Let $\alpha$ be a root of $f(x)=x^3+11x+4$. I'd like to compute the field polynomial of $\gamma=(\alpha+\alpha^2)/2$ (that is, the polynomial whose roots consist of the conjugates of $\gamma$).
First, it's not hard to see that $f(x)$ is irreducible over $\mathbb{Q}$ by the rational roots test. So $\mathbb{Q}[\alpha]$ has degree $3$ over $\mathbb{Q}$ and so the field polynomial for $\gamma$ can have degree at most $3$.
First, we have that $\gamma^2=(\alpha^2+2\alpha^3 +\alpha^4)/2=-5\alpha^2-13\alpha -4$ (reducing exponents using the fact that $\alpha$ is a root of $f(x)=0$.
Now, I'm not sure where to go for here, or if there is perhaps a better to solve this than guess and check. How should I proceed?
Best Answer
Rewriting the symmetric polynomial $\,f(x)f(y)\,$ in terms of $\,xy\,$ and $\,x\!+\!y\ (=-1)\,$ yields
$x\!+\!y = -1,\, f = x^3\!+\!11x\!+\!4\,\Rightarrow\, f(x)f(y) = (xy)^3\!-\!22(xy)^2\!+\!144(xy)\!-\!32 := g(xy)$
so $\,\color{#c00}{-\frac{1}2}xy = x(x\!+\!1)/2\,$ is a root of $\,-g(\color{#c00}{-2}z)/8 = \bbox[5px,border:1px solid #c00]{z^3+11z^3+36z+4}\,\pmod{\!f(x)}$
Remark $\ $ For completeness, below are details of the simple symmetric rewriting
$\qquad \begin{align} f(x)f(y) = (xy)^3 + 4(\!\underbrace{x^3\!+\!y^3}_{\textstyle \color{#c00}{3xy-1}}\!)+11xy(11\!+\!\underbrace{x^2\!+\!y^2}_{\textstyle \color{#0a0}{1-2xy}}\!)+ 44(\underbrace{x\!+\!y}_{\textstyle -1})+16 \end{align}$
$\begin{align}{\rm using}\ \ x^3\!+\!y^3 &= (x\!+\!y)^3\!-\!3(x\!+\!y)xy,\\ &=\ \ \ \color{#c00}{{-}1\ \, -\, \ 3\,(-1)\,xy}\end{align}$ $ \begin{align}{\rm \&}\ \ x^2\!+\!y^2 &= (x\!+\!y)^2\!-\!2xy,\\ &=\ \ \ \ \ \color{#0a0}{1\ -\ 2xy}\end{align} $