Find a representation of the field $\mathbb{F}_{16}$ as $\frac{\mathbb{Z}_2}{(f)}$

Question

I have to find representation of the field $\mathbb{F}_{16}$ as $\frac{\mathbb{Z}_2}{(f)}$ where $f \in \mathbb{Z}_2[X]$ a monic irreducible polynomial over $\mathbb{Z}$.

I don't really have a clue how to solve this, but I think it has to do with finding the irreducible polynomials that divide $X^{16}-X$ and have a degree that divide 4.

I also have to find a unique subfield of $\mathbb{F}_{16}$ with 4 elements starting from a representation that I found, which I only can do after finding the specific representation, so I'm also stuck on this one.

Thanks in advance!

Answer 1

$x^2+x+1$ is the unique irreducible polynomial of degree $2$ over $\Bbb F_2$. $x^4+x+1$ has no roots, so no irreducible factor of degree $1$, it is also distinct from $(x^2+x+1)^2=x^4+x^2+1$, so it's not the product of two irreducible factors of degree $2$ either. Thus $x^4+x+1 \in \Bbb F_2[x]$ is irreducible and we have $\Bbb F_2[x]/(x^4+x+1) \cong \Bbb F_{16}$

Answer 2

Well, referring to the comment, the decomposition of $X^{16}-X$ into irreducibles over $GF(2)$ is

$X\cdot (X+1) \cdot (X^2+X+1)\cdot (X^4+X+1) \cdot (X^4+X^3+1)\cdot (X^4+X^3+X^2+X+1)$$

where $X^4+X+1$ and $X^4+X^3+1$ are primitive and conjugate, while $X^4+X^3+X^2+X+1$ is not primitive as it divides $X^5+1$.