Conjugates of an element in algebraic number fields Let $K$ be an algebraic number field of degree $n$. Then by primitive element theorem there exist a $\theta \in K$ such that $K = \mathbb{Q}(\theta)$. Let $\theta_1 = \theta, \theta_2, \dots \theta_n$ be the conjugates of $\theta$ over rational numbers (i.e. roots of minimal polynomial in $\mathbb{Q}[x]$ which is satisfied by $\theta$). Since any $\alpha \in K$ can be written uniquely as
$$ \alpha = c_0  + c_1 \theta + \cdots c_{n-1} \theta ^ {n-1}$$ for some rational numbers $c_i ( i =1, 2, ...,n)$. Let
$$ \alpha_k = c_0 + c_1 \theta_k + \cdots + c_{n-1}\theta_k^{n-1}$$ for $k = 1,2,\dots ,n$
We call $\{ \alpha_k \, \vert k = 1 , 2 \dots , n \}$ as $K$-conjugates of $\alpha $.
Our claim is that

$K$-conjugates of $\alpha $ does not depend on $\theta$ for all $k =1,2, \dots n$

Since there exist $\phi \ne \theta $ in $K$ such that $K = \mathbb{Q}(\phi)$, there are unique rational numbers $d_1, \dots d_{n-1}$ such that $\alpha = d_0 + \cdots d_{n-1} \phi^{n-1} $. Now let $\phi_1 = \phi , \phi_2 ,\dots \phi_n$ be conjugates of $\phi$ over rational numbers. Let $\beta_1 = \alpha$ and $$ \beta _ k = d_0 + d_1 \phi_k + \cdots + d_{n-1} \phi_k^{n-1} $$
for $k = 2, \dots , n$.
If we can show that $\beta_k$ are permutation of $\alpha_k$ we are done.
This is easy to see in some examples, consider $K = \mathbb{Q}(\sqrt 2)$. Let $\theta = \sqrt 2, \phi = 5 + \sqrt 2$ and $\alpha = 1 + \sqrt 2$. Since the $\sqrt 2$ satisfies the equation $x^2 - 2 = 0$ we have $\theta_1 = \sqrt 2, \theta _ 2 = - \sqrt 2$. Similarly $\phi_1 = 5 + \sqrt 2 $ and $\phi_2 = 5 - \sqrt 2$.
Now we have $\alpha_1 = 1 +\sqrt 2$ and $\alpha_2 =  1 - \sqrt 2$. Since we can write
$$ \alpha = -4 + (5 + \sqrt 2)$$
we have $\beta_1 =  \alpha_1 = 1 + \sqrt 2$ and $\beta_2 = -4 + (5 - \sqrt 2) = 1 - \sqrt 2 = \alpha_2 $ which fits with out hypothesis.
Attempt to prove:
Initially I thought expressing each $\phi_k ^ m $ as a linear combination of $1, \theta_k, ^2  \dots , \theta_k ^ {n-1}$ and then try to write $\beta_k$ in terms of power of $\theta_k$ and finally show that corresponding coefficients and $c_i$ are equal. But I could not finish this.
I also tried another method which I include as an answer because it is too long to fit here.
But in the above example it appears that $\alpha_k = \beta_k$ for all $k$.I am not sure whether this is true in general. If any other approaches are known it will be helpful. Hints are also welcome.
 A: Let $f_{\alpha}(x)$ be the monic, minimal polynomial of $\alpha$ over $\mathbb{Q}[x]$. Consider two polynomials $f(x)$ and $g(x)$ such that
$$f(x) = (x - \alpha_1) \cdots (x - \alpha_n)$$ and $$g(x) = (x - \beta_1) \cdots (x - \beta_n)$$
We claim that $f(x) \in \mathbb{Q}[x]$. We first observe that $f(x)$ is a symmetric polynomial in terms of $\theta_1, \dots \theta_n$. By fundamental theorem of symmetric polynomials we know that $f(x)$ is a symmetric polynomial in terms of elementary symmetric polynomials in terms of $\theta_1, \dots \theta_n$. But elementary symmetric polynomials are coefficients of minimal polynomial of $\theta$ over $\mathbb{Q}$ so they are rational numbers. Hence $f(x)$ is a polynomial with rational coefficients. A similar argument shows that $g(x)$ is a polynomial over rational numbers.
We observe that $f(\alpha) = 0$ so that $f_{\alpha}(x) \vert f(x)$. So there exist a $h(x) \in \mathbb{Q}[x]$ such that
$$f(x) = f_{\alpha}(x) ^ s h(x)$$ since $\mathbb{Q}[x]$ is an UFD where $h(x)$ does not divide $f_{\alpha}(x)$ and $s$ is a positive integer .
We claim that $h(x)$ is constant polynomial.
If it is not a constant polynomial then there exist a $k$ such that $h(\alpha_k) = 0$. Consider the polynomial $r(x) = c_0 + c_1x + \cdots + c_{n-1} x^{n-1}$ so that $\alpha = r(\theta)$. Now we see that $h(r(\theta_k)) = h(\alpha_k) = 0$ so that $f_{\theta_k}(x)$ divides $h(r(x))$. But it is well-known that $f_{\theta_k}(x) = f_{\theta}(x)$. So that $h(r(\theta)) = h(\alpha) = 0$ which implies that $h(x)$ divides $f_{\alpha}(x)$ which is a contradiction. So $h(x)$ is a constant. Since $f_{\alpha}(x)$ and $f(x)$ are monic and so is $h(x)$. So we have $h(x) = 1$. So
$$ f(x) = f_{\alpha}(x)^s$$
Similarly there exist a positive integer $t$ such that
$$g(x) = f_{\alpha}(x)^t$$
Since $\deg f = \deg g = n$ we must have $s =t $. So we have
$$ (x- \alpha_1) \cdots (x-\alpha_n) = (x - \beta_1) \cdots (x- \beta_n)$$ which is sufficient to conclude that $\alpha_i$s are permutation of $\beta_i$s since $\mathbb{C}[x]$ is a UFD.
Any improvements to this post are welcome.
A: I'd think of this as follows. Write $\sigma_1, \ldots, \sigma_n\colon K\hookrightarrow\mathbb C$ for the field embeddings of $K$ into $\mathbb C$. If we fix $\sigma_1$ to be an initial embedding (so that we can view $K\subset\mathbb C$), and let $\theta\in K\subset\mathbb C$ be a primitive element, then it's clear that the elements $\sigma_k(\theta)$ all have the same minimal polynomial as $\theta$. Thus, up to reordering, we can take $\theta_k = \sigma_k(\theta)$.
If $\alpha = \sum c_i\theta^i$, then $\alpha_k$ is just $\sigma_k(\alpha)$. Since the definition of $\sigma_k$ has nothing to do with $\theta$, we see that $\alpha_k=\sigma_k(\alpha)$ does not depend on $\theta$.
The upshot is that you should think of the $K$-conjugates of an element $\alpha\in K$ as being the $\sigma_k(\alpha)$ rather than elements of $\mathbb C$ with the same minimal polynomial as $\alpha$. Both definitions are equivalent, but the former allows you to use field arithmetic!
