Understanding $\mathbb{C} \otimes_{\mathbb{Q}} \mathbb{C}$ I am really confused about understanding $\mathbb{C} \otimes_{\mathbb{Q}} \mathbb{C}$ as a $\mathbb{C}$-vector space and as a $\mathbb{C}$-algebra. I was under the impression that $\mathbb{C} \cong \mathbb{C} \otimes_{\mathbb{Q}} \mathbb{C}$ as $\mathbb{C}$-algebras. However after reading on the internet (https://www.math.ru.nl/~bmoonen/CatHomAlg/TensorProd.pdf), I believe that $\mathbb{C} \otimes_{\mathbb{Q}} \mathbb{C}$ viewed as a
$\mathbb{C}$-algebra is 
isomorphic to an uncountable product of copies of $\mathbb{C}$. However, I believe I have constructed a proof that at least $\mathbb{C} \cong \mathbb{C} \otimes_{\mathbb{Q}} \mathbb{C}$ as $\mathbb{C}$-vector spaces, but I never use that the tensor is over $\mathbb{Q}$ at all. Here is the proof:
Proof: Since tensor products are unique, I will show that $\mathbb{C}$ satisfies the universal property for the tensor product $\mathbb{C} \otimes_{\mathbb{Q}} \mathbb{C}$, which will imply that they are isomorphic as
$\mathbb{C}$-vector spaces. 
To this end, let $\tau: \mathbb{C} \times \mathbb{C} \to \mathbb{C}$ where
$\tau((c_1, c_2)) = c_1 \cdot c_2$. 
It is easy to check that this is $\mathbb{C}$-bilinear. 
Then let $M$ be any $\mathbb{C}$-module and let $f$ be any 
$\mathbb{C}$-bilinear map from $\mathbb{C} \times \mathbb{C} \to M$. 
Then we can choose $g: \mathbb{C}  \to M$
such that $g(c) = f(c, 1)$. 
Then this map is a $\mathbb{C}$-module homomorphism and the diagram commutes
since $g \circ \tau ((c_1, c_2)) = g(c_1 \cdot c_2) = f(c_1 \cdot c_2, 1)
= f(c_1, c_2)$ since $f$ is $\mathbb{C}$-bilinear. 
Thus $\mathbb{C}$ satisfies the definition of the tensor product.
Hence $\mathbb{C} \cong \mathbb{C} \otimes_{\mathbb{Q}} \mathbb{C}$ 
as a $\mathbb{C}$-vector space (or at least as an abelian group). 
Can someone please help me understand if 
(1) $\mathbb{C} \cong \mathbb{C} \otimes_{\mathbb{Q}} \mathbb{C}$ as $\mathbb{C}$-vector spaces or is there an error in my proof (why does my proof not use that the tensor is over $\mathbb{Q}$ anywhere) and
(2) what is $\mathbb{C} \otimes_{\mathbb{Q}} \mathbb{C}$ isomorphic to as a $\mathbb{C}$-algebra and how do we prove this?
Thank you so much in advanced!
 A: Regarding (1). I don’t know why your proof fails because it doesn’t even work syntactically. You have to clean up the mess first and then see what’s really left of your proof. Let me go into detail.


Then let $M$ be any $ℂ$-module and let $f$ be any $ℂ$-bilinear map $ℂ × ℂ → M$. 

If you want to consider the tensor product $ℂ \otimes_ℚ ℂ$ and verify its universal property for $ℂ$, then you should only consider $ℚ$-linear modules $M$ and $ℚ$-bilinear maps $ℂ × ℂ → M$. So you have less to work with.

Then we can choose $g\colon ℂ \otimes_ℚ ℂ → M$ such that $g(c_1\otimes c_2)=c_1⋅c_2$.

What do you mean? You have $c_1·c_2 ∈ ℂ ≠ M$ in general. Maybe you do mean “$f(c_1,c_2)$” instead of “$c_1·c_2$”? Don’t you want to construct a map $ℂ → M$? If so, how would you choose it?

Then this map is a $ℂ$-module homomorphism and the diagram commutes since $g∘τ((c_1,c_2))=g(c_1⋅c_2)=f(c_1⋅c_2,1)=f(c_1,c_2)$ since $f$ is $ℂ$-bilinear. 

