Given a minimal polynomial of $a$, calculate the minimal polynomial of $a^2$ 
Let $a\in\mathbb C$ be a root of $f=X^3-X-1\in\mathbb Q[X]$.
a) Show that $f$ is irreducible.
b) Show that $b:=2a^2-a-2\neq 0$.
c) Compute the minimal polynomial of $a^2$.

Now a) and b) are trivial and I want to show c). Let $g$ be the minimal polynomial of $a^2$. We have that $\mathbb Q(a^2)$ is a subfield of $\mathbb Q(a)$ which is of degree $3$ by a), so either $g=X-a^2$ or $g$ is of degree 3. The first case can't be since $a^2\notin\mathbb Q$, so $g$ is of degree 3. Is there a smart way to find it? Probably it has something to do with $b$ but I can't see it yet.
 A: Let $b=a^2$.
\begin{align*}
\text{Then}\;\;&a^3-a-1=0\\[4pt]
\implies\;&ab-a-1=0\\[4pt]
\implies\;&a=\frac{1}{b-1}\\[4pt]
\implies\;&\left(\frac{1}{b-1}\right)^3-\left(\frac{1}{b-1}\right)-1=0\\[4pt]
\implies\;&\frac{b^3-2b^2+b-1}{b-1}=0\\[4pt]
\implies\;&b^3-2b^2+b-1=0\\[4pt]
\end{align*}
A: Hint:
Met $a, b,c$ the roots of $f$. Use Vieta's relations: you have to determine $\sum_a a^2$, $\sum_{a\ne b} a^2b^2$ and $a^2b^2c^2$, knowing that
$$\sum_a a =0,\quad\sum_{a\ne b} ab=-1 \quad\text{and}\quad abc=-(-1)=1.$$
Now of course $a^2b^2c^2=(abc)^2=1$,
$$\sum_a a^2=\Bigl(\sum_a a)^2-2\sum_{a\ne b} ab=2$$
and $$\sum_{a\ne b} a^2b^2=\Bigl(\sum_{a\ne b} ab\Bigr)^2-2\sum_{a\ne b\ne c} (ab)(bc)=(-1)^2-2\sum_{a\ne b\ne c} ab^2c=1-2abc\sum_a a=1,$$
so the minimal polynomial of $\:a^2, b^2, c^2\:$ is
$$X^3-2X^2+X-1.$$
A: First note that the degree of the extension $\Bbb Q(a)$ over $Q$ is $3$ so there are no intermediate field extensions so either $\Bbb Q(a^2)=\Bbb Q$ or $\Bbb Q(a^2)=\Bbb Q(a)$. The former is absurd because then $\Bbb Q(a)$ would be a degree $2$ extension with irreducible polynomial $x^2-a^2$.
So you know the degree of the extension $\Bbb Q(a^2)$ over $\Bbb Q$ is $3$ so let's find an irreducible polynomial $p$ of degree $3$ with $p(a^2)=0$
$$p(x)=x^3+b_2x^2+b_1x+b_0$$ using the relation in $\Bbb Q(a)$ that $a^3=a+1$ substitute 
$$\begin{split}p(a^2)&=a^6+b_2a^4+b_1a^2+b_0\\
&=(a+1)^2+b_2a(a+1)+b_1a^2+b_0\\
&=a^2(1+b_2+b_1)+a(2+b_2)+(b_0+1)\\
&=0
\end{split}$$
So setting $b_0=-1$ $b_2=-2$ and $b_1=1$ gives
$p(x)=x^3-2x^2+x-1$
$p$ is monic and $p(a^2)=0$ since no smaller degree polynomial $f$ can have $f(a^2)=0$ (this would give an intermediate extension) $p$ is minimal
A: You can just reduce $a^6$ using the relation for $a^3$ that you know.
A: Hint: Suppose a nonconstant polynomial $p(x)$ of degree $n$ has roots
$\alpha_1,\ldots,\alpha_n$ in some algebraic closure of the coefficient field,
that is
$$p(x) = p_n(x - \alpha_1)\cdots(x - \alpha_n)$$
Now set
$$\begin{align}
    q(x^2) &= (-1)^n p(x)\,p(-x)\tag{1}
\\  &= p_n^2(x^2 - \alpha_1^2)\cdots(x^2 - \alpha_n^2)
\\\therefore\quad
    q(y) &= p_n^2(y - \alpha_1^2)\cdots(y - \alpha_n^2)
\end{align}$$
Thus $q$ has roots $\alpha_1^2,\ldots,\alpha_n^2$.
If $p$ is monic, so is $q$. However, $q$ might not be irreducible.
Note that the coefficients of $q$ can be computed from the coefficients of $p$
using $(1)$; thus all computations take place in the coefficient field.
Generalization: Given $p$ and the $\alpha_i$ as before and a polynomial $s$,
the resultant $\operatorname{Res}_x\bigl(p(x),y-s(x)\bigr)$ yields a polynomial in $y$
whose roots are the $s(\alpha_i)$.
A: Can't you show that $\mathbb{Q}(a^2) \simeq \mathbb{Q}(a)$?  Therefore the root should be another cubic polynomial. Let $\phi:a \mapsto a^2$ To see this isomorphism, we have for one thing that.
\begin{eqnarray*}
 1 &\stackrel{\phi}{\mapsto}& 1 \\ a &\mapsto& a^2 \\ a^2 &\mapsto& a^4 = a(a+1)= a^2 + a \\ a^3 &\mapsto& a^6 = (a^3)^2 = (a+1)^2 
\end{eqnarray*}
Is my algebra wrong?  There might even be a quadratic relation.  We have:
$$ (a^2)^3 - 2 (a^2)^2 + a^2 - 1 =  \big(a^2 + 2a+1\big)- 2\big(a^2 + a\big) + a^2  - 1  = 0 $$
This looks about write.  We could summarize:
$$ \mathbb{Q}(a^2) \simeq \mathbb{Q}(x)/(x^3 - 2x^2 + x - 1) $$
Your job would be to find an inverse map $\phi^{-1}$.  Here's the map for $\phi$ as a $4 \times 4$ matrix over $\mathbb{Q}$:
$$
\left[ 
\begin{array}{c} 
1 \\ a^2 \\ a^4 
\end{array}\right] = \left[
\begin{array}{cccc}  0 & 0 & 1 \\
1 & 0 & 0 \\
 1 & 1 & 0  \end{array} \right]\,
\left[ 
\begin{array}{c} 
1 \\ a \\ a^2
\end{array}\right]
 $$
