Polynomial: Is there a theorem that can save my proof when $K$ doesn't include $\mathbb C$ 
Suppose $f(x),g(x)\in K[x]$ ($K$ a number field), let $f(x)=x^{3m}+x^{3n+1}+x^{3p+2}$, where $m,n,p\in\mathbb N$, and let $g(x)=x^2+x+1$, prove:
  $$g(x)\mid f(x)$$

I think this problem is not very easy unless I specify $K$ to be $\mathbb C$. If in $\mathbb C$, then it is plain to see 
$$g(x)=(x-\omega_1)(x-\omega_2)$$
where $\omega_{1}=e^{i2\pi/3}$ and $\omega_{2}=e^{-i2\pi/3}$. And it's also obvious after simple calculation, that
$$f(\omega_1)=f(\omega_2)=0$$
By the Factor Theorem, since $\omega_{1,2}$ are distinct,
$$(x-\omega_1)(x-\omega_2)\mid f(x)$$
i.e.
$$g(x)\mid f(x)$$
Nevertheless, all I have done so far is based on the assumption that $K$ is (or includes) $\mathbb C$, without which I think it will be ridiculous to write something like $(x-\omega_{1,2})$ as polynomials. However, I feel strongly that my approach, although based on a not quite reasonable assumption, is in fact a quick-and-easy way, whatever $K$ be.
So, maybe there is a theorem unknown to me which can justify my method when $K$ doesn't include $\mathbb C$, say, $K=\mathbb R$? Of course, I know I am not allowed to factorize like this when $K=R$, but $g(x)\mid f(x)$ is still alright. And I think $g(x)\mid f(x)$ always holds whatever $K$ be, even though the factorization is not always true.
So am I wrong in not giving up my proof? Or is there really a god-sent theorem that will save my proof?  Need some help. Best regards here.
 A: You can work in the extension $K[\omega_1,\omega_2]$, where the argument works. It is just $K[t]/(t^2+t+1)$, so you need no embedding in the complex numbers (if the polynomial $x^2+x+1$ has no roots in $K$, otherwise there's nothing to worry about).
However, the division algorithm between two polynomials in $K[x]$ only uses rational operations on the coefficients. Hence the quotient is in $K[x]$, since $g(x)$ divides $f(x)$ and both are in $K[x]$.
A: The conclusion $g(x)\mid f(x)$ follows directly from
$$
f(x) - g(x) = (x^{3m} - 1) + (x^{3n}-1)\,x +  (x^{3p} -1)\, x^2 
$$
since all the polynomials $x^{3m} - 1$, $x^{3n}-1$ and $x^{3p} -1$
are divisible by $g$:
$$
  x^{3m} - 1 = (x^3 -1)(1 + x^3 + x^6 + \dots + x^{3m-3}) \\
=  g(x) \, (x-1) \, (1 + x^3 + x^6 + \dots + x^{3m-3})
$$
A: It is easy to see $\,g\mid f\,$ in $\,\Bbb Z[x]\,$ so $\,f = gh\,$ in $\,\Bbb Z[x],\,$ which remains true in any commutative ring, i.e. it is a universal identity of commutative rings. The proof that $\,g\mid f\,$ is straightforward
$$g\mid x^3\!-\!1\,\Rightarrow\, {\rm mod}\ g\!:\,\ \color{#c00}{x^3\equiv 1}\,\Rightarrow\, (\color{#c00}{x^3})^m\!+x(\color{#c00}{x^3})^n\!+x^2(\color{#c00}{x^3})^p\equiv 1+x+x^2\equiv g\equiv 0\quad $$
A: If $K$ is a number field, then any element of $K$ is algebraic over $\mathbb{Q}$, so we can say $\mathbb{Q} \subset K \subset \bar{\mathbb{Q}} \subset \mathbb{C}$.
