How Does E Relate to Q[x]/(x²+x+1) in Complex Algebra?

[mathjam0990]

Let, E={a+bw : a,b in ℚ) ⊆ ℂ
w = -1/2 + [√(3)/2]*i ∈ C

Prove: E is closed under addition, subtraction, multiplication and division (by non zero elements)

Prove: E ≅ Q[x]/(x²+x+1)

Is the goal to show that for any two elements in E, all 4 operations can be performed on those two elements and the result would still be within E?

Is every element of Q[x]/(x²+x+1) in the form (a+bi)(x²+x+1) which would lead to showing why E ≅ Q[x]/(x²+x+1) ?

I'm not even sure of my statements are correct so it is hard to proceed forward. If anyone could provide a detailed answer as to how to solve this that would be most helpful. Thanks!

 

[Deveno]

The elements of $\Bbb Q[x]/\langle x^2 + x + 1\rangle$ are all cosets of the form:

$a + bx + \langle x^2 + x + 1\rangle$

This is because we can write any element of $\Bbb Q[x]$ as $q(x)(x^2 + x + 1) + r(x)$, where the degree of $r$ is less than 2, and $q(x)(x^2 + x + 1) \in \langle x^2 + x + 1\rangle$.

The above might give a hint as to what the possible isomorphism between $E$ and $\Bbb Q[x]/\langle x^2 + x + 1\rangle$ might be.

It suffices to show that $E$ is closed under subtraction and division (for non-zero elements for division). This is because:

$a+b = a-(-b)$, and $ab = \dfrac{a}{\frac{1}{b}}$

The particular complex number $\omega$ is, isn't all that important, what IS important is that:

$\omega^3 = 1$, and $\omega \neq 1$.

(Why? well it turns out that $x^2 + x + 1 = \dfrac{x^3 - 1}{x-1}$).