Minimal polynomial of $\alpha + \beta$ over $\mathbb{Q}$ Let $\alpha,\beta \in \mathbb{C}$ such that $\alpha^3+\alpha+1=0$ and $\beta^2+\beta-3=0$ . Find the minimal polynomial of $\alpha+\beta$ over $\mathbb{Q}$.
I was trying the usual trick for this kind of problems, that is
Let $\gamma = \alpha+\beta$, and therefore
\begin{align*}
(\gamma - \beta)^3 &= \alpha^3 = -\alpha - 1 \\
\gamma^3 - 3\gamma^2\beta+3\gamma\beta^2-\beta^3 &= -\alpha - 1
\end{align*}
and using $\beta^2+\beta-3=0$, i get $\gamma^3 - 3\gamma^2\beta-3\gamma\beta + 9\gamma+3-4\beta = -\alpha - 1$.
I think the minimal polynomial of $\alpha+\beta$ has degree $6$, then i was trying raise to the $2^{nd}$ power the last equation, but i don't get nothing to obtain the minimal polynomial of $\alpha + \beta$.
 A: The way to compute this is as follows: if $A$ is a matrix which has $\alpha$ as an eigenvalue and $B$ is a matrix that has $\beta$ as an eigenvalue, then $A \otimes I + I \otimes B$ is a matrix that has $\alpha + \beta$ as an eigenvalue. This is because if $Av = \alpha v$ and $Bw = \beta w$ then
$$ (A \otimes I + I \otimes B)(v \otimes w) = (Av \otimes Iw) + (Iv \otimes Bw) = (\alpha + \beta)(v \otimes w). $$
To get $A$ and $B$, you can use the companion matrix of a polynomial:
$$ \mathrm{Char. Poly.} \begin{pmatrix}
0 & 0 & \dots & -a_0 \\
1 & 0 & \dots & -a_1 \\
0 & 1 & \dots & -a_2 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \dots & -a_{n-1}
\end{pmatrix} = x^n + a_{n-1}x^{n - 1} + \dots + a_0. $$
If you do this computation, you'll find that the characteristic polynomial of
$$ \begin{pmatrix} 0 & 0 & -1 \\ 1 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix} \otimes I_2 + I_3 \otimes \begin{pmatrix} 0 & 3 \\ 1 & -1 \end{pmatrix} $$
is $x^6 + 3x^5 - 4x^4 - 11x^3 + 25x^2 + 52x -39$ and that's the minimal polynomial of $\alpha + \beta$.
A: Here’s another method:
The roots of the second (quadratic) polynomial are $\beta$ and $\overline\beta=-1-\beta$ (because the sum of the roots has to be the negative of the next-highest coefficient).
Now take the polynomial $f(X)=X^3+X+1$, which is the minimal polynomial for $\alpha$, and write $g(X)=f(X-\beta)$, which is a polynomial with $\alpha+\beta$ for a root, coefficients in $\Bbb Q(\beta)$. Also, write $\overline g(X)=f(X-\overline\beta)$, just what you get from $g(X)$ by replacing $\beta$ by $\overline\beta$. If we’ve done everything right, we’ll get $g\overline g$ to have rational coefficients. In fact,
\begin{align}
g(X)&=X^3-3\beta X^2 +(10-3\beta)X  +4-5\beta\\
\overline g(X)&=X^3+(3+3\beta)X^2+(13+3\beta)X+9+5\beta\\
g\overline g(X)&=X^6 +3X^5-4X^4-11X^3+25X^2+52X-39\,,
\end{align}
as desired.
(This is easily done by hand, if you work carefully. I confess that I was feeling lazy, and used a symbolic-computation package to do it. If working by hand, you may find it easier to use what you know about $\beta$, namely that it’s equal to $\frac{-1+\sqrt{13}}2$, when $\overline\beta$ is the same thing with a minus sign in front of the radical.)
A: Note that

 
*
 
*The polynomial $a^3+a+1$ has three distinct roots, exactly one of which is real.$\\[4pt]$
 
*The polynomial $b^2+b-3$ has two distinct roots, both of which are real.
 
 
 By pairing one of the non-real roots of $a^3+a+1$ with each of the roots of $b^2+b-3$, it follows that the minimal polynomial of $\gamma$ has at least two distinct non-real roots.
 