Now, $g(c_1·c_2)$ doesn’t make any sense, as $c_1·c_2$ is no element in $ℂ \otimes_ℚ ℂ$. I also don’t see why “$g(c_1·c_2) = f(c_1·c_2,1)$” if not as “$g(c_1\otimes c_2) = f(c_1·c_2,1)$” through
$$g(c_1\otimes c_2) = f(c_1,c_2) = f(c_1·c_2,1),$$
so that chain of equalities seems to be mixed up to me, in a wrong order. Also, you probably don’t have the $ℂ$-bilinearity of $f$ to work with, see above.

Thus $ℂ$ satisfies the definition of the tensor product. Hence $ℂ \cong ℂ \otimes_ℚ ℂ$ as a $ℂ$-vector space (or at least as an abelian group).

Even if you had constructed a linear map  $ℂ → M$ so that $f$ factorizes through it and multiplication $ℂ × ℂ → ℂ$, you would still need to show that it is unique.

Not knowing how you mean to prove $ℂ = ℂ \otimes_ℚ ℂ$ in particular because your proof so far doesn’t make sense to me syntactically, I suspect that you think that $ℂ \otimes_ℚ ℂ = \{c_1 \otimes c_2;~c_1, c_2 ∈ ℂ\}$. Put otherwise: I suspect that you think that all elements of $ℂ \otimes_ℚ ℂ$ are elementary tensors – elements of the form “$c_1\otimes c_2$”. However, the elements of $ℂ \otimes_ℚ ℂ$ are sums of such elementary tensors and do not need to be elementary tensors themselves.
For example, since multiplication $ℂ × ℂ → ℂ$ has no zero divisors in $ℂ$, the corresponding map
$$μ_ℂ\colon ℂ \otimes_ℚ ℂ → ℂ,~c_1 \otimes c_2 ↦ c_1·c_2$$
has no nontrivial elementary tensors in the kernel, yet $x = \mathrm i \otimes \mathrm i + 1 \otimes 1 ∈ \ker μ_ℂ$ and as $1, \mathrm i$ are $ℚ$-linearly independent in $ℂ$, $1 \otimes 1$ and $\mathrm i \otimes \mathrm i$ are $ℚ$-linearly independent in $ℂ \otimes_ℚ ℂ$, so $x ≠ 0$.

Regarding (2). I don’t know of any nicer description of $ℂ\otimes_ℚ ℂ$ as a $ℚ$-Algebra, but here are two points regarding its structure.


*

*$ℂ \otimes_ℚ ℂ$ as a $ℚ$-linear space is isomorphic to $\bigoplus_{c ∈ \mathfrak c} ℂ$, where $\mathfrak c$ is the cardinality of the continuum. This is because, as a $ℚ$-linear space $ℂ = \bigoplus_{c ∈ \mathfrak c} ℚ$, $ℂ \otimes_ℚ ℚ = ℂ$ and the tensor product commutes with direct sums.

*$ℂ \otimes_ℚ ℂ$ is not an integral domain. For example,
$$(1 \otimes \mathrm i + \mathrm i \otimes 1)(1 \otimes \mathrm i - \mathrm i \otimes 1) = 1 \otimes (-1) - ( (-1) \otimes 1) = 0.$$
So certainly, $ℂ \otimes_ℚ ℂ \not\cong ℂ$.
A: The universality of the candidate has to be shown for any $\Bbb Q$-bilinear morphism, i.e. that the candidate is a factorizing in between station for any such morphism.
Your error in thinking is that you pick a wrong candidate, show the factorization for only one (relatively simple) bilinear morphism. Note that the structure of $\Bbb C$ as a $\Bbb Q$-vectorspace is very complicated. And $\Bbb Q$-linear morphisms can also be very complicated. Here is an example. By the axiom of choice, there exists a basis $B$ of the vector space $\Bbb C$ over $\Bbb Q$. Put some order on it. Let $b$ be the first element in $B$. We define $f:\Bbb C\to \Bbb Q$ as follows. For $x\in \Bbb C$ we write it as a linear combination w.r.t. $B$ and pick the coefficient of $b$ in it. (It is zero if $b$ "does not appear".)
Now consider the map from $\Bbb C\times \Bbb C\to\Bbb Q$ defined as
$$
(x,y)\to f(x)f(y)\ .
$$
Can you factor through $(x,y)\to xy$? If not, your candidate is not universal.
Just a final comment. Try to understand (as a vectorspace and as an algebra) the tensor product of fields
$$ \Bbb Q[i]\otimes_{\Bbb Q} \Bbb Q[i]\ .$$
You will immediately realize the problem in your argumentation. (The tensor product is not the field $\Bbb Q[i]$, although we can write down some product map.)
