# Irreducible polynomial of product of roots

Let $\alpha_1,\alpha_2,\alpha_3 \in \mathbb{C}$ be the roots of the polynomial $f(x)=x^3+x+1$. What is the irreducible polynomial of $\beta=\alpha_1\alpha_2$ over $\mathbb{Q}$?

I think that if $K$ is the splitting field of $f$ over $\mathbb{Q}$, then $Gal(K/\mathbb{Q})$ is the symmetric group $S_3$. However, I don't know where to go from here.

• Hint: $\beta = -1 / \alpha_3\,$. – dxiv May 11 '17 at 22:11

$\beta=\frac{-1}{\alpha_3}$. So we have $\frac{-1}{\beta}$ satisfies $X^3+X+1$. And hence $\beta$ satisfies $X^3-X^2-1$. To show this is the irreducible polynomial we observe that $Q(\alpha_3):Q=3$. Therefore $Q(\beta):Q=Q(\frac{-1}{\alpha_3}):Q=Q(\alpha_3):Q=3$ and the proof is complete.
Compute the elementary symmetric functions of $\beta_1$, $\beta_2$, $\beta_1$: \begin{align} &\beta_1+\beta_2+\beta_3 = \alpha_1\alpha_2+\alpha_2\alpha_3+\alpha_3\alpha_1=1\\ &\beta_1\beta_2+\beta_2\beta_3+\beta_3\beta_1 =\alpha_1\alpha_2^2\alpha_3+\alpha_1\alpha_2\alpha_3^2+\alpha_1^2\alpha_2\alpha_3=\alpha_1\alpha_2\alpha_3(\alpha_1+\alpha_2+\alpha_3)=-1\cdot0=0\\ &\beta_1\beta_2\beta_3=(\alpha_1\alpha_2\alpha_3)^2=1, \end{align} so the minimal polynomial is $\;x^3-x^2-1$.