Extending base vector field via tensorproduct from $\mathbb{Q}$ to $\mathbb{R}$ Let $L$ be a finite dimensional vector space over some field $\mathbb{F}$ ($\mathrm{char}(\mathbb{F})=0$), $H\subset L$ a subspace of $L$ with $\mathrm{dim}_{\mathbb{F}}(H)=h$ and let $H^*$ be its dual space. Now let $\Phi\subset H^*$ be a root system$^1$ (aka $\Phi$ spans $H^*$) and define
$$E_{\mathbb{Q}}:=\mathrm{span}(\Phi\mid \text{with coefficients in }\mathbb{Q}),$$
so we have $E_{\mathbb{Q}}\subset H^*$. We can now define 
$$\mathbb{E}:=\mathbb{R}\otimes_{\mathbb{Q}}E_{\mathbb{Q}}.$$ 
I'm having a hard time relating this space $\mathbb{E}$ to $H^*$. I think its basis should be $\{1\otimes\alpha\}_{\alpha\in E_{\mathbb{Q}}}$ and every element of this space is a linear combination of this basis with coefficients from $\mathbb{Q}$ (thats at least how I interpret $\otimes_{\mathbb{Q}}$). 
Questions:


*

*In the lecture it was mentioned that $\mathbb{E}$ "is obtained by extending the base field from $\mathbb{Q}$ to $\mathbb{R}$". This sounds to me as if $\mathbb{E}\cong E_{\mathbb{R}}$, is this true?

*Assuming $\mathbb{F}=\mathbb{C}$, is $H^*\cong \mathbb{E} \oplus i \mathbb{E}$?



1) I initially left this information out, since I didn't think it was relevant for the question, but since it may be here is a bit more context: the question came up when I was studying section 8.5 of "Introduction to Lie Algebras and Representation Theory", J. E. Humphreys.
 A: I'll try to answer all of your questions, step by step. First of all, contrary to what is claimed in the OP as of now, a root system is much more than just a set that spans $H^*$. This will have consequences in the answer to question 4 below.
To view the mathemical objects more clearly, you need to consider bases. In general, $\Phi$ will not be a ${\mathbb Q}$-basis of $E_{\mathbb Q}$, but we can extract a basis $\Psi \subseteq \Phi$ from it. We also need a $\mathbb Q$-basis $B$ of $\mathbb F$.
Then, we know that $\mathbb{E}(\mathbb{F})=\mathbb{F}\otimes_{\mathbb{Q}}E_{\mathbb{Q}}$ is a ${\mathbb{Q}}$-vector space with basis $\lbrace b\otimes_{\mathbb{Q}} \psi \ | \ b\in B,\psi \in \Psi \rbrace$.
Question 1: How is $\mathbb{E}(\mathbb{F})$ related to $H^*$ ?
Answer: Consider the map $f : B \times \Psi \to H^*$, defined by $f(b,\psi)=b\psi$. Then $f$ can be uniquely extended to a $\mathbb Q$-bilinear map $f^{\sim}: \mathbb{E}(\mathbb{F}) \to H^*$. Because both $B$ and $\Psi$ are bases in their respective spaces, $f^{\sim}$ is bijective. This allows us to identify $\mathbb{E}(\mathbb{F})$ with $f^{\sim}({\mathbb{E}}(\mathbb{F}))$ which is a $\mathbb Q$-subpace of $H^*$. Note that, since $f^{\sim}({\mathbb{E}}(\mathbb{F}))$ is stable by multiplication by an element of $B$ or $\mathbb Q$, it is also stable by multiplication by an element of $\mathbb F$, so it is a $\mathbb F$-subpace of $H^*$ as well.
Question 2: Is a ${\cal B}=\{1\otimes\alpha\}_{\alpha\in E_{\mathbb{Q}}}$ a basis of $\mathbb E(\mathbb{F})$ ?
Answer: No. It is not a basis  because it is not linearly independent : for a non-zero $\alpha \in E_{\mathbb{Q}}$, if you put $v_1=1\otimes\alpha$, $v_2=1\otimes(-\alpha)$, then $v_1$ and $v_2$ are two distinct vectors of $\cal B$ but $v_1+v_2=0$.
On the other hand, it is a generating set over $\mathbb F$ (but not over $\mathbb Q$ in general).
Question 3: Does $\mathbb{E}\cong E_{\mathbb{R}}$ ?
Answer: (here ${\mathbb F}=\mathbb R$) Yes, because if you recall the $f^{\sim}$ from question 1, the image $f^{\sim}({\mathbb{E}})$, coincides with the $\mathbb R$-space of linear combinations with real coefficients of vectors in $\Psi$ . So $f^{\sim}({\mathbb{E}}) \cong E_{\mathbb{R}}$.
Question 4:  Assuming $\mathbb{F}=\mathbb{C}$, is $H^*\cong \mathbb{E} \oplus i \mathbb{E}$?
Answer: (here ${\mathbb F}=\mathbb C$) Yes to the sum decomposition part, No to the direct sum part (although the sum is indeed direct when $\Phi$ is a root system). Since $\Phi$ generates $H^*$ over $\mathbb C$ and $\Psi$ generates $\Phi$ over $\mathbb Q$, it follows that $\Psi$ generates $H^*$ over $\mathbb C$ also. So the elements of $H^*$ are exactly the sum of elements of the form $v=(a+ib)\psi$, with $a,b\in{\mathbb R},\psi \in \Psi$. We can then write $v=a\psi+ib\psi$ ; the first summand is in $\mathbb E$ while the second is in $i\mathbb E$ ; this shows that $H^*= \mathbb{E} + i \mathbb{E}$.
However, there is no reason for the sum $\mathbb{E} + i \mathbb{E}$ to be direct : if, for example, you have a $\phi$ such that $\Phi$ contains both $\phi$ and $i\phi$, then $\phi \in \mathbb{E} \cap i \mathbb{E}$.
On the other hand, when $\Phi$ is a root system, its members live in a Euclidean space with real coordinates, so the sum is obviously direct.
