2
$\begingroup$

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 $\mathbb Q$?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Nov 2 '15 at 22:07
3
$\begingroup$

$(1)$ 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: $$ \det\begin{pmatrix}0&0&1\\ 1&0&1 \\ 3&1&1\end{pmatrix} \neq 0. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.