Find a field extension $E/\mathbb{F}_7$ of degree $3$

Question

Find a field extension $E/\mathbb{F}_7$ of degree $3$.

Let $F=\mathbb{F}_7$.

Option A:

$$E=\mathbb{F}_7^3$$ where we look at the isomorphism $F\cong \{(a,0,0)\in E:a\in F\}$. But then $(0,1,0)\in E$ is not algebric in $F$ because we can't zereing it with the elements of $\{(a,0,0)\in F^3:a\in F\}$.

Option B: $$E=\mathbb{F}_{343}=\mathbb{F}_{7^3}$$ But the probelm in that case is that I don't find a suitable isoomorphism such that $F"\subseteq "E$

Answer 1

Step 1: Find an irreducible polynomial of degree $3$ over $\Bbb F_7$. For instance, the polynomial function $x\mapsto x^3$ can only take the values $0,1$ and $-1$, so $f(x)=x^3+2$ has no roots and is therefore irreducible.

Step 2: $\Bbb Z_7[x]/(f(x))$ is the field you're after.

Answer 2

What is $\mathbb{F_{p^n}}$ in general? This is a splitting field of the polynomial $x^{p^n}-x\in\mathbb{F_p}[x]$. Hence it is indeed an extension of $\mathbb{F_p}$. This is the only extension of $\mathbb{F_p}$ of degree $n$ up to isomorphism.