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 $F$.
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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 $f(1)=1$.
$F\cong \mathbb{Z}/2\mathbb{Z}[X]/(f)$ 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.
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:
