I'm taking a course in algebraic number theory and we are starting to discuss lattices and Minkowski theory. We were given this introductory exercise:
Let $K$ be a field of numbers of order $n$ and embbedings $\tau_1, ..., \tau_n:K\to\mathbb{C}$. Show that $\phi:K\otimes_{\mathbb{Q}}\mathbb{C}\to \prod_{\tau}\mathbb{C}$ defined by $\phi(a\otimes z)=(z\,\tau_1(a), ..., z\,\tau_n(a))$ is an isomorphism between $\mathbb{C}$-vector spaces.
I've already seen a solution, which uses linear independence of characters (which I'm familiar with) to prove injectivity and then uses a dimension argument to conclude bijectiviy.
I know I'm able to follow the outline of that proof, but actually my problem is more basic: I just don't know what the sets $K\otimes_{\mathbb{Q}}\mathbb{C}$ and $\prod_{\tau}\mathbb{C}$ actually mean.
I'm familiar with the tensor product $U\otimes V$, when $U$ and $V$ are both $\mathbb{F}$-vector fields (i.e., both have the same base field), so when I see the notation $U\otimes_{\mathbb{Q}} V$, I assume implicitly that $U$ and $V$ are both $\mathbb{Q}$-vector spaces, but that is not the case with $K\otimes_{\mathbb{Q}}\mathbb{C}$. That is really confusing me: take for example $K=\mathbb{Q}(i)$. What is the canonical basis for $\mathbb{Q}(i)\otimes_{\mathbb{Q}} \mathbb{C}$? How am I supposed to write $1\otimes \sqrt{2}$ in the canonical basis? Like $\sqrt{2}(1\otimes 1)$? But am I allowed to "pull out" an irrational number from the second coordinate like that? Then what does $\mathbb{Q}$ in "$\otimes_{\mathbb{Q}}$" stand for? I can't make sense of it.
Second, I thought $\prod_{\tau}\mathbb{C}$ should be the set of $n$-uples $(\tau_1(a), ..., \tau_n(a))$ where $a\in K$. But then how do I know that $\phi(a\otimes z)=(z\,\tau_1(a), ..., z\,\tau_n(a))$ is of the form $(\tau_1(b), ..., \tau_n(b))$ for some $b\in K$? If there is no such $b$, $\phi$ would not be well defined...
Any help would be useful, thanks in advance!
In general, if $V$ is a vector space over a field $K$ and $L$ is a field extension of $K$, then $V\otimes_KL$ is a vector space over $L$. This operation is called extension of scalars.
We want to formally define scalar multiplication by $L$ in a way that is compatible with the multiplication by $K$. The tensor product allows us to do this. For $v\otimes \lambda\in V\otimes_KL$ and $\mu\in L$, we have $$\mu(v\otimes\lambda) = v\otimes\mu\lambda.$$ And if $\mu\in K$, then this is just $\mu v\otimes\lambda$. You can think of this as taking the $L$-span of a $K$-basis of $V$. So if $\{e_i\}$ is a $K$-basis of $V$, then $\{e_i\otimes 1\}$ is an $L$-basis of $V\otimes_KL$.
So in your case, $K\otimes_\mathbb Q\mathbb C$ is a $\mathbb C$-vector space of dimension $[K:\mathbb Q]$. If $K=\mathbb Q(i)$, then a $\mathbb C$-basis of $K\otimes_\mathbb Q\mathbb C$ is $\{1\otimes 1,i\otimes 1\}$.
The product $\prod_\tau\mathbb C$ is just $\mathbb C^n$.