Is it true that $\mathbb{C}\otimes_\mathbb{Z}\mathbb{C}=\mathbb{C}$? Is it true that $\mathbb{C}\otimes_\mathbb{Z}\mathbb{C}=\mathbb{C}$?
This is not the homework. I just want to know what it is and I cannot find it any text I have.
 A: $\def\tensor{\otimes}\def\C{{\mathbb C}}\def\Z{{\mathbb Z}}\def\Q{{\mathbb Q}}$Let $G$ be an abelian group and $b\colon \C \times \C \to G$ $\Z$-bilinear. Then $b$ is $\Q$-bilinear, as for $q=\frac mn \in \Q$, $z,w \in C$:
\[
  b(qz,w) = b\left(\frac mnz, w\right) = b\left(\frac 1n z, m\frac nn w\right)
   = b\left(\frac nn z, \frac mn w\right) = b(z,qw)
\] 
So $b$ induces an unique homomorphism $\beta\colon\C \otimes_\Q \C \to G$. That is, as abelian groups $\C \otimes_\Z\C \cong \C \otimes_\Q \C$. But the latter is isomorphic to $\C$ as $\Q$-vector space (having a basis of cardinality $2^{\aleph_0} \cdot 2^{\aleph_0} = 2^{\aleph_0}$), hence as abelian group. So $\C \tensor_\Z \C \cong \C$ as abelian groups.
A: It is the abelian group whose elements are pairs of complex numbers, where you identify $(kz,w)$ with $(z,kw)$ if $k\in \mathbb Z$. 
In effect, this also identifies $(qz,w)$ with $(z,qw)$ if $q\in\mathbb Q$.
In a way you have an independent copy of $\mathbb C$ for each residue class of $\mathbb C/\mathbb Q$.
A: To the perfectly correct answer by Hagen one might add that $\mathbb{C}\otimes_\mathbb{Z}\mathbb{C}$ is a horrible monster that you will want to avoid dealing with if you possibly can. It is an uncountable dimensional vector space over $\Bbb C$, but one that cannot be identified with a space of functions without invoking the axiom of choice (since it would require choosing a representative in each one of the uncontably many classes of $\Bbb C/\Bbb Q$).
