Is $\mathbb{C}\otimes _{\mathbb{Z}} \mathbb{C}$ isomorphic to $\mathbb{C}\otimes _{\mathbb{Q}} \mathbb{C}$ as ring?

Question

Is the natural ring morphism $\mathbb{C}\otimes_{\mathbb{Z}}\mathbb{C}\to\mathbb{C}\otimes_{\mathbb{Q}}\mathbb{C}$ an isomorphism?

In other words, is there a $\mathbb Z$-linear map $f:\mathbb{C}\otimes_{\mathbb{Q}}\mathbb{C}\to\mathbb{C}\otimes_{\mathbb{Z}}\mathbb{C}$ such that $$ f(z\otimes w)=z\otimes w $$ for all $z,w\in\mathbb{C}$? (Note that the two occurrences of $z\otimes w$ in the above display have different meanings.)

Answer 1

If $U$ and $V$ are vector spaces over $\Bbb Q$, then $U\otimes_\Bbb Z V$ and $U\otimes_\Bbb Q V$ are always isomorphic. To see this, $$U\otimes_\Bbb Z V\cong U\otimes_\Bbb Q\Bbb Q\otimes_\Bbb Z V \cong U\otimes_\Bbb Q V.$$ As the last stage we use $\Bbb Q\otimes_\Bbb Z\Bbb Q\cong\Bbb Q$ and the fact that $V$ has a basis as a $\Bbb Q$-vector space and that direct sums preserve tensor products.

Answer 2

Hint:

 $$\mathbf C\otimes_{\mathbf Z }\mathbf C\simeq(\mathbf C\otimes_{\mathbf Z}\mathbf Q)\otimes_{\mathbf Q}\mathbf C. $$ Now $\;\mathbf C\otimes_{\mathbf Z}\mathbf Q\simeq \mathbf C$ because of the universal property of rings of fractions.

Answer 3

In this post $U,V,X$ and $Y$ are $\mathbb Q$-vector spaces, $a$ is an integer, $b$ is a nonzero integer, and $u,v$ and $x$ are vectors of $U,V$ and $X$ respectively.

$(\star)$ A $\mathbb Z$-linear map $f:X\to Y$ is automatically $\mathbb Q$-linear.

Proof of $(\star)$: We have $$ f\left(\frac abx\right)=af\left(\frac 1bx\right)=\frac ab\ b\ f\left(\frac 1bx\right)=\frac ab\ f(x).\ \square $$ We equip the $\mathbb Z$-module $U\otimes_{\mathbb Z}V$ with the $\mathbb Q$-vector space structure defined by $$ \frac ab\ (u\otimes v):=\left(\frac abu\right)\otimes v. $$ Consider the $\mathbb Z$-bilinear maps $$ g:U\times V\to U\otimes_{\mathbb Q}V,\quad(u,v)\mapsto u\otimes v, $$ $$ h:U\times V\to U\otimes_{\mathbb Z}V,\quad(u,v)\mapsto u\otimes v. $$ The map $g$ induces a $\mathbb Z$-linear map $$ g':U\otimes_{\mathbb Z}V\to U\otimes_{\mathbb Q}V, $$ which is $\mathbb Q$-linear by $(\star)$.

We claim that $h$ is $\mathbb Q$-bilinear.

The $\mathbb Q$-linearity in the first variable is clear. The $\mathbb Q$-linearity in the second variable follows from $(\star)$.

As $h$ is $\mathbb Q$-bilinear, it induces a $\mathbb Q$-linear map $$ h':U\otimes_{\mathbb Q}V\to U\otimes_{\mathbb Z}V. $$ It is easy to see that $g'$ and $h'$ are mutual inverses.