# Abstract Nonsense

## Q and Q^2 are not Isomorphic as Groups

Point of Post: In this post we prove that $\mathbb{Q}\not\cong\mathbb{Q}^2$ as abelian groups, and discuss a generalization.

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Motivation

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As I was trying to go to sleep last night, I was pondering the contents of my post the other day. In particular, I was thinking about the statement that I made that “one should always ask the obvious questions”, whatever that’s supposed to mean. It hit me then, that while there are many ‘obvious’ questions that could follow from the fact that $\mathbb{R}\cong\mathbb{C}\cong\mathbb{R}^2$ as abelian groups, I didn’t ask perhaps the most obvious one. Namely, is $\mathbb{Q}\cong\mathbb{Q}^2$ as groups? I was pretty sure the answer was no, but it took me a little bit to come up with a proof. Of course, as one is supposed to, one sees that the argument that proves this really only has to do with the fact that $\mathbb{Q}$ as a group, is the localization of some commutative unital ring and the fact that we’re discussing the localization as a module over the original ring. So, in the spirit of quicker, simpler post I record this fact here.

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The Theorem

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Let’s get right down to it. Why is $\mathbb{Q}\not\cong\mathbb{Q}^2$ as abelian groups? Roughly because we shall show that any $\mathbb{Z}$-linear map $\mathbb{Q}\to\mathbb{Q}^2$ is actually $\mathbb{Q}$-linear, from where the result will clearly follow. Indeed, the key is the following lemma:

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Lemma: Let $f:\mathbb{Q}\to\mathbb{Q}$ be a $\mathbb{Z}$-linear map, then $f$ is $\mathbb{Q}$-linear.

Proof: We must merely note that if $n\in\mathbb{Z}-\{0\}$ then

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$\displaystyle nf\left(\frac{1}{n}\right)=f\left(\frac{n}{n}\right)=f(1)$

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Thus, if $\displaystyle \frac{p}{q}\in\mathbb{Q}$, then

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$\displaystyle f\left(\frac{p}{q}\right)=pf\left(\frac{1}{q}\right)=\frac{p}{q}f(1)$

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Thus, $f(x)=f(1)x$ for all $x\in\mathbb{Q}$, and so clearly $f$ is $\mathbb{Q}$-linear. $\blacksquare$

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Incidentally, this also proves:

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Corollary: $\text{End}_\mathbb{Z}(\mathbb{Q})\cong\mathbb{Q}$ as left $\mathbb{Q}$-modules.

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Now onto the titular theorem:

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Theorem: Let $\lambda$ be a cardinal. Then, $\mathbb{Q}^{\oplus\lambda}\cong\mathbb{Q}$ as abelian groups if and only if $\lambda=1$.

Proof: Suppose that $f:\mathbb{Q}\to\mathbb{Q}^{\oplus\lambda}$ is a group isomorphism. Note then that since $\pi_i\circ f:\mathbb{Q}\to\mathbb{Q}$ are group homomorphisms for each $i$, we may conclude by the previous lemma that $f(x)=(a_j x)$ for each $x\in\mathbb{Q}$ and so, in particular, $f$ is a $\mathbb{Q}$-map. Since then $f$ is a $\mathbb{Q}$-isomorphism and commutative rings have the IBN property we may conclude that $\lambda=1$ as desired. $\blacksquare$

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Of course, a natural question then is to ask whether $\mathbb{Q}^{\oplus\lambda}\cong\mathbb{Q}^{\oplus\kappa}$ for $\kappa\ne\lambda$. The answer for infinite cardinals is a resounding yes, since, as we have proven in the quest to prove that infinite dimensional vector spaces aren’t isomorphic to their dual since, since $\mathbb{Q}$ is countable and $\lambda,\kappa$ infinite we have that $\#\left(\mathbb{Q}^{\oplus\lambda}\right)=\lambda$ and $\#\left(\mathbb{Q}^{\oplus\kappa}\right)=\kappa$. So, if the two are isomorphic as groups then they must be equipotent, and so $\lambda=\kappa$. As for the finite case, I’m not quite sure as of now. I’ll give it a thought, anyone who has a lead/knows the answer, feel free to leave a comment saying as much!

