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Suppose we have two fields $F_1$ and $F_2$ of order $9$ where both groups of units are cyclic, i.e. $$F_1=\{0\}\cup\{\alpha^i\,|\,0\leq i\leq 7\},\qquad F_2=\{0\}\cup\{\beta^i\,|\,0\leq i\leq 7\}$$ for some $\alpha\in F_1$ and $\beta\in F_2$. We know that those two fields have to be isomorphic as they have the same number of elements. A concrete isomorphism would be given by $\varphi: F_1\to F_2$ where $\varphi(\alpha):=\beta$. It is easy to see that $\varphi(xy)=\varphi(x)\varphi(y)$ for every $x,y\in F_1$ due to the representation of the fields as powers of the generators $\alpha$ and $\beta$; the fact that $\varphi$ is bijective follows immediately as well.

It is not clear to me, however, why $\varphi(x+y)=\varphi(x)+\varphi(y)$ for every $x,y\in F_1$. Is there any elementary way to see this?

Edit: As metamorphy noted in the answer below, $\varphi$ is not generally an isomorphism. Is there any way, then, to choose another suitable generator $\beta'\in F_2^\times$ such that $\varphi$ is actually an isomorphism?

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    $\begingroup$ Find the minimal polynomial of $\alpha$ which is $f(X)=\prod_{j=0}^{d-1} (X-\alpha^{p^j})$, here $p=3,d=2$, this polynomial is in $\Bbb{F}_p[X]$, then the ring homomorphisms (necessarily field isomorphisms) are those sending $\alpha$ to a root of $f$. $\endgroup$
    – reuns
    Commented Jan 29, 2021 at 14:30
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    $\begingroup$ There's a quick way to see that your original claim couldn't be true, namely if it were there would be $\phi(p^n-1)$ automorphisms of $\mathbb{F}_{p^n}$, but we know there are only $n$ of them. (More precisely they're powers of the Frobenius automorphism, which answers your question exactly.) $\endgroup$ Commented Jan 29, 2021 at 23:01

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$\varphi(\alpha):=\beta$ (i.e. $\varphi(\alpha^i):=\beta^i$) does not necessarily define an isomorphism.

The automorphisms of $\mathbb{F}_q$, for $q=p^n$ and $p$ prime, are $x\mapsto x^{p^k}$ for $0\leqslant k<n$. In particular, for $q=9$, the only nontrivial automorphism is $x\mapsto x^3$. Hence, for $F_1=F_2=\mathbb{F}_9$ and $\beta=\alpha^5$ (say), the corresponding $\varphi$ is not an isomorphism.

As for the final subquestion, the choice of $\beta'$ is determined by the action of $+$ (in $F_1$ and $F_2$) or, as noted in the comments, by the action of $x\mapsto x+1$. For $\mathbