Express the other roots with the powers of $\alpha$.

Question

Let $f(x)=x^4+x+1 \in Z_2[x]$ and $\alpha$ one of its roots'. Express the other roots using the powers of $\alpha$.

I believe I have to find the exact elements of its Galois group, and use them to express the other roots. The problem is, I could only find the identity and complex conjugation. I'm not even sure what its splitting field is, but i know that it has 2 complex conjugate root-pairs. Can this be done without exactly calculating the roots?

Answer 1

Hint: Let $\alpha$ be a root of $x^4+x+1$ in $GF(16)$, the splitting field of the polynomial over $GF(2)$. Then $\alpha^4=\alpha+1$ in $GF(16)$ and so each nonzero element of $GF(16)$ can be expressed as a power of $\alpha$:

$$1, \alpha, \alpha^2, \alpha^3,\alpha^4=\alpha+1, \alpha^5=\alpha^2+\alpha, \alpha^6=\alpha^3+\alpha^2, \alpha^7 = \alpha^3+\alpha+1, \alpha^8=\alpha^2+1,\alpha^8=\alpha^3+\alpha,\alpha^{10} = \alpha^2+\alpha+1, \alpha^{11} = \alpha^3+\alpha^2+\alpha, \alpha^{12}=\alpha^3+\alpha^2+\alpha+1,\alpha^{13}=\alpha^3+\alpha^2+1,\alpha^{14}= \alpha^3+1,\alpha^{15}= 1.$$

Addendum:

$\alpha,\alpha^2,\alpha^4,\alpha^8$ are the roots of $x^4+x+1$.

• The zeros of an irreducible polynomial over $GF(2)$ are the powers of the Frobenius isomorphism: $x\mapsto x^2$. The zeros form a socalled conjugacy class.

$\alpha^7,\alpha^{14},\alpha^{13},\alpha^{11}$ are the roots of $x^4+x^3+1$, the conjugate polynomial of $x^4+x+1$.

• The zeros of the reciprocal polynomial contain $\alpha^{-1}=\alpha^{14}$.

$\alpha^3,\alpha^6,\alpha^{12},\alpha^9$ are the roots of $x^4+x^3+x^2+x+1$ which divides $x^5+1$.

• These zeros are 5th roots of unity.

$\alpha^5,\alpha^{10}$ are the roots of $x^2+x+1$ which divides $x^3+1$.

• These zeros are contained in $GF(4)$.

$1$ is the root of $x+1$.