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I would like to prove that

$$\mathbb{C} \otimes_{\mathbb{Z}} \mathbb{Q}$$

is not zero.

First I thought that I can compute the above tensor product using the fact that $\mathbb{C}$ is isomorphic to $\frac{R[x]}{(x^2+1)}$ and use the fact that $\frac{R[x]}{(x^2+1)} \otimes \mathbb{Q}$ is isomorphic to $\mathbb{R} \otimes \frac{\mathbb{Q}[x]}{(x^2+1)}$. But I guess this doesn't lead me to anything. Perhaps I should only find a non-zero element of the tensor product. To this end, the only thing that comes to my mind is taking the numerators of rational numbers to the other side of the tensor product but this does not show anything (trying to mimic the method we use to prove that $\mathbb{Z_{n}} \otimes_{\mathbb{Z}} \mathbb{Q}$ is zero). However, here we actuallty want to prove it is not zero, so this shouldn't work. Also, $\mathbb{Q}$ is not a free $\mathbb{Z}$-module, so we cannot use its direct sum in terms of $\mathbb{Z}$ and use the property of tensor product wrt direct sum.

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3 Answers 3

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One way to do this, if you've seen this before and don't mind hitting an ant with a sledgehammer, is to use the fact that homomorphisms $A \otimes B \rightarrow C$ correspond to $\mathbb{Z}$-billinear maps $ A \times B \rightarrow C$.

$(z, q) \mapsto qz$ is a non-zero billinear map $\mathbb{C} \times \mathbb{Q} \rightarrow \mathbb{C}$. Therefore, there is at least one non-zero homomorphism $\mathbb{C} \otimes \mathbb{Q} \rightarrow \mathbb{C}$.

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    $\begingroup$ There is no sledgehammer in this answer. You merely use the definition of the tensor product. $\endgroup$ Commented Jan 3 at 1:29
  • $\begingroup$ @chessanator I know the universal property but didn't think of it in this case, thanks! Plus, I do agree with Martin. Although, what possibly Chessanator meant was we may alternatively need to go through the proof of the universal property. $\endgroup$
    – Tim
    Commented Jan 3 at 2:37
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Hint: If $A$ is an abelian group, then its tensor product with $\mathbb{Q}$ is isomorphic to the localization $(\mathbb{Z} \setminus \{0\})^{-1} A$. And if $A$ has the structure of a vector space over $\mathbb{Q}$ (which means that $A$ is uniquely divisible), this is just $A$. So your tensor product is just ...

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  • $\begingroup$ Localization isn't really my strong suit, so I need to get back to your hint later, as I like this technique as well. $\endgroup$
    – Tim
    Commented Jan 3 at 2:40
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If the tensor product of two $ \mathbb Z $-modules $ A $ and $ B $ is zero then every $ \mathbb Z $-linear map $$ g\colon A\times B\to C $$ into another $ \mathbb Z $-module $ C $ is the zero map. This follows from the universal property of $ A\otimes_{\mathbb Z} B $.

Can you provide a counterexample then?

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