Quotient Rings example I have a field = $F_2$ and a polynomial $x^2 + x + 1$ over that field.
I understand that since the polynomial is degree 2, it has no root in $F_2$, so it is irreducible. But why is $F_2[X]/(x^2+x+1)$ a field with 4 elements? (0, 1, x, x+1)
Also when I do the multiplication table of the elements, why is xx = (x+1) and why is x(x+1) = 1?
 A: $\newcommand{\FF}{\mathbb{F}}$
Concretely, the field has an explicit presentation as equivalence classes of polynomials in $F := \FF_2[x]$ modulo the relation: $f(x)\equiv g(x)$ if $f(x)-g(x)\equiv 0\mod x^2+x+1$.
It's clear that $0,1,x,x+1$ are inequivalent elements of $F$. On the other hand, any other element is a polynomial of degree at least 2, and thus it starts $x^n + \cdots$. But modulo $x^2+x+1$, you know that $x^2 \equiv -x-1 \equiv x+1$. This is because in $\FF_2$, $-1 = 1$, and thus
$$x^2 - (x+1) = x^2 - x - 1 = x^2 + (-1)x + (-1) = x^2 + x + 1\equiv 0 \mod x^2+x+1$$
This shows that $x^2 - (x+1)\equiv 0\mod x^2+x+1$, and so $x^2\equiv x+1\mod x^2+x+1$.
Thus, any polynomial that starts $x^n + \cdots$ is equivalent to $x^{n-2}(x+1) + \cdots$ mod $x^2+x+1$, which has degree $n-1$. By induction this shows that any polynomial of degree at least 2 is equivalent to one with degree less than 2. This shows that $0,1,x,x+1$ are a complete set of representatives for elements of $F$.
As an example, consider $f = x^3+x^2+x+1$. Using $x^2 \equiv x+1$, you get
$$f \equiv x(x+1) + (x+1) + x + 1 = x^2 + x + x + 1 + x + 1 = x^2 + x \equiv (x+1) + x\equiv 1\mod x^2+x+1$$
A: If $K$ is a field and $f \in K[X]$ is irreducible, then $L=K[X]/(f)$ has degree $2$ over $K$. If $K$ has $2$ elements, then $L$ has $2^2=4$ elements.
In $L=F_2[X]/(X^2+X+1)$, we certainly have $0,1,x$, and these are all different. Since it is a ring, $x+1$ is there too, and it cannot be one of the others. Hence, all four elements are $0,1,x,x+1$.
In $L$, we have $x^2=-x-1=x+1$ because $L$ has characteristic $2$.
Finally, $x(x+1)=x^2+x=x+1+x=2x+1=1$, for the same reason.
