I've been solving problems from my Galois Theory course, and I'm not sure how to answer one of the questions of this one. It says:
Given $f(X)=X^4+X+1$, $g(X)=X^4+X^3+X^2+X+1 \in \mathbb Z_2[X]$.
- Prove both polynomials are irreducible over $\mathbb Z_2[X]$.
- Find an explicit isomorphism between $L_1 = \mathbb{Z}_2[X]/\langle f(X)\rangle$ and $L_2=\mathbb{Z}_2[X]/\langle g(X)\rangle$.
- Find a field $M$ extension of $L_1$ such that $[M:L_1]=2$.
This is the work I've done:
- Proving this is very easy, since $f(X)$ and $g(X)$ don't have roots in $\mathbb Z_2$, so it's unique possible factorisation is being a product of 2 irreducibles of degree 2, but there's only one irreducible of degree 2 inside $\mathbb Z_2[X]$, it's $X^2+X+1$ and neither $f$ nor $g$ are it's square, so it's proven they're both irreducible.
- Here's where I have trouble. I know they are isomorphic ($L_1$ and $L_2$ are both splitting fields of the polynomial $X^{2^4}-X$ over $\mathbb Z_2$), in fact they're both simple extensions for being finite fields, so $\exists\alpha,\beta$ roots of $f$ and $g$ respectively such that $L_1=\mathbb Z_2(\alpha)$ and $L_2=\mathbb Z_2(\beta)$. My problem is, how do I find a way to express $\beta$ in terms of $\alpha$? I know about Frobenius automorphism to find the rest of the roots of $f$ and $g$, but don't know if there's a way to find how to write $\beta$ in terms of $\alpha$ instead of just randomly trying. After finding that, I guess the automorphism must be
the one that does $\sigma(\alpha)=\beta$ (am I correct?). - This one's very easy. Being $M$ splitting field of $X^{2^8}-X$, then it's verified.
How can I do that second question? Is the rest of my reasoning correct? Any help will be appreciated, thanks in advance.
Best Answer
For each $\alpha \in \mathbb{Z}_2[X]/\langle f(X)\rangle$ exits $p_{\alpha}\in \mathbb{Z}_2[X]$ such that $\alpha = p_{\alpha}(x) +\langle f(x)\rangle$, so $$\alpha \to p_{\alpha}(x)+\langle g(x)\rangle$$ is such isomorphism.