# Conjugate of element of the form $e = aω^2+bω+c$

If $e$ is a polynomial like element of the form $aω^{n-1}+a_2ω^{n-2}+...+a_{n-1}ω+a_n$ where $ω$ is a root of a polynomial $P(x)$ of degree $n$, then what is the complex conjugate of $e$?

For instance, if $P(x)=x^2-r$, and if $ω$ is a root of $P(x)$, then $ω^2=r$. In the following field $K$ of roots of $P(x)$, each element has the form $e=aω+b$. If $ex=b^2-ra^2=s$ is an integer, then $x=-aω+b$. Therefore, $-aw+b$ is the complex conjugate of an element of the form $aw+b$.

Now suppose we are looking at $P(x)=x^3-r$, and if $ω$ is a root of $P(x)$, then $ω^3=r$. In the following field $K$ of roots of $P(x)$, each element has the form $e=aω^2+bω+c$. If $ex=s$ and $s$ is an integer, then what is $x$?

Also, for any polynomial $P(x)$ of degree $n$, with integer coefficients (does not have to be of the form $x^n-r$), and $ω$ is a root of $P(x)$, then what is $ω^n$ in terms of $ω$ or $r$?

For instance, suppose we have the polynomial $P(x)=x^4+x^3+x^2+x+1$ and $ω$ is a root of $P(x)$. In the field of roots of $P(x)$, which happens to be the field of the $5$th roots of unity, each element has the form $e=aω^3+bω^2+cω+d$.

Given the information above,

What is $ω^4$ in terms of $ω$ and $r$?

If $ex=s$ is an integer, then what is $x$? In other words what is the complex conjugate of $e=aω^3+bω^2+cω+d$?

A PARI/GP script can be used to evaluate products of elements (in the field $F$, roots of $P(x)$) of the forms $(ay+b)(cy+d)$ where $y$ is a root of the polynomial $P(x)=x^2-s$, then $y^2=s$

(20:21) gp > W(a,b,y,c,d)=(a*d + b*c)*y + (a*c*s + b*d)
%125 = (a,b,y,c,d)->(a*d+b*c)*y+(a*c*s+b*d)
(20:21) gp >


For instance, if one sets $s=-1$, then the user defined function W(a,b,y,c,d) is the same as the product of the complex numbers $(ai+b)(ci+d)$ where $i$ is the imaginary unit.

(20:21) gp > W(a,b,y,-a,b)
%126 = -a^2*s + b^2
(20:25) gp >


If $(ay+b)(cy+d) = b^2 - sa^2$, then $c=-a$ and $d=b$.

(20:29) gp > s=-1
%127 = -1
(20:29) gp > W(a,b,y,c,d)=(a*d + b*c)*y + (a*c*s + b*d)
%128 = (a,b,y,c,d)->(a*d+b*c)*y+(a*c*s+b*d)
(20:29) gp > W(a,b,y,-a,b)
%129 = a^2 + b^2
(20:29) gp >


When $s=-1$, $P(x)=x^2+1$, and $F$ is the field of roots of $P(x)$, or $4$th roots of unity. As in the code, W(a,b,y,a,-b) is $(ay+b)(-ay+b)=a^2+b^2$.

Two questions regarding the code,

If $(ay+b)(-ay+b)=a^2+ab+b^2$, then what does one need to set $s$ to (in the code)?

If $P(x)$ is any polynomial of degree $n$, and $y$ is a root of $P(x)$, in the field of roots of $P(x)$, with elements $e=ay^{n-1}+a_2y^{n-2}+a_3y^{n-3}+...+a_{n-2}y^2+a_{n-1}y+a_n$ and $e_2=by^{n-1}+b_2y^{n-2}+b_3y^{n-3}+...+b_{n-2}y^2+b_{n-1}y+b_n$, then what is $y^n$ defined to be? That is, using the W function (like the way it is used for W(a,b,y,c,d) explained in the code above) W(a,a_2,a_3,..a_{n-1},y,b,b_2,b_3,...b_{n-1}) in PARI/GP, which gives the product of $e$ and $e_2$, what should one set $s$ variable to?

If you have an example for this, please refer to the polynomial $P(x)=x^4+x^3+x^2+x+1$, the $5$th cyclotomic polynomial, $s$ its roots. Thank you for help, explanations, and other helpful references and guides.

With the polynomial $P(x)=x^3-r$, $r$ a rational number, it often happens that the extension field $K=\Bbb{Q}(\omega)$, $P(\omega)=0$, does not contain all the zeros of $P(x)$. For example, if $\omega$ is a non-real cubic root of two, then its complex conjugate $\overline{\omega}$ is NOT an element of $K$ at all. OTOH, if we consider, instead of $K$ the splitting field $L=\Bbb{Q}(\root3\of2,e^{2\pi i/3})$ of $x^3-2$, then that field is a degree six extension of $\Bbb{Q}$. The field $L$, however, is stable under complex conjugation.
Things are markedly simpler, if $K$ is a Galois extension of $\Bbb{Q}$. In that case $K$ is automatically stable under complex conjugation. Any automorphism in the Galois group $Gal(K/Q)$ then maps a zero of $P(x)$ to some zero. The question is then to identify the usual complex conjugation among the automorphisms. You asked about $P(x)=x^4+x^3+x^2+x+1$ specifically. Because $P(x)(x-1)=x^5-1$ any zero $s$ of $P(x)$ satisfies also the equation $s^5=1$. As a complex number $s$ is on the unit circle, so $\overline{s}=1/s$. But also because $s^5=1$ we have $1/s=s^4$. So in this case $\overline{s}=s^4=-1-s-s^2-s^3$. Similarly we see that $\overline{s^2}=1/s^2=s^3$. Consequently \begin{aligned} \overline{a_0+a_1s+a_2s^2+a_3s^3}&=a_0+a_1s^4+a_2s^3+a_3s^2\\ &=(a_0-a_1)-a_1s+(a_3-a_1)s^2+(a_2-a_1)s^3. \end{aligned}