Is there an isomorphism of fields between $\mathbb{F}_{3^{2}}$ and $\mathbb{F}=\{a+bi; a,b \in \mathbb{F}_{3}\}$?

Question

if $\mathbb{F}=\{a+bi; a,b \in \mathbb{F}_{3}\}$ where $i=\sqrt{2}=\sqrt{-1}$ and we define $(a+bi)+(c+di):=(a+c)+(b+d)i$ and $(a+bi)\ast (c+di):=(ac-bd)+(ad+bc)i$

Is there an isomorphism of fields between $\mathbb{F}_{3^{2}} and\ \mathbb{F}$ such that:

$\phi(0_{\mathbb{F}_{3^2}})=0_{\mathbb{F}}$

$\phi(1_{\mathbb{F}_{3^2}})=1_{\mathbb{F}}$

$\phi(x+y)=\phi(x)+\phi(y)$ and

$\phi(x\ast y)=\phi(x)\ast\phi(y)$

Answer 1

The polynomial $p(x)=x^2+1$ is irreducible in $\mathbb F_3[x]$. Hence $\mathbb F \cong \mathbb F_3[x]/(p)$ is a field with 9 elements. Therefore isomorphic to $\mathbb F_{3^2}$ (as finite fields are unique up to isomophism).