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.

  • 1
    $\begingroup$ 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 $L_1$. $\endgroup$ Commented Jun 14, 2021 at 18:03

1 Answer 1

  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.

  • $\begingroup$ So, given $p(x)+\langle f(x) \rangle \in L_1$, the isomorphism is $$\sigma(p(x)+\langle f(x) \rangle) = p(x)+\langle g(x) \rangle ?$$ $\endgroup$ Commented Jun 14, 2021 at 14:08
  • $\begingroup$ Yes, you have to check if it is well defined, bijective and homomorphism. $\endgroup$
    – nonuser
    Commented Jun 14, 2021 at 14:13
  • $\begingroup$ Ok, I'll try to do it myself. Thank you very much! $\endgroup$ Commented Jun 14, 2021 at 14:13
  • $\begingroup$ 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). $\endgroup$ Commented Jun 14, 2021 at 17:56
  • 1
    $\begingroup$ @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. $\endgroup$ Commented Jun 14, 2021 at 20:18

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