Let $\mathbb K$ be a subfield of $\mathbb F$. Suppose that $V$ is a vector space over $\mathbb K$ and set: $V_{\mathbb F}:= \mathbb F \otimes_{\mathbb K}V$. Show that there exists a unique function $\mathbb K$ -bilinear $\mu: \mathbb F \times V_{\mathbb F}\rightarrow V_{\mathbb F}$ satisfying: $\mu(\lambda,1\otimes v) = \lambda \otimes v$ for every $\lambda \in F, v \in V$. Then, prove that $(V_{\mathbb F},+,\mu)$ is a vector space over $\mathbb F$.
I'm having trouble trying to justify the existence and uniqueness of the $\mathbb K$- bilinear map. I'll post here what i've done and what is actually bothering me.
Since $V_{\mathbb F}$ is tensor product over $\mathbb K$ to $\mathbb F, V$, if we consider $\alpha = (v_i)_{i\in I}$ basis to $V$ and $\{1\} $ basis to $\mathbb F$, then $(1\otimes v_i)_{i\in I}$ is a basis to $V_{\mathbb F}$. Using the fact that every bilinear map is totally determined by how it is defined in a basis (existence and uniqueness), we define $\mu$ by the following rule: $$\mu(1,1\otimes v_i) = 1\otimes v_i$$ for every $v_i\in \alpha.$ Now we prove that $\mu$ satisfy the desired property. In fact, given $\lambda\in F, v\in V$, we write $v = \sum_ia_iv_i .$ It follows that: $$\mu(\lambda,1\otimes v)= \mu (1.\lambda,1\otimes \sum_ia_iv_i) = \lambda\sum_ia_i\mu(1,1\otimes v_i)=\\=\lambda\sum_ia_i(1\otimes v_i) = \lambda (1\otimes \sum_ia_iv_i) = \lambda(1\otimes v) = \lambda\otimes v. $$
So what bothers me is: how do I find a basis to $V_{\mathbb F} = \mathbb F \otimes_{\mathbb K} V?$ In this 'proof' i've utilized that $\{1\}$ is basis to $\mathbb F$ over $\mathbb K$, which is false in general (as we see by: $\{1,i\}$ is basis to $\mathbb C$ over $\mathbb R$). I think that the idea of defining $\mu$ over elements of the basis is the correct approach, but i'm not sure how those elements are.
Any help would be much appreciated, and thanks in advance!
Given any $\mu\in \mathbb F$, define: $F_\mu: \mathbb F \times V \rightarrow V_{\mathbb F}$ by $(\lambda,v)\mapsto \mu \otimes \lambda v$. This map is bilinear, hence by the universal property there exists a unique linear map $g_\mu: V_{\mathbb F} \rightarrow V_{\mathbb F}$ satisfying: $g_\mu(\lambda \otimes v) = \mu \otimes \lambda v$ for any $\lambda \in \mathbb F$ and $v\in V$.
Define $\phi:\mathbb F \rightarrow \mbox{Hom}(V_\mathbb{F},V_\mathbb{F})$ by $\phi(\mu) = g_\mu$, which is linear. Now we are done, because we can simply consider $\mu: \mathbb F \times V_{\mathbb F}$ by $\mu(a,w) = \phi(a)(w)$ for any $a\in \mathbb F,v\in V_{\mathbb F}$.
In particular, we verify that: $\mu(\lambda,1\otimes v) = \phi(\lambda)(1\otimes v) = g_{\lambda}(1\otimes v) = \lambda\otimes1v = \lambda\otimes v$ as required.