Constructing an isomorphism between two finite fields of order 25.

Question

The fields in question are \begin{equation*} \mathbb{F}_5[x]/(x^2+x+1),\ \mathbb{F}_5(\sqrt{2}). \end{equation*} I know that there's an isomorphism between the above fields as they are finite fields of the same order. My idea was to find a generator of the group of units of each field, and construct an isomorphism by mapping one generator to the other.

I found that $x+2$ generates $(\mathbb{F}_5[x]/(x^2+x+1))^{\times}$ and $1+\sqrt{2}$ generates $\mathbb{F}_5(\sqrt{2})^{\times}.$ Then, calling the map $\varphi$, I send $x+2$ to $1+\sqrt{2}$ which gives, after rearranging, $\varphi(x)=\sqrt{2}+4$ where I also used that any isomorphism shall fix the base field $\mathbb{F}_5$. The problem is that the map \begin{align*} \varphi:&\mathbb{F}_5[x]/(x^2+x+1)\longrightarrow \mathbb{F}_5(\sqrt{2})\\ &a+bx \mapsto a+4b+b\sqrt{2} \end{align*} doesn't satisfy $\varphi(fg)=\varphi(f)\varphi(g)$ for all $f,g \in \mathbb{F}_5[x]/(x^2+x+1).$ Is this down to the general approach being incorrect?

Answer 1

We notice that $\omega$, a primitive third root of unity, has as minimum polynomial $f(x)=x^2+x+1 \in \mathbb{F}_5[x]$. As $\omega=\frac{-1+\sqrt{-3}}{2},$ this gives the following isomorphism $\varphi:$ \begin{align*} \varphi: \mathbb{F}_5[x]/(x^2+x+1) &\longrightarrow \mathbb{F}_5(\frac{-1+\sqrt{-3}}{2})\\ g(x)&\longmapsto g(\frac{-1+\sqrt{-3}}{2}). \end{align*} However, $-3=2 \in \mathbb{F}_5$ and $\mathbb{F}_5(\frac{-1+\sqrt{-3}}{2})=\mathbb{F}_5(\sqrt{-3})$ so \begin{equation*} \mathbb{F}_5[x]/(x^2+x+1) \cong \mathbb{F}_5(\frac{-1+\sqrt{-3}}{2}) = \mathbb{F}_5(\sqrt{2}). \end{equation*}