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EDIT: Let me now resolve the dispute about whether or not $\mathbb{Q}^{\oplus \lambda}\cong\mathbb{Q}^{\oplus\kappa}$ as abelian groups for any cardinals $\kappa,\lambda$. The basic idea (secretly very similar to Yaghoub’s, as below) is to show that if they are isomorphic as $\mathbb{Z}$-modules they are isomorphic as $\mathbb{Q}$-modules. What we show, more generally, is that if $R$ is an integral domain then $\text{Frac}(R)^{\oplus\lambda}\cong\text{Frac}(R)^{\oplus\kappa}$ as $R$-modules implies that $\lambda=\kappa$. The key lemma is the following:

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Theorem: Let $R$ be an integral domain. Then, $\text{Frac}(R)\otimes_R\text{Frac}(R)\cong\text{Frac}(R)$ as $\text{Frac}(R)$-spaces.

Proof: We know that $\text{Frac}(R)\otimes_R\text{Frac}(R)$ is a $\text{Frac}(R)$-space from basic theory. It clearly is not zero since $\text{Frac}(R)\times\text{Frac}(R)\to\text{Frac}(R)$ given by $(x,y)\mapsto xy$ is a non-zero $R$-bilinear map. That said, note that for any simple tensor $\displaystyle \frac{r}{s}\otimes\frac{p}{q}$ is equal to $\displaystyle \frac{pr}{qs}(1\otimes 1)$ and thus $\text{Frac}(R)\otimes_R\text{Frac}(R)$ is singly generated as a $\text{Frac}(R)$-space. The conclusion follows. $\blacksquare$

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We can now prove our desired theorem:

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Theorem: Let $R$ be an integral domain. Then, $\text{Frac}(R)^{\oplus \lambda}\cong\text{Frac}(R)^{\oplus\kappa}$ if and only if $\lambda=\kappa$.

Proof: Suppose that $\text{Frac}(R)^{\oplus\lambda}\cong\text{Frac}(R)^{\oplus\kappa}$ as $R$-modules. Then,

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$\text{Frac}(R)\otimes_R\text{Frac}(R)^{\oplus\lambda}\cong\text{Frac}(R)\otimes_R\text{Frac}(R)^{\oplus\kappa}$

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as $\text{Frac}(R)$-spaces. But, then we see that

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\begin{aligned}\text{Frac}(R)^{\oplus\lambda} &\cong\left(\text{Frac}(R)\otimes_R\text{Frac}(R)\right)^{\oplus\lambda}\\ &\cong \text{Frac}(R)\otimes_R \text{Frac}(R)^{\oplus\lambda}\\ &\cong \text{Frac}(R)\otimes_R\text{Frac}(R)^{\oplus\kappa}\\ &\cong \left(\text{Frac}(R)\otimes_R\text{Frac}(R)\right)^{\oplus\kappa}\\ &= \text{Frac}(R)^{\oplus\kappa}\end{aligned}

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as $\text{Frac}(R)$-spaces, which implies that $\lambda=\kappa$ as desired. $\blacksquare$

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It may seem, at first, that this implies that $\mathbb{R}\not\cong\mathbb{R}^2$ as $\mathbb{Z}$-modules, for it seems the same argument applies. Let’s analyze where it breaks down. Basically we, using the same method, get to the point that $(\mathbb{R}\otimes_\mathbb{Z}\mathbb{R})^{\oplus\lambda}\cong(\mathbb{R}\otimes_\mathbb{Z}\mathbb{R})^{\oplus\kappa}$ as $\mathbb{R}$-spaces. Now, we know that $\mathbb{R}\otimes_\mathbb{Z}\mathbb{R}$ is an $\mathbb{R}$-space an and so looks like $\mathbb{R}^{\oplus\nu}$ for some cardinal $\nu$. So, all we can really conclude is that $\lambda\nu=\kappa\nu$. So, if $\nu$ is infinite this is no contradiction if, say, $\lambda=1$ and $\kappa=2$. In fact, since we know it can’t be a contradiction for finite $\lambda,\kappa$ we MUST have that $\nu$ is infinite.

January 31, 2012 -

1. If $\mathbb{Q}^m \cong \mathbb{Q}^n,$ as abelian groups, then
$M_n(\mathbb{Q}) \cong End_{\mathbb{Z}}(\mathbb{Q}^n) \cong End_{\mathbb{Z}}(\mathbb{Q}^m) \cong M_m(\mathbb{Q}),$
as $\mathbb{Q}$-algebras. Now take dimensions of both sides, as vector spaces over $\mathbb{Q},$ to get $m=n.$ 🙂