Given a minimal polynomial of $a$, calculate the minimal polynomial of $a^2$

Question

Let $a\in\mathbb C$ be a root of $f=X^3-X-1\in\mathbb Q[X]$.

a) Show that $f$ is irreducible.

b) Show that $b:=2a^2-a-2\neq 0$.

c) Compute the minimal polynomial of $a^2$.

Now a) and b) are trivial and I want to show c). Let $g$ be the minimal polynomial of $a^2$. We have that $\mathbb Q(a^2)$ is a subfield of $\mathbb Q(a)$ which is of degree $3$ by a), so either $g=X-a^2$ or $g$ is of degree 3. The first case can't be since $a^2\notin\mathbb Q$, so $g$ is of degree 3. Is there a smart way to find it? Probably it has something to do with $b$ but I can't see it yet.

Answer 1

First note that the degree of the extension $\Bbb Q(a)$ over $Q$ is $3$ so there are no intermediate field extensions so either $\Bbb Q(a^2)=\Bbb Q$ or $\Bbb Q(a^2)=\Bbb Q(a)$. The former is absurd because then $\Bbb Q(a)$ would be a degree $2$ extension with irreducible polynomial $x^2-a^2$.

So you know the degree of the extension $\Bbb Q(a^2)$ over $\Bbb Q$ is $3$ so let's find an irreducible polynomial $p$ of degree $3$ with $p(a^2)=0$

$$p(x)=x^3+b_2x^2+b_1x+b_0$$ using the relation in $\Bbb Q(a)$ that $a^3=a+1$ substitute

$$\begin{split}p(a^2)&=a^6+b_2a^4+b_1a^2+b_0\\ &=(a+1)^2+b_2a(a+1)+b_1a^2+b_0\\ &=a^2(1+b_2+b_1)+a(2+b_2)+(b_0+1)\\ &=0 \end{split}$$ So setting $b_0=-1$ $b_2=-2$ and $b_1=1$ gives

$p(x)=x^3-2x^2+x-1$

$p$ is monic and $p(a^2)=0$ since no smaller degree polynomial $f$ can have $f(a^2)=0$ (this would give an intermediate extension) $p$ is minimal

Answer 2

You can just reduce $a^6$ using the relation for $a^3$ that you know.

Answer 3

Hint:

Met $a, b,c$ the roots of $f$. Use Vieta's relations: you have to determine $\sum_a a^2$, $\sum_{a\ne b} a^2b^2$ and $a^2b^2c^2$, knowing that $$\sum_a a =0,\quad\sum_{a\ne b} ab=-1 \quad\text{and}\quad abc=-(-1)=1.$$ Now of course $a^2b^2c^2=(abc)^2=1$, $$\sum_a a^2=\Bigl(\sum_a a)^2-2\sum_{a\ne b} ab=2$$ and $$\sum_{a\ne b} a^2b^2=\Bigl(\sum_{a\ne b} ab\Bigr)^2-2\sum_{a\ne b\ne c} (ab)(bc)=(-1)^2-2\sum_{a\ne b\ne c} ab^2c=1-2abc\sum_a a=1,$$ so the minimal polynomial of $\:a^2, b^2, c^2\:$ is $$X^3-2X^2+X-1.$$

Answer 4

Let $b=a^2$. \begin{align*} \text{Then}\;\;&a^3-a-1=0\\[4pt] \implies\;&ab-a-1=0\\[4pt] \implies\;&a=\frac{1}{b-1}\\[4pt] \implies\;&\left(\frac{1}{b-1}\right)^3-\left(\frac{1}{b-1}\right)-1=0\\[4pt] \implies\;&\frac{b^3-2b^2+b-1}{b-1}=0\\[4pt] \implies\;&b^3-2b^2+b-1=0\\[4pt] \end{align*}

Answer 5

Can't you show that $\mathbb{Q}(a^2) \simeq \mathbb{Q}(a)$? Therefore the root should be another cubic polynomial. Let $\phi:a \mapsto a^2$ To see this isomorphism, we have for one thing that. \begin{eqnarray*} 1 &\stackrel{\phi}{\mapsto}& 1 \\ a &\mapsto& a^2 \\ a^2 &\mapsto& a^4 = a(a+1)= a^2 + a \\ a^3 &\mapsto& a^6 = (a^3)^2 = (a+1)^2 \end{eqnarray*} Is my algebra wrong? There might even be a quadratic relation. We have: $$ (a^2)^3 - 2 (a^2)^2 + a^2 - 1 = \big(a^2 + 2a+1\big)- 2\big(a^2 + a\big) + a^2 - 1 = 0 $$ This looks about write. We could summarize: $$ \mathbb{Q}(a^2) \simeq \mathbb{Q}(x)/(x^3 - 2x^2 + x - 1) $$ Your job would be to find an inverse map $\phi^{-1}$. Here's the map for $\phi$ as a $4 \times 4$ matrix over $\mathbb{Q}$:

$$ \left[ \begin{array}{c} 1 \\ a^2 \\ a^4 \end{array}\right] = \left[ \begin{array}{cccc} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 0 \end{array} \right]\, \left[ \begin{array}{c} 1 \\ a \\ a^2 \end{array}\right] $$

Answer 6

Hint: Suppose a nonconstant polynomial $p(x)$ of degree $n$ has roots $\alpha_1,\ldots,\alpha_n$ in some algebraic closure of the coefficient field, that is $$p(x) = p_n(x - \alpha_1)\cdots(x - \alpha_n)$$ Now set $$\begin{align} q(x^2) &= (-1)^n p(x)\,p(-x)\tag{1} \\ &= p_n^2(x^2 - \alpha_1^2)\cdots(x^2 - \alpha_n^2) \\\therefore\quad q(y) &= p_n^2(y - \alpha_1^2)\cdots(y - \alpha_n^2) \end{align}$$ Thus $q$ has roots $\alpha_1^2,\ldots,\alpha_n^2$. If $p$ is monic, so is $q$. However, $q$ might not be irreducible. Note that the coefficients of $q$ can be computed from the coefficients of $p$ using $(1)$; thus all computations take place in the coefficient field.

Generalization: Given $p$ and the $\alpha_i$ as before and a polynomial $s$, the resultant $\operatorname{Res}_x\bigl(p(x),y-s(x)\bigr)$ yields a polynomial in $y$ whose roots are the $s(\alpha_i)$.