# Find the Irreducible Polynomial of This Element over $\mathbb Q$

Let $K = \mathbb Q(a)$ where $a$ is a root of $x^3 - x - 1$. Find the irreducible polynomial of $c = 1 + a^2$ over $\mathbb Q$.

I found the answer by brute force. Write: \begin{align} a^3 & = a + 1 \\ a^4 & = a^2 + a \\ a^5 = a^3 + a^2 & = a^2 + a + 1 \\ a^6 = a^3 + a^2 + a & = a^2 + 2a + 1 \\ \end{align}

With these equalities in place, we have $(1 + a^2)^3 - 5(1 + a^2)^2 + 8(1 + a) - 5 = 0$. Clearly, the minimal polynomial of $c$ over $\mathbb Q$ is not linear. Suppose that there exist $\beta, \gamma \in \mathbb Q$ such that: $$(1 + a)^2 + \beta(1 + a^2) + \gamma = 0$$ Then, we have: $$a^4 + (2+ \beta)a^2 + (1 + \beta + \gamma) = (3 + \beta)a^2 + a + (1 + \beta + \gamma) = 0$$ That is not possible because the middle term of the polynomial is $a \ne 0$.

Here are my questions:
(1) We need to know that $[ \, \mathbb Q(a) : \mathbb Q \, ] = 3$ to confirm that $\{ \, 1, a, a^2 \, \}$ is linearly independent. How do I confirm that $x^3 - x - 1$ is irreducible over $\mathbb Q$? Once in a while I can "shift" the indeterminant, like $(x + n)^3 - (x + n) - 1$, and apply Eisenstein. This time it did not work ....
(2) Is there some way to put an upper limit on the degree of the minimal polynomial of $c$ over $\mathbb Q$?

• Irreducibility can be shown for quadratics and cubics by showing there is no rational root, which is easy in this case by the Rational Roots Theorem. – André Nicolas Nov 2 '15 at 22:07

$(1)$ The irreducibility of $p(x)=x^3-x-1$ is granted by the rational root theorem. $p(x)$ does not vanish over $\pm 1$, hence it irreducible; in this particular case, the irreducibility over $\mathbb{Q}$ also follows from the fact that $p(x)$ is irreducible over $\mathbb{F}_2$ (or $\mathbb{F}_3$, or $\mathbb{F}_{13}$);
$(2)$ If $\alpha$ is an algebraic number of degree $d$ over $\mathbb{Q}$ and $q(x)\in\mathbb{Q}[x]$, $q(\alpha)$ is an algebraic number of degree $\leq d$, since $1,q(\alpha),q(\alpha)^2,\ldots,q(\alpha)^d$ are $d+1$ elements in a $d$-dimensional vector space.
In our case, we have $\alpha^3=\alpha+1$, hence: $$1=1,\quad \alpha^2+1 = \alpha^2+1,\quad (\alpha^2+1)^2 = \alpha^4+2\alpha^2+1 = 3\alpha^2+\alpha+1,$$ $$(\alpha^2+1)^3 = \alpha^6+3\alpha^4+3\alpha^2+1 = (\alpha+1)^2+3(\alpha^2+\alpha)+3\alpha^2+1 = 7\alpha^2 + 5\alpha + 2$$ and by Gaussian elimination, if we set $\beta=\alpha^2+1$ we get: $$\beta^3 - 5\beta^2 + 8\beta - 5 = 0.$$ $x^3-5x^2+8x-5$ is the minimal polynomial of $\beta$ since, always by Gaussian elimination, there is no polynomial in $\mathbb{Q}_{\leq 2}[x]$ that vanishes over $\beta$, due to: $$\det\begin{pmatrix}0&0&1\\ 1&0&1 \\ 3&1&1\end{pmatrix} \neq 0.$$