Constructing a multiplication table for a finite field 
Let $f(x)=x^3+x+1\in\mathbb{Z}_2[x]$ and let $F=\mathbb{Z}_2(\alpha)$, where $\alpha$ is a root of $f(x)$. Show that $F$ is a field and construct a multiplication table for $F$.

Can you please help me approach this problem? I've tried searching around, but I don't really know what I'm looking for!
Thanks.
 A: By the division algorithm, any polynomial $g\in\mathbb{Z}_2[x]$ can be uniquely written as
$$g=a_0+a_1x+a_2x^2+qf$$
for some $q\in\mathbb{Z}_2[x]$ and some $a_0,a_1,a_2\in\mathbb{Z}_2$ (depending on $g$, of course). Thus, the quotient ring $\mathbb{Z}_2[x]/(f)$ consists precisely these eight cosets (corresponding to each possible choice of the $a_i$):
$$\begin{array}{cc}
0 + (f) &\quad 1 + (f) \\
x + (f) &\quad 1 + x + (f) \\
x^2 + (f) &\quad 1 + x^2 + (f) \\
x + x^2 + (f) &\quad 1 + x + x^2 + (f) \\
\end{array}$$
Use the definition of addition and multiplication in a quotient ring to construct the multiplication table. For example,
$$\begin{align*}
\biggl[x + (f)\biggr]\cdot \biggl[x^2 + (f)\biggr]&=x^3 + (f)\\\\
&= \biggl[0 +(f)\biggr] + \biggl[x^3+(f)\biggr]\\\\
&= \biggl[f +(f)\biggr] + \biggl[x^3+(f)\biggr]\\\\
&= \biggl[1 + x + x^3+(f)\biggr] + \biggl[x^3+(f)\biggr]\\\\
&= 1 + x + 2x^3+(f)\\\\
&=1+x+0x^3+(f)\\\\
&=1+x+(f)
\end{align*}$$
You can prove that $F\cong\mathbb{Z}_2[\alpha]\cong\mathbb{Z}_2[x]/(f)$ is a field because: $\mathbb{Z}_2[x]$ is a PID, hence a non-zero ideal of $\mathbb{Z}_2[x]$ is maximal iff it is prime iff it is generated by an irreducible element, so $\mathbb{Z}_2[x]/(f)$ is a field iff $f$ is irreducible, and you can either check directly that $f$ doesn't factor non-trivially, or observe that since $\deg(f)\leq 3$ it suffices to check that $f$ has no roots in $\mathbb{Z}_2$, which it doesn't because $f(0)=1$ and $f(1)=1$.
A: $F\cong \mathbb{Z}/2\mathbb{Z}[X]/(f)$ since $f$ is irreducible in $\mathbb{Z}/2\mathbb{Z}[X]$. Multiplation in $F$ is modulo $f(\alpha)$ and of characteristic $2$ (on the coefficients of $\alpha$). There should be $8$ elements as $2^3=8$ (this is a degree $3$ field extension) and as Gamamal has said, the nonzero ones will form a cyclic group.

Example:
Take $p(X)\in \mathbb{Z}/2\mathbb{Z}[X]$ and by the division algorithm, find the remainder after division by $f(X)$. This will be a polynomial of degree less than $3$. Since our field is finite, we can write all $8$ of these remainders. Wherever there is an $X$, put an $\alpha$ (as this is it's image under the mapping).
Choose two elements, say $\alpha^2+1$ and $\alpha$. We have $$(\alpha^2+1)\alpha=\alpha^3+\alpha$$
Also, $\alpha$ is a root of $f$ so it satisfies the relation $\alpha^3+\alpha +1=0$. Subtraction by $0$ gives $$(\alpha^3+\alpha)-0$$$$=(\alpha^3+\alpha)+(-\alpha^3-\alpha-1)$$$$=-1$$$$=1$$
Remembering which field the coefficients are in.
A: Using repeatedly the relation $\,\alpha^3=\alpha+1$, knowing a basis of $\mathbf Z_2[\alpha]$ over $\mathbf Z_2$ is $\{1,\alpha,\alpha^2\}$ we have the following table of multiplication:

