I am going through some lecture notes in commutative algebra. I am struggling with one basic example which I want to understand fully before going further. The example is the following:
We take $\mathbb{C}$-module $\mathbb{C}^n.$ We restrict scalars to $\mathbb{R}$ and obtain $\mathbb{R}^{2n}.$ Then we extend scalars to $\mathbb{C}$ and obtain $\mathbb{C}\bigotimes_\mathbb{R} \mathbb{R}^{2n}\cong \mathbb{C}^{2n}.$
I guess I understand the first (restriction) part of this example. We think of the natural embedding $f: \mathbb{R}\to\mathbb{C}.$ That allows us to view the $\mathbb{C}$-module $\mathbb{C}^n$ as a $\mathbb{R}$-module via the action $r(c_1,\ldots,c_n):=(f(r)c_1,\ldots,f(r)c_n)=(rc_1,\ldots,rc_n)$ for $\forall r\in\mathbb{R},~(c_1,\ldots, c_n)\in \mathbb{C}^n.$ This $\mathbb{R}$-module is isomorphic to $\mathbb{R}^{2n}$ via the $\mathbb{R}$-linear map $(a_1+b_1i,\ldots,a_n+b_ni)\mapsto (a_1,b_1,a_2,b_2,\ldots, a_n,b_n).$
Now I want to take this $\mathbb{R}$-module $\mathbb{R}^{2n}$ and extend scalars to $\mathbb{C}$ using the same embedding $f.$ In order to do that I endow the tensor product $\mathbb{C}\bigotimes_\mathbb{R} \mathbb{R}^{2n}$ with a $\mathbb{C}$-module structure via the action $c\left(d\otimes (r_1,\ldots,r_{2n})\right):=(cd)\otimes (r_1,\ldots,r_{2n})$ for $\forall c,d\in\mathbb{C},~(r_1,\ldots, r_{2n})\in \mathbb{R}^{2n}.$ How should I immediately see that this $\mathbb{C}$-module is isomorphic to the $\mathbb{C}$-module $\mathbb{C}^{2n}$ (and what is this isomorphism precisely)? When I think about an arbitrary element $\sum\limits_{i=1}^{k} c_i\otimes (r_{i,1},\ldots, r_{i,n})\in\mathbb{C}\bigotimes_\mathbb{R} \mathbb{R}^{2n},$ I don't see how it resembles an element of $\mathbb{C}^{2n}$ even if we can multiply by complex numbers from the left.
Thank you!
When you work with tensor products, always use the universal Property. First define a bilinear map $\mathbb{C}\times\mathbb{R}^{2n}\to\mathbb{C}^{2n}$. The map is very natural, simply $(\lambda, (x_1,...,x_{2n}))\to(\lambda\cdot x_1,...,\lambda\cdot x_{2n})$. It is clearly bilinear over $\mathbb{R}$, and so induces a homomorphism $\varphi:\mathbb{C}\otimes_{\mathbb{R}}\mathbb{R}^{2n}\to\mathbb{C}^{2n}$ of $\mathbb{R}$-modules. By a direct computation, we see that actually it is even a map of $\mathbb{C}$-modules. (note, you don't get that from the universal property, this is just something you check by hand)
In order to show $\varphi$ is an isomorphism, we define an inverse map. This is also quiet natural. Given an element $(a_1+ib_1,...,a_{2n}+ib_{2n})=(a_1,...,a_{2n})+i(b_1,...,b_{2n})$ of $\mathbb{C}^{2n}$, send it to the following element:
$1\otimes (a_1,...,a_{2n})+i\otimes(b_1,...,b_{2n})$
It's very easy to check that this is an inverse of $\varphi$. Note that an inverse is automatically a $\mathbb{C}$-linear map.
Note: If we denote the inverse by $\psi$, note that when you check that $\psi\circ\varphi$ is the identity, it is enough to check that on simple tensors. (so no need to write complicated sums) That is because the composition is clearly an additive map, and every element is a sum of simple tensors.