Explicit construction of a finite field with $8$ elements 
Give an explicit contruction of the finite field $K$ containing $8$ elements, as a quotient of an appropriate polynomial ring. Include the multiplication table of the group $K^{*}=K\setminus \{0\},$ and write $K^{*}=\langle \alpha \rangle$ for some $\alpha \in K.$

I have no idea how to approach this problem. Can anyone guide me in the right direction? Thanks.
 A: Start with a field $\mathbf{F}$ with $2$ elements. A field with $8$ elements must contain $\mathbf{F}$ and be an extension of degree $3$, by size considerations.
Do you know how to get an extension of degree $3$ of a given field? Once you have such a field, the rest of the problem will follow by simply staring at your field long enough.
A: Your textbook should explain you the passage from a cubic irreducible polynomial with coefficients in a given field $F$ (here $\mathbb{Z}/2\mathbb{Z}$) to an extension field $K$ of degree three (of the same field $F$). Then use the method described in your previous question for finding such a polynomial.
In your case $K^*$ will have seven elements. What do you know about groups of seven elements? In terms of having a generator?
A: As a warm-up, how about we try to write down a field $k$ with $4$ elements? This is a degree $2$ extension of $\mathbf F_2 = \mathbf Z/2\mathbf Z$, so we need to find an irreducible polynomial of degree $2$ in $\mathbf F_2[X]$. We quickly find that $f(X) = X^2 + X + 1$ is the only one, so let $k = \mathbf F_2[X]/(f(X))$. 
Using $\alpha$ to denote the image of $X$ in $k$, the set $\{1, \alpha\}$ is a basis for $k$ over $\mathbf F_2$. To perform multiplication, use the relation $\alpha^2 =\alpha + 1$ imposed by $f$. For example, $$(1 + \alpha)\alpha = \alpha + \alpha^2 = \alpha + \alpha + 1 = 1.$$
