Construction of field extension for $[E:\mathbb F_{11}]=3$

Question

Let $\mathbb F_{11}\subset E$. Construct a field extension $E$ of $\Bbb{F}_{11}$ such that $[E:\mathbb F_{11}]=3$

Answer: Let $f(x)=x^3+1 $ be a polynomial in $\mathbb F_{11}[x]$ with $deg(f)=3$. This is also an irreduible in $\mathbb F_{11}$, so $$E=\mathbb F_{11}[x]/(f)$$ is a field extension with $[E:\mathbb F_{11}]=deg(f)=3$.

Do you agree?

Answer 1

$x^3+a$ with $a \in \mathbb F_{11}$ is always reducible because $x \mapsto x^3$ is a bijection in $\mathbb F_{11}^\times$ since $\gcd(3,10)=1$.

$x^3+x^2+2$ is irreducible over $\mathbb F_{11}$.

Answer 2

What about $\Bbb{F}_{11^3}\approx \Bbb{F}_{11}^3$ the isomorphism being an $\Bbb{F}_{11}$-algebra isomorphism. The uniqueness up to (field) isomorphism of a finite field of cardinal $p^q$ shows that $\forall P$ irreducible in $\Bbb{F}_{11}[X]$ of degree $3$

$$\Bbb{F}_{11}[X]/(P)\approx \Bbb{F}_{11^3}$$