(scattered notes)

Homework #6

Question: 4

First step, multiply the polynomial by $x-1$

$x^6+x^5+x^4+x^3+x^2+x+1=0$ over $F$, $|F|=2^6=64$

$x^7 -1$

$x^7=1\implies \text{ord}_{F^x}(x)\mid 7$

$x=1\iff 1, 7$

$F^x\cong \Z_{63}$ ← ring

$|F^x|=2^6-1=63$

Legrange thm→ $7\mid 63\implies \exists$ an element of order 7, $\alpha \in F^x$

$F^x$ is cyclic, so $F^x=\braket{g}$

$\alpha ^7=1$

$0=\alpha^7-1=(\alpha -1)(\alpha^6+\cdots + \alpha +1)$

$(\alpha -1)\neq 0$ because $\alpha$ is of order 7 and 1 is of order 1

So the other factor must be 0! So $\alpha$ must be a root of $(\alpha^6+\cdots + \alpha +1)$

if $d\mid n \text{ in } \Z_n$

Elements of order d

Practice Exam #2