# Abstract Nonsense

## R and C are Isomorphic as Groups

Point of Post: In this post we prove that $\mathbb{R}$ and $\mathbb{C}$ are isomorphic as abelian groups, as well as discussing a slightly more general theorem concerning when infinite abelian groups are isomorphic to finite powers of themselves.

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Motivation

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I rarely post about random theorems/facts/ideas that come up in my daily mathematical life. I think this type of post is actually very helpful since, from a blog-reader point of view, these are often the most entertaining posts. To this end I’d like to discuss a problem that came up in discussion with a friend today. The problem actually came up when we were discussing a different, still interesting (I hope to discuss it on here), question altogether. In particular, say a category $\mathcal{C}$ has the Schroeder-Bernstein property if whenever $x$ and $y$ are objects of $\mathcal{C}$ such that there exists monomorphisms $x\overset{\displaystyle \longleftarrow}{\longrightarrow}y$ then $x\cong y$. The classic Schroeder-Bernstein theorem then reduces to the statement that $\mathbf{Set}$ enjoys the Schroeder-Bernstein property. A natural question then comes as to which of the common categories have the Schroeder-Bernstein property. A quick check shows that $\mathbf{FinGrp}$, $\mathbf{FinRing}$, and $\mathbf{FinVect}_k$ all do, where the first two categories are finite groups and finite rings respectively. Ok, so ‘finite’ things tend to have the Schroeder-Bernstein property, so what are some examples of categories that don’t? I was aware of the fact that $\mathbf{Ab}$ did not have the Schroeder-Bernstein property but that the category of finitely generated abelian groups did.  So, me and my friend set out to find the examples in $\mathbf{Ab}$ which would violate the Schroeder-Bernstein property. So, being the good little math students that we are, we jumped right at the first non-finitely generated abelian group that came to mind, $\mathbb{R}$. Then, looking for another non-finitely generated abelian group (besides $\mathbb{R}$) to throw $\mathbb{R}$ against we immediately jumped at $\mathbb{C}$. There is an obvious embedding $\mathbb{R}\hookrightarrow\mathbb{C}$, and we started to wonder about embeddings the other way, when the obvious thing hit us. This is all moot if $\mathbb{R}\cong\mathbb{C}$ as abelian groups? Of course, such an ‘obvious’ question should have an obvious answer, right? Well, embarrassing as it is, neither of us knew the answer off hand. It took me the rest of walk to the car, and half the ride home to figure out the answer–and a simple answer it is–yes, they are isomorphic. The proof is actually a one-liner. This is just a reminder to always ask the ‘obvious’ questions, because sometimes we forget to.

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The Proof

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Theorem: The real numbers $\mathbb{R}$ and the complex numbers $\mathbb{C}$ are isomorphic as abelian groups.

Proof: Since $\mathbb{Q}$ is a subfield of $\mathbb{R}$ we know that $\mathbb{R}$ is a $\mathbb{Q}$-vector space in the obvious way. We have proven before on this blog that $\dim_\mathbb{Q}(\mathbb{R})=\#(\mathbb{R})$, and so we see that $\dim_\mathbb{Q}\mathbb{R}^2=2\#(\mathbb{R})=\#(\mathbb{R}). Thus,$latex \dim_\mathbb{Q}\mathbb{R}=\dim_\mathbb{Q}\mathbb{R}^2\$ and so $\mathbb{R}\cong\mathbb{R}^2$ as $\mathbb{Q}$-vector spaces. In particular though, this implies that $\mathbb{R}\cong\mathbb{R}^2$ as abelian groups. Since, $\mathbb{C}\cong\mathbb{R}^2$ the conclusion follows. $\text{ }$

Of  course, the above readily generalizes to show the following:

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Theorem: Let $A$ be an abelian group such that there exists a field $k$ such that $A$ can be made into an infinite dimensional $k$-space, say of dimension $\lambda$. then, $A\cong A^{\oplus\kappa}$ for any $\kappa\leqslant\lambda$.

Proof: We have that $\dim_k A=\lambda$, and $\dim_k A^{\oplus\kappa}=\kappa\cdot\lambda=\min\{\kappa,\lambda\}=\lambda$. Thus, $A$ and $A^{\oplus\kappa}$ have the same dimension as $k$-spaces, and thus are isomorphic as such. In particular, $A$ and $A^{\oplus\lambda}$ are isomorphic as abelian groups. $\blacksquare$

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Of course, all of the generality of the above can make the usefulness/neatness slightly opaque. A corollary of this theorem, which is the more ‘obvious’ generalization of our first theorem is the following:

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Theorem: Let $k$ be an uncountable field. Then, $k\cong k^n$ as abelian groups for every $n\in\mathbb{N}$ and $k\cong k^{\oplus\mathbb{N}}$.

Proof: Whatever the characteristic of $k$ we know that it’s prime subfield $k_0$ is countable. But, we know that $k$ is a $k_0$-vector space (in the obvious way) and it must be infinite dimensional, in fact, as we have proven, it is $\#(k)$ dimensional.  Thus, the previous theorem applies to show that $k\cong k^{\oplus\lambda}$ for any $\lambda\leqslant\aleph_0$ from where the conclusion follows. $\blacksquare$

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As a perhaps still more interesting consequence we see that the above gives:

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Corollary: If $k$ is an uncountable field, then $k\cong k[x]$ as abelian groups.

Proof: It’s trivial that as abelian groups $k[x]\cong k^{\oplus\mathbb{N}}$ from where the previous theorem kills it. $\blacksquare$

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In fact, this corollary gives us another corollary!

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Corollary: Let $k$ be an uncountable field. Then, $k\cong k[x_1,\cdots,x_n]$ as abelian groups for any $n\in\mathbb{N}$.

Proof: Using the previous corollary we have that $k\cong k[x_1]$ and then by induction

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$k[x_1,\cdots,x_{n+1}]\cong k[x_1,\cdots,x_n][x_{n+1}]\cong k[x_{n+1}]\cong k$

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from where the result follows. $\blacksquare$