minimal polynomial of $\alpha+\alpha^2$ where $\alpha^5=1$ and $\alpha\neq 1$ I have to find the minimal polynomial of of $\alpha+\alpha^2$ where $\alpha^5=1$ and $\alpha\neq 1$ in $\mathbb{Q}$. First I found the minimal polynomial of $\alpha$: $X^5-1=(X-1)(X^4+X^3+X^2+X+1)$ so $X^4+X^3+X^2+X+1$ must be the minimal polynomial of $\alpha$ since its irreducible and has $\alpha$ as a root. I didn't come much further than this. I really need help. Thanks.
 A: Let $\beta=\alpha+\alpha^2$.
Then 
$$ \beta^2=\alpha^4+2\alpha^3+\alpha^2=\alpha^3-\alpha-1$$
$$ \beta^3=\alpha^6+3\alpha^5+3\alpha^4+\alpha^3=3\alpha^4+\alpha^3+\alpha+3=-2\alpha^3-3\alpha^2-2\alpha$$
$$ \beta^4=\alpha^8+4\alpha^7+6\alpha^6+4\alpha^5+\alpha^4=\alpha^4+\alpha^3+4\alpha^2+6\alpha+4=3\alpha^2+5\alpha+3$$
and of course $\beta^0=1$. This gives us five vectors expressed as linear combinations of $1,\alpha,\alpha^2,\alpha^3$, which are therefore linearly dependent. That is you can find nontrivial coefficients $a_4,a_3,a_2,a_1,a_0\in\mathbb Q$ such that $a_4\beta^4+a_3\beta^3+a_2\beta^2+a_1\beta+a_0=0$, that is some nonzero polynomial that has $\beta$ as root.
A: Each row of
$$
M=\left[\begin{array}{r}
1&0&0&0\\
0&1&1&0\\
-1&-1&0&1\\
0&-2&-3&-2\\
3&5&3&0\\
\end{array}\right]
$$
is $(\alpha+\alpha^2)^k$ using the basis $(1,\alpha,\alpha^2,\alpha^3)$ and relation $\alpha^4+\alpha^3+\alpha^2+\alpha+1=0$.
Using the higher dimensional extension of the cross product, we find the vector perpendicular to all the columns of $M$. That is, take the determinant of the matrix deleting successive rows, then alternating signs, we get
$$
\left[\begin{array}{r}
1&3&4&2&1
\end{array}\right]
\left[\begin{array}{r}
1&0&0&0\\
0&1&1&0\\
-1&-1&0&1\\
0&-2&-3&-2\\
3&5&3&0\\
\end{array}\right]=0
$$
Since the determinant of the first four rows is not $0$, the minimal polynomial of $\alpha+\alpha^2$ is $x^4+2x^3+4x^2+3x+1=0$.

Verification
$$
\begin{align}
&(x^2+x)^4+2(x^2+x)^3+4(x^2+x)^2+3(x^2+x)+1\\
&=x^8+4x^7+8x^6+10x^5+11x^4+10x^3+7x^2+3x+1\\
&=(x^4+3x^3+4x^2+2x+1)(x^4+x^3+x^2+x+1)\\
&\equiv0\pmod{x^4+x^3+x^2+x+1}
\end{align}
$$

Computation of $M$
The action of multiplying by $\alpha$ is given by the matrix
$$
A=\left[\begin{array}{r}
0&1&0&0\\
0&0&1&0\\
0&0&0&1\\
-1&-1&-1&-1
\end{array}\right]
$$
Multiplication by $\alpha^2+\alpha$ is given by the matrix
$$
\begin{align}
B
&=A^2+A\\[12pt]
&=\left[\begin{array}{r}
0&1&1&0\\
0&0&1&1\\
-1&-1&-1&0\\
0&-1&-1&-1\\
\end{array}\right]
\end{align}
$$
The $k+1^\text{st}$ row of $M$ is
$$
\left[\begin{array}{r}
1&0&0&0
\end{array}\right]\,B^k
$$
A: Since $\mathbb{Q}[\beta]\subseteq\mathbb{Q}[\alpha]$, the degree of $\beta$ over $\mathbb{Q}$ is $1$, $2$ or $4$. It can't be $1$, because $\alpha$ has degree $4$ and satisfies a degree two polynomial with coefficients in $\mathbb{Q}[\beta]$.
Let's see if the degree can be $2$, that is, $\beta$ is a root of $X^2+qX+r$, with $q,r\in\mathbb{Q}$:
$$
0=\beta^2+q\beta+r=\alpha^2+2\alpha^3+\alpha^4+q\alpha+q\alpha^2+r=
\alpha^4+2\alpha^3+(1+q)\alpha^2+q\alpha+r
$$
The polynomial $X^4+2X^3+(1+q)X^2+qX+r$ should belong to the ideal generated by the minimal polynomial of $\alpha$ and this is impossible, because this ideal contains only one monic polynomial, namely $X^4+X^3+X^2+X+1$.
Therefore the degree of $\beta$ is $4$, so writing a non trivial zero linear combination of $\beta^0,\beta^1,\beta^2,\beta^3,\beta^4$ will give the minimum polynomial.
