Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
2  
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

3 Answers 3

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

share|improve this answer
1  
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 $\mathbb C/\mathbb Q$.

share|improve this answer
1  
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 $\Bbb C/\Bbb Q$).

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.