<s>[INST] 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 $\alpha + \beta$.
 [/INST] 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
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