# 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.

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What do you mean with "what is..."? Do you know what $\mathbb{C}$, $\mathbb{Z}$ and the tensor product '$\otimes$' are? Be a bit more specific! – Sh4pe Oct 24 '12 at 8:19
@Sh4pe I think the user wants a nice description of this group. – Olivier Bégassat Oct 24 '12 at 8:23
Right, I changed the question. – user9384023 Oct 24 '12 at 8:23
The two are definitely not equal in a set-theoretic sense, but then this is hardly ever the case (the non-negative integers are not equal to $\Bbb N$ either, strictly speaking). So you are asking about some kind of isomorphism; as the answers show it depends on what kind of isomorphism (sets, abelian groups, $\Bbb C$-vector spaces) you are interested in. – Marc van Leeuwen Oct 24 '12 at 9:59

$\def\tensor{\otimes}\def\C{{\mathbb C}}\def\Z{{\mathbb Z}}\def\Q{{\mathbb Q}}$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.

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I think you should add "as $\Bbb Q$ vector spaces" at the end, even though it has been said before, because just $\C \tensor_\Z \C \cong \C$ can also be interpreted as a statement about (either left or right) $\C$ vector spaces, in which case it is patently false. I know you do not intend this, but this is bound to confuse some readers. Also note that OP asked about equality (which is certainly false) not isomorphism. – Marc van Leeuwen Oct 24 '12 at 9:51
@MarcvanLeeuwen I added "as abelian groups". – martini Oct 24 '12 at 11:51

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 $\mathbb C/\mathbb Q$.

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The first sentence is wrong. – Martin Brandenburg Oct 24 '12 at 9:04
@MartinBrandenburg: Do you mean that it should say "whose elements are classes of pairs..."? That is true, but this seems implied by "where you identify...". – Marc van Leeuwen Oct 24 '12 at 9:54

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 $\Bbb C/\Bbb Q$).

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Remark: "It is an uncountable dimensional vector space over C" does not contradict martinis proof that it is isomorphic to C, since this isomorphism is only Q-linear. – Martin Brandenburg Oct 24 '12 at 9:06