# Proving that $K\otimes_{\mathbb{Q}}\mathbb{C}\simeq \prod_{\tau}\mathbb{C}$ [duplicate]

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!

## marked as duplicate by punctured dusk, Misha Lavrov, B. Mehta, jgon, Robert ZNov 28 '17 at 8:20

• But $K$ and $\mathbb{C}$ are quite obviously $\mathbb{Q}$-vector spaces, aren't they? So what is wrong with your tensor $1 \otimes \sqrt{2}$? And no, you are not allowed to "pull out" anything but rational numbers. – Torsten Schoeneberg Jan 25 '17 at 0:31
• And $\prod_\tau \mathbb{C}$ should be the set of tuples $(a_1, ..., a_n)$ with $a_i \in \mathbb{C}$, in other words, $\mathbb{C}^n$. Then the $\phi$ you define looks like a perfectly valid map to me, and it should be easy to see that it is an isomorphism. – Torsten Schoeneberg Jan 25 '17 at 0:35
• Perhaps what is confusing you is that $\mathbb{C}$ is infinite-dimensional as a $\mathbb{Q}$-vector space, so there is no "nice" basis. In particular, you could take a basis for $\mathbb{C}$ that has $\sqrt{2}$ as a basis vector, so that $1 \otimes \sqrt{2}$ is already a basis vector of $K \otimes_\mathbb{Q} \mathbb{C}$. – André 3000 Jan 25 '17 at 2:20
• @Mathmo123 Well, if we are very formal, there is a need to consider $\mathbb{C}$ as a vector space over $\mathbb{Q}$, in order to define the tensor product (say, as additive group) in the first place. Then you show, as in your answer, that it has a natural $\mathbb{C}$-vector space structure. – Torsten Schoeneberg Jan 25 '17 at 20:55
• Agreed. On the other hand, $K$ does not have a canonical $\mathbb{Q}$-basis either. The example $K = \mathbb{Q}(i)$ is a bit misleading because $\lbrace 1, i \rbrace$ looks very canonical, although $\lbrace 1, 2+\frac{1}{3}i \rbrace$ from an algebraic viewpoint is not better or worse. The one "canonical" thing about $K$ is its embeddings into $\mathbb{C}$, and using those as "basis" instead of some arbitrarily chosen basis vectors (@AguirreK: there is no "canonical basis") presumably is what the whole exercise is about. – Torsten Schoeneberg Jan 25 '17 at 23:09

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 $\mathbb C^n$.