$X_1X_2$ in $Z_2[X_1] /(X_1^2+X_1+X_1)[X_2]/(X_2^2 +X_2 + X_1)$ Let $K = Z_2[X_1] /(X_1^2+X_1+1)[X_2]/(X_2^2 +X_2 + X_1)$
I’ve been trying to make the multiplication table for $K$ and haven’t managed to figure out what $X_1 X_2$ equals.
Any help?
 A: $x^2 + x + 1$ is the unique irreducible quadratic polynomial over $\mathbb{F}_2$, so the first stage of your construction gives the unique presentation of the finite field $\mathbb{F}_4 = \mathbb{F}_2[x]/(x^2 + x + 1)$. I will follow Jyrki in renaming this element, but I'll name it $\omega$ because it's a primitive third root of unity. Hence the elements of $\mathbb{F}_4$ are $\{ 0, 1, \omega, \omega^2 \}$. The second stage of your construction is
$$K = \mathbb{F}_4[x]/(x^2 + x + \omega).$$
To figure out what this should be we want to figure out how $x^2 + x + \omega$ factors over $\mathbb{F}_4$. Note that since we're working in characteristic $2$ the quadratic formula isn't available. Fortunately $\mathbb{F}_4$ only has $4$ elements so it's possible to just search exhaustively through all possible factorizations. The only possible factorizations given the linear term are
$$(x + \omega)(x + \omega^2) = x^2 + x + 1$$
and
$$(x + 1)x = x^2 + x$$
so we conclude that $x^2 + x + \omega$ is irreducible. Hence $K = \mathbb{F}_4[x]/(x^2 + x + \omega)$ is the unique quadratic extension of $\mathbb{F}_4$, namely $\mathbb{F}_{16}$. Accordingly it has a basis over $\mathbb{F}_4$ given by $\{ 1, x \}$ and so a basis over $\mathbb{F}_2$ given by $\{ 1, \omega, x, \omega x \}$.
Note that polynomials of the form $x^2 + x + a$ in characteristic $2$ are Artin-Schreier polynomials. Adjoining roots of such polynomials is a characteristic $2$ analogue of adjoining square roots.
