minimal polynomial of $1+\alpha^2$ where $\alpha$ is a root of $x^3-x-1$

Question

Suppose $\alpha$ is a root of $x^3-x-1$, and $\gamma=1+\alpha^2$.

What is the minimal polynomial of $\gamma$ over $\mathbb Q$?

Answer 1

Hints:

Observe that $\;\alpha^3=\alpha+1\;$ , and from here

$$\gamma^2=1+2\alpha^2+\alpha^2+\alpha=\gamma+3(\gamma-1)\;\ldots$$

Answer 2

Hint:

Find the minimal polynomial $p(x)$ of $\alpha^2$, using Vieta's relations between the roots of $x^3-x-1$.

The minimal polynomial of $1+\alpha^2$ will be $q(x)=p(x-1)$.

Some details:

Explicitely, denotin $\beta$ and $\gamma$ the other roots of $x^3-x-1$, you have to calculate $$S=\alpha^2+\beta^2+\gamma^2,\quad S_2=\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2,\quad P=\alpha^2\beta^2\gamma^2,$$ knowing $$s=\alpha+\beta+\gamma,\quad s_2=\alpha\beta+\beta\gamma+\gamma\alpha,\quad p=\alpha\beta\gamma.$$ This is possible by Newton's theorem, since $S, S_2$ and $P$ are symmetric functions in $\alpha,\beta,\gamma$.

Example: $$S=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)=s^2-2s_2$$

Answer 3

Write the matrix of the map $x \mapsto \gamma x$ in the basis $1,\alpha,\alpha^2$.

The minimal polynomial of this matrix is the minimal polynomial of $\gamma$.

Answer 4

A bit of linear algebra makes this very straightforward. To expand on the answer by lhf, the minimal polynomial over$\def\Q{\Bbb Q}~\Q$ of an element $\beta$ is also the minimal polynomial of the $\Q$-linear map of mulitpliciation by$~\beta$. The (sub)field $\Q[\alpha]$ is isomorphic to $\Q[X]/(X^3-X-1)$, and in that $\Q$-vector space the matrix, on the basis $[1,\alpha,\alpha^2]$ of images of $1,X,X^2$ is the companion matrix of $X^3-X-1$, which is $$ M=\pmatrix{0&0&1\\1&0&1\\0&1&0}. $$ The minimal polynomial of $\gamma=\alpha^2+1$ is of degree$~3$ (the only other possibility of an element of $\Q[\alpha]$ is degree$~1$, but that clearly is not the case here since $X^2+1$ is not divisible by $X^3-X-1$), so it will also be the characteristic polynomial of multiplication by$~\gamma$, which has matrix $$ M^2+I=\pmatrix{1&1&0\\0&2&1\\1&0&2}.$$ The characteristic polynomial of that matrix is easily computed to be $X^3-5X^2+8X-5$.