 By pairing the real root of $a^3+a+1$ with each of the roots of $b^2+b-3$, it follows that the minimal polynomial of $\gamma$ has at least two distinct real roots.

Thus the minimal polynomial of $\gamma$ has degree at least $4$, so $[\mathbb{Q}(\gamma):\mathbb{Q}]\ge 4$.
 
 Continuing the algebra from where you left off . . .
 \begin{align*}
 &
 \gamma^3 - 3\gamma^2\beta-3\gamma\beta + 9\gamma+3-4\beta = -\alpha - 1
 \\[4pt]
\implies &
\gamma^3 - 3\gamma^2\beta-3\gamma\beta + 9\gamma+3-4\beta = -(\gamma-\beta) - 1
\\[4pt]
\implies &
\beta=\frac{\gamma^3+10\gamma+4}{3\gamma^2+3\gamma+5}
\\[4pt]
\implies &
\left(\frac{\gamma^3+10\gamma+4}{3\gamma^2+3\gamma+5}\right)^2
+
\left(\frac{\gamma^3+10\gamma+4}{3\gamma^2+3\gamma+5}\right)
-
3
=
0
\\[4pt]
\implies &
\frac{\gamma^6+3\gamma^5-4\gamma^4-11\gamma^3+25\gamma^2+52\gamma-39}{\left(3\gamma^2+3\gamma+5\right)^2}=0
\\[4pt]
\implies &
\gamma^6+3\gamma^5-4\gamma^4-11\gamma^3+25\gamma^2+52\gamma-39=0
\\[4pt]
\end{align*}
so $\gamma$ is a root of the polynomial
$$p(x)=x^6+3x^5-4x^4-11x^3+25x^2+52x-39$$
hence $4\le [\mathbb{Q}(\gamma):\mathbb{Q}]\le6$.

But from the equation
$$\beta=\frac{\gamma^3+10\gamma+4}{3\gamma^2+3\gamma+5}$$
it follows that $\beta\in\mathbb{Q}(\gamma)$, hence $[\mathbb{Q}(\gamma):\mathbb{Q}]$ must be a multiple of $3$.

Thus we must have $[\mathbb{Q}(\gamma):\mathbb{Q}]=6$, hence $p(x)$ is the minimal polynomial of $\gamma$.
A: Let $\alpha=\alpha_1$ be one of the roots, $\alpha_i,i=1,2,3,$ of $x^3+x+1$ and let $\beta=\beta_1$ be one of the roots, $\beta_j,j=1,2,$ for $x^2+x-3$. Consider the polynomial $$f(x)=\prod_{i,j}(x-(\alpha_i+\beta_j)).$$ Then by the relation between coefficients of a polynomial and its roots, one sees that $f(x)\in {\mathbb Q}[x]$ and $f(\alpha+\beta)=0$. To find $f(x)$, it just takes some algebraic manipulations: $$f(x)=\prod_j\prod_i((x-\beta_j)-\alpha_i)$$
$$=\prod_j((x-\beta_j)^3+(x-\beta_j)+1)$$
$$=\prod_j((x^3-3\beta_jx^2+3\beta_j^2x-\beta_j^3)+(x-\beta_j)+1)$$
$$=\prod_j((x^3+10x+4)-\beta_j(3x^2+3x+5)),\quad (1)$$ where one uses the relations $\beta_j^2=3-\beta_j,\beta_j^3=-3+4\beta_j$ derived from $\beta_j^2+\beta_j-3=0$. Note that $$\beta_1\beta_2=-3,\beta_1+\beta_2=-1.\quad (2)$$ Now expanding (1) and using (2), one sees that $$f(x)=(x^3+10x+4)^2-(\beta_1+\beta_2)(x^3+10x+4)(3x^2+3x+5)+\beta_1\beta_2(3x^2+3x+5)^2$$
$$=x^6+3x^5-4x^4-11x^3+25x^2+52x-39.$$
To show that $f(x)$ is the minimal polynomial of $\alpha+\beta$, reduce its coefficients mod $2$ to get $$\overline{f(x)}=x^6+x^5+x^3+x^2+1,$$ which is irreducible over ${\mathbb F}_2$, since it is not divisible by $x,x+1,x^2+x+1,x^3+x^2+1$ and $x^3+x+1$. It follows that $f(x)$ irreducible over ${\mathbb Q}$. QED
