Determine if the subset $E$ determine a subfield of the Complex Numbers Let $E$ be the set of all $a+bu$, where $a,b$ are rational numbers and $u = \frac{ (-1+\sqrt{-3})}{2}$, Hint: Notice that $u^2 + u + 1 = 0$, therefore, $u^3 = 1$.
In the book Im reading, the consired conditions that a subset need to determine a subfield, is to have at least two elements, be closed under subtraction and for any two elements $x,y$, $(x,y \in E \land y\neq 0) \Rightarrow x/y \in E$.
The first and second condition are clearly true, for the first we can fix b=0 and just let a=1 and a=0, then we have two distintic elements in $E$.
For the second condition just let $x=a+bu$ and $y=c+du$ for $a,b,c,d \in \mathbb{Q}$, then $x-y = (a+bu) - (c+du) = (a-c) + (b-d)u \in E$.
But I got stucked in the last condition, I dont know how can I show if the quotient of all elements in $E$ still in $E$, the first thing I tried was just to show that $E$ is closed under multiplication, starting from $(a+bu) \times (c+du)$ from this I got $(ac-bd)+(ad+bc-bd)u$, which is an element of $E$, thus the set is closed under multiplication.
Then I tried to assume that $c+du \neq 0$ and show that $\frac{(a+bu)}{(c+du)}$ is an element of $E$, I wrote it as $(a+bu)\times(c+du)^{-1}$, but since I dont know if the inverse of $(c+du)$ is in the set I give my last try assuming $(p+qu) = (c+du)^{-1}$, because if its true, I will can solve the following equation:
$$(c+du)\times(p+qu) = 1$$
I just managed to got $(cp-dq)+(cq+dp-dq)u = 1$, and then using the hint,
$$(cp-dq)+(cq+dp-dq)u = u^3$$
But I dont know how to proceed to show if $E$ is closed under taking the multiplicative inverses, So any tip will be appreciated.
 A: You have correctly (as far as I can see) gotten to
$$ (c+du)(p+qu) = (cp-dq)+(cq+dp-dq)u $$
You want this to equal $1$ (which was a good initial instinct), but we can be slightly more general and want it to be $a+bu$ such that we get to $\frac{a+bu}{c+du}$ in a single step. Setting the expression above equal to $a+bu$ gives us
$$ \begin{align} cp-dq &= a \\ cq+dp-dq &= b \end{align} $$
because $1$ and $u$ are obviously linearly independent over $\mathbb Q$ (or even over $\mathbb R$). These equations happen to be linear in $p$ and $q$, so we can also write them as
$$ \begin{bmatrix} c & -d \\ d & c-d \end{bmatrix} \begin{bmatrix} p \\ q \end{bmatrix} = \begin{bmatrix} a \\ b \end{bmatrix} $$
and we know from linear algebra that this has a solution for $p$ and $q$ if the $2\times 2$ matrix on the left is non-singular. For our present purposes we don't need to find that solution, only to know that it exists -- Cramer's rule tells us that it will have rational $p$ and $q$ as we need it to.
Can you take it from here?
