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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?

carraig
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    Hint: In $\mathbb{F}_5,$ $$x^2+x+1=(x-2)^2+2$$ – Bumblebee Jan 01 '21 at 19:18
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    Another hint: the discriminator of $x^2+x+1$ is just $1-4=-3=2$ in $\Bbb F_5$. – Berci Jan 01 '21 at 19:22
  • What other said. Adjoining a root of $x^2+x+1$ (a root of unity of order three) gives $\sqrt{-3}$ and in this field $-3=2$, so you have $\sqrt2$. With very little extra work you can construct more root of unity familiar from the complex domain :-) – Jyrki Lahtonen Jan 01 '21 at 20:14
  • So as $\frac{-1+\sqrt{-3}}{2}$ is a root of $x^2+x+1$, we find $$\mathbb{F}_5/(x^2+x+1)\cong \mathbb{F}_5(\frac{-1+\sqrt{-3}}{2})=\mathbb{F}_5(\sqrt{-3})=\mathbb{F}_5(\sqrt{2})$$ because $-3=2 \in \mathbb{F}_5$. To go from $\mathbb{F}_5/(x^2+x+1)$ to $\mathbb{F}_5(\frac{-1+\sqrt{-3}}{2})$, we just evaluate the polynomial elements of the former field at $\frac{-1+\sqrt{-3}}{2}$. Does this give the desired isomorphism? – carraig Jan 02 '21 at 02:45
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    Correct, @carraig! I would like to encourage you to write that down as an answer. That way you get more feedback on the details (and the question off the list of unanswered ones). – Jyrki Lahtonen Jan 02 '21 at 07:36

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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*}

carraig
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