I have following, probably really trivial question.
Lets take $\mathbb{Z}^d \otimes \mathbb{R}$. Consider this as vector space over $R$ by defining
$\mu (v\otimes \lambda) = v \otimes (\lambda \mu)$
Then claim is that dimension of this real vector space is $d$. Clearly it's not bigger then $d$:
Take set $e_1 \otimes 1,e_2 \otimes 1,.., e_d \otimes 1$, any element of tensor product is linear combination of those ($e_i$ are of course basis of the lattice) But how to prove that this set is linearly independent? It seems really obvious but what's formal proof?
Define the homomorphism $\psi_k:\mathbb{Z}^d\otimes \mathbb{R}\to\mathbb{R}$ by $\psi_k(\sum n_i e_i\otimes r_i)=n_k r_k$ where the $e_i$ are the basis for $\mathbb{Z}^d$. All you need to do now is to show that this is well defined (using the definition of the tensor product) to conclude that if $x=\sum e_i\otimes r_i=0$ then the $\psi_i(x)=r_i$ are zero, namely the $e_i$ are linearly independent over $\mathbb{R}$.