Galois extension $\frac{\mathbb Z_2[x]}{\langle 1+x+x^3 \rangle}:\mathbb Z_2$

Question

Let $\frac{\mathbb Z_2[x]}{\langle 1+x+x^3 \rangle}:\mathbb Z_2$ be a extension.

Is it a Galois extension? Is it a normal extension?

Solution:

Let's see if it is a Galois extension. We have that $[\mathbb Z_2[x]/\langle 1+x+x^3 \rangle:\mathbb Z_2]=3$. Elements of the field are of the form $a(1+I)+b(x+I)+c(x^2+I)$ where $I = \langle 1 + x + x^3 \rangle$ and $a, b, c \in \mathbb Z_2$

Now let $T\in G(\mathbb Z_2[x]/\langle 1+x+x^3 \rangle:\mathbb Z_2)$. Then $$T(a(1+I)+b(x+I)+c(x^2+I))=a(1+I)+bT(x+I)+c(T(x+I))^2. $$ Therefore $T$ will be completely determined if we know what is $T(x+I)$.

I know that $\alpha_1=x+I$ is a root of $p(x)=1+x+x^3,$ and I'd like to find another 2 roots to then conclude that $G(\mathbb Z_2[x]/\langle 1+x+x^3 \rangle:\mathbb Z_2)$ has only 3 elements.

How to find the other 2 roots?

Also is there a similar method to test if the extension is normal?

Answer 1

There is a general structure theorem to finite fields.

Theorem 1: A field has $p^n$ elements if and only if it is the splitting field of $x^{p^n} - x$ over $\mathbf{F}_p$.

It follows from Theorem 1 that all finite extensions of finite fields are normal.

Theorem 2: The Galois group of $\mathbf{F}_{p^n}/\mathbf{F}_p$ is generated by the Frobenius automorphism $\sigma : x \mapsto x^p$.

Now let us pretend we haven't proved this yet. We can still use the theorem. Here's how:

Let $\mathbf{F}_2[\alpha]$ be an extension of $\mathbf{F}_2$ where $\alpha$ is a root of $1 + x + x^3$. We know the roots of $1 + x + x^3$ are going to be generated by applying to $\alpha$, all Galois automorphisms of the normal closure of $\mathbf{F}_2[\alpha]$. Now by cheating and using Theorem 2. We "guess" that the roots are $\alpha, \alpha^2 = \sigma(\alpha)$ and $\alpha^4 = \sigma^2(\alpha)$. Now let's prove this.

Note that \begin{align} 0 &= (1 + \alpha + \alpha^3)^2 = 1^2 + \alpha^2 + (\alpha^2)^3 \\ 0 &= (1 + \alpha + \alpha^3)^4 = 1^4 + \alpha^4 + (\alpha^4)^3 \end{align} using the result that $(x + y)^p = x^p + y^p$ in characteristic $p$. Hence, as claimed, $$1 + x + x^3 = (x - \alpha)(x - \alpha^2)(x - \alpha^4).$$

Therefore $\mathbf{F}_2[\alpha]/\mathbf{F}_2$ is normal (it is the splitting field of $1 + x + x^3$) and $\mathbf{F}_2[\alpha]/\mathbf{F}_2$ is Galois (its Galois group is $\{\operatorname{id}, \sigma, \sigma^2\}$).