$ is a non-zero map from $\mathbb Q$ or $\mathbb R$ to $\mathbb Z$. Then $f(a)=n\neq0$ for some $a$. What can we say about the image of $a/2n$? – Andrew Hubery Apr 24 '25 at 11:53
  • If $F$ is a free module, then it has a basis $b_i$. Fix some index $j$ and define a map $F\to\mathbb Z$ sending $\sum_im_ib_i$ to $m_j$. – Andrew Hubery Apr 24 '25 at 11:55
  • Regarding your first comment, I'm like completely stumped. It's also an integer, but I don't really now where I should be going with this. Likewise with the second comment. Should you conclude something from how it's not injective (if i see that correctly).

    Sorry for these stupid answers, but I think I might not to see the wood for the trees here. If you could please give another hint or further guidance, I'd appreciate it.

     – Kartöffelchen Apr 24 '25 at 15:58
  • @Kartöffelchen We have $$(2n)f(a/(2n))=f(a)=n.$$ If $n\neq 0$ we get $$2 f(a/(2n))=1.$$ However, $f(a/(2n))$ is an integer (by definition of $f$). This is a contradiction, as $2$ does not have a multiplicative inverse in $\mathbb{Z}$. – Severin Schraven Apr 25 '25 at 20:08
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An other approach is through divisibility. For every $n\neq 0$ and $x\in \mathbb{Q}$, there exist some $y\in\mathbb{Q}$ such that $ny=x$. We say that $\mathbb{Q}$ is divisible. As $\mathbb{Z}$ is not divisble, free $\mathbb{Z}$ are not divisble. Thus $\mathbb{Q}$ is not a free $\mathbb{Z}$-module. The same goes for $\mathbb{R}$.

Note : Over a PID, submodule of a free modulea re again free, done here Submodule of free module over a p.i.d. is free even when the module is not finitely generated? . Thus as $\mathbb{Q}$ is not free as a $\mathbb{Z}$-module, it is automatic that $\mathbb{R}$ is not either.

Enguerrand Moulinier
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