Splitting field of polynomial

Question

We consider the polynomial $f(x)=x^2+1 \in \mathbb{Z}_7[x]$ and let $E \subseteq \overline{\mathbb{Z}}_7$ be the splitting field.

1. Let $F \subseteq \overline{\mathbb{Z}}_7$ the splitting field of the polynomial $x^{48}-1 \in \mathbb{Z}_7[x]$. Show that $F=E$.
2. Let $a$ a solution of $f(x)$. Show that $w=2a+2$ is a solution of the polynomial $x^8-1 \in \mathbb{Z}_7[x]$.
3. Show that the splitting field of $x^8-1 \in \mathbb{Z}_7[x]$ is $\mathbb{Z}_7(w)$.

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Is the splitting field of $f(x)$ $E=\mathbb{Z}_7(i)$??

$\mathbb{Z}_7$ has characteristic $7$, so $x^7=x$.

Therefore $x^{48}=(x^7)^6 \cdot x^6=x^6 \cdot x^6=x^7 \cdot x^5=x^6$

$x^{48}-1=x^6-1$

So, is the splitting field $F=\mathbb{Z}(\omega_6)$, where $\omega_6=e^{\frac{2 \pi i}{6}}$ ??

We have that $f(a)=0 \Rightarrow a^2+1=0$

$x^8-1=x^2-1: w^2-1=(2a+2)^2-1=4(a^2+2a+1)-1=4a^2+8a+4-1=4a^2+a+3$

How could I continue to show that $w$ is a solution of $x^8-1$??

Answer 1

I think, you are a bit confused. The numbers $i, e^{\frac{2 \pi i}{6}}$ are complex numbers, so they do not belong to the algebraic closure of $\mathbb{Z}_7$.

Since $f=x^2+1 \in \mathbb{Z}_7[x]$ has no roots in $\mathbb{Z}_7$, then call $a \in \overline{\mathbb{Z}_7}$ a root of $f$. Then also $-a$ is a root of $f$, hence the splitting field $F$ of $f$ over $\mathbb{Z}_7$ is $\mathbb{Z}_7(a,-a) = \mathbb{Z}_7(a)$.

Clearly this field is isomorphic to the field $\mathbb{Z}_7[x]/(x^2 +1)$ which has $49$ elements, so $F$ is just the field with $49$ elements.

Note that all elements of $F\setminus \{ 0\}$ are roots of the polynomial $x^{48}-1$ since $F\setminus \{ 0\}$ is a group of $48$ elements (all elements have order dividing $48$). Since $x^{48}-1$ cannot have more than $48$ roots, all elements of $F\setminus \{ 0\}$ are all of its roots. So $F$ is the splitting field of $x^{48}-1$.

Let $w=2a+2 \in F$. Then $$w^8 = (2a+2)^8 = 2^8(a+1)(a+1)^7=4(a+1)(a^7+1^7)=$$ $$=4(a^8+a^7+a+1)=4((a^2)^4+a(a^2)^3+a+1) =4(1-a+a+1) = 8 = 1$$

so $w$ is a root of $x^8-1$.

Finally, the splitting field of $x^8-1$ is contained in the splitting field of $x^{48}-1$ (which is $F$) since $x^8-1$ divides $x^{48}-1$. But $F=\mathbb{Z}_7(w)$, and $w \notin \mathbb{Z}_7$. Since there are no intermediate fields between $\mathbb{Z}_7$ and $F$, the splitting field must be $F$.