Question about Constructing Finite Fields

Question

The following example showed up in my textbook, and I am struggling to understand a certain part of it.

Theorem 17.5 simply states that if $F$ is a field and $f(x)$ exists in $F[x]$, and is of degree 2 or 3, then $f(x)$ is reducible if and only if $f(x)$ has a zero in $F$. The corollary then says $F[x]/<f(x)>$ is a field.

The part I do not follow is:

$Z_2[x]/<x^3 + x + 1>$ = { $ax^2 + bx + c + <x^3 + x + 1> | a,b,c \epsilon Z_2$ }

I understand that $x^3 + x + 1$ is irreducible over $Z_2[x]$ and therefore it is a field(directly from the theorem and corollary), and I understand that given the equality, the field has 8 elements, but I do not understand the equailty itself.

I am still getting a grasp of quotient rings and polynomial rings so I apologize in advance if I don't understand something that is rather obvious.

Thanks for any help.

edit: Perhaps I do not fully understand the reasoning behind picking a cubic polynomial either. But I think it's because if we picked a quadratic polynomial, then our quotient ring wouldn't contain(all?) quadratic polynomials, and our field would be too small, is that correct?

Answer 1

$\mathbb Z_2[x]/\langle x^3+x+1 \rangle$ has, as its underlying set of elements, the set of cosets of the ideal $\langle x^3+x+1\rangle$.

Such a coset can always be written in the form $$ a_0+a_1x+\cdots +a_nx^n + \langle x^3+x+1\rangle $$ (where the bit on the left is an arbitrary element of $\mathbb Z_2[x]$). By polynomial long division, we may write $$ a_0+a_1x+\cdots+a_nx^n = q(x)(x^3+x+1)+r(x) $$ where $q(x)$ is some polynomial and $r(x)=ax^2+bx+c$ is a polynomial of degree less than $3$. But now we have $$ q(x)(x^3+x+1)\in \langle x^3+x+1\rangle $$ and so \begin{align} a_0+a_1x+\cdots +a_nx^n + \langle x^3+x+1\rangle&=q(x)(x^3+x+1)+r(x)+\langle x^3+x+1\rangle\\&=r(x)+\langle x^3+x+1\rangle\\&=ax^2+bx+c+\langle x^3+x+1\rangle \end{align} which is of the desired form.

Answer 2

$\mathbb{F}:=\mathbb{Z}_2[X]/\langle X^3+X+1\rangle$ is a vector-space over $\mathbb{Z}_2$. Using euclidean divisions by $X^3+X+1$, it has dimension $3$, a basis being $\{1+\langle X^3+X+1\rangle,X+\langle X^3+X+1\rangle,X^2+\langle X^3+X+1\rangle\}$. Regarding its cardinality, one has $\mathbb{F}\cong{\mathbb{Z}_2}^3$ as vector-spaces, which implies $\#\mathbb{F}=2^3=8$.