# Find isomorphism between $L_1$ and $L_2$ finite fields

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 $$f(X)=X^4+X+1$$, $$g(X)=X^4+X^3+X^2+X+1 \in \mathbb Z_2[X]$$.

1. Prove both polynomials are irreducible over $$\mathbb Z_2[X]$$.
2. Find an explicit isomorphism between $$L_1 = \mathbb{Z}_2[X]/\langle f(X)\rangle$$ and $$L_2=\mathbb{Z}_2[X]/\langle g(X)\rangle$$.
3. Find a field $$M$$ extension of $$L_1$$ such that $$[M:L_1]=2$$.

This is the work I've done:

1. 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.
2. 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?).
3. 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.

• Hint: The zeros of $g(x)$ are exactly the fifth roots of unity whereas zeros of $f(x)$ are roots of unity of order fifteen (see the middle part of this old answer I prepared with referrals like this in mind). So if $\alpha=x+\langle f(x)\rangle$ then you should be able to check that $\alpha^3$ is a zero of $g(x)$. As $\beta=x+\langle g(x)\rangle$ is a zero of $g(x)$ in $L_2$, and $g(x)$ has coefficients in the prime field, an isomorphism must map $\beta$ to a zero of $g(x)$ in $L_1$. Commented Jun 14, 2021 at 18:03

1. Find an explicit isomorphism between $$L_1 = \mathbb{Z}_2[X]/\langle f(X)\rangle$$ and $$L_2=\mathbb{Z}_2[X]/\langle g(X)\rangle$$.
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.
• So, given $p(x)+\langle f(x) \rangle \in L_1$, the isomorphism is $$\sigma(p(x)+\langle f(x) \rangle) = p(x)+\langle g(x) \rangle ?$$ Commented Jun 14, 2021 at 14:08
• I'm afraid this answer is wrong. It does not preserve the multiplicative structure. For much the same reason that $a+b\sqrt2\mapsto a+b\sqrt3$ is not an isomorphism between the fields $\Bbb{Q}(\sqrt2)$ and $\Bbb{Q}(\sqrt3)$ (unlike in the present case those two fields are not isomorphic at all). Commented Jun 14, 2021 at 17:56
• @Aqua I don't know for sure whether you solved that problem yourself. I would do it by projecting everything modulo the ideal $I=\langle 1-\omega\rangle$ that can easily be shown to be an index two ideal of the ring $\Bbb{Z}[\omega]$. Basically the first term is odd, and the other term is even, so their sum cannot be even. Commented Jun 14, 2021 at 20:18