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Show that $\mathbb{C} \otimes_\mathbb{Z} \mathbb{C} \cong \mathbb{C} \otimes_\mathbb{Q} \mathbb{C}$

This is not homework, it is part of an answer of Show that $\mathbb{A}_\mathbb{C}^2 \ncong \mathbb{A}_\mathbb{C}^1 \times_{Spec(\mathbb{Z})} \mathbb{A}_\mathbb{C}^1$.

How can I prove that?

Thanks!

Community
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Leafar
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    Use the universality property of the tensor product and the fact that a $\mathbb{Z}$-linear function is also $\mathbb{Q}$-linear. – anomaly May 18 '15 at 16:15

1 Answers1

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In fact $a\otimes b\mapsto a\otimes b$ is an isomorphism $A\otimes_DB\leftrightarrow A\otimes_FB$ as spaces or $D$-algebras, for any spaces/algebras $A$ and $B$ over a field $F$ which is the fraction field of a domain $D$. The reason is that

$$\begin{array}{ll} \displaystyle \color{Blue}{\frac{1}{y}}a\otimes b & \displaystyle =\frac{1}{y}a\otimes \color{Green}{y}\frac{1}{y}b \\ & \displaystyle =\color{Green}{y}\frac{1}{y}a\otimes \frac{1}{y}b \\ & \displaystyle =a\otimes\color{Blue}{\frac{1}{y}}b. \end{array}$$

Using the fact that $A$ and $B$ are divisible, we can move inverses across the $\otimes_D$ symbol by simply moving the denominator across from the other side. Hence, any fraction in $F$ can be moved.

(I'll leave you to turn these thoughts into a proof with the desired level of rigor.)

anon
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  • This is an isomorphism of $F$ algebras? How does on e view the left side as one? –  May 24 '15 at 21:08
  • @user114539 Even though when I wrote the parenthetical I was only thinking of the isomorphism as being of $D$-algebras, one can also interpret it as true of $F$-algebras too - the way to make scalars in $F$ act on $A\otimes_DB$ should be obvious (at least after knowing $ra\otimes b=a\otimes rb$ for all $r\in F$ from this answer). – anon May 24 '15 at 22:50