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Let $x = \cos(2π/3)$.
I need to prove that $F$ = {$a+bx+cx^2 \mid a, b, c ∈ \mathbb Q$} with the standard addition and multiplication is a field.
I'm trying to prove there is a multiplicative inverse, with no progress.

user26857
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fgv
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    This question seems (to some extent) related: http://math.stackexchange.com/questions/1429965/mathbb-qe-frac2-pi-i3-cong-mathbb-qx-x2x1 – Martin Sleziak Nov 04 '15 at 19:04
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    $x = \cos(2π/3) = -1/2$. Are you sure you don't want $\cos(2\pi/3) + i\sin(2\pi/3) = (-1 + \sqrt{3}i)/2$? – Viktor Vaughn Nov 05 '15 at 00:35
  • For the general case, see http://math.stackexchange.com/questions/1198326/for-f-alpha-as-a-finite-field-extension-of-f-is-1-alpha-in-f-alpha. – lhf Nov 05 '15 at 00:45

2 Answers2

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First show that $x$ is a root of the polynomial $X^2 + X + 1$. Another way of saying this is that $x^2 = -x -1$. Then when you are given $a + bx + cx^2$, substitute $-x-1$ for $x^2$, and reduce to the case $a + bx$. Now I claim $X^2 + X +1$ is irreducible over $\mathbb{Q}$; therefore $a+ bX$ is coprime to $X^2 + X + 1$ and we may use the Euclidean Algorithm to produce polynomials $P$, $Q$ such that

$$P(X)(a + bX) + Q(X)(X^2+X+1) = 1.$$

Then $P(x)$ will be an inverse for $a+bx$.

You'll notice from this process that there can be different $a$, $b$, $c$ corresponding to the same element. For instance, $(0,0,1)$ is the same element as $(-1,-1,0)$.

(How did I get that polynomial $X^2 + X + 1$? I know that $\operatorname{cis} 2\pi/3$ is a third root of unity, so I started with $X^3 - 1$ and factored out $X-1$.)

Eric Auld
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If you have proved that $F$ is a vector space over $\mathbb Q$ of finite dimension ($3$ in this case), then for $z\in F$, $z\ne0$, the map $w \mapsto zw$ is an injective linear transformation and so is surjective, by the rank-nullity theorem. Therefore, $zw=1$ for some $w\in F$ and every non-zero element in $F$ has a multiplicative inverse.

lhf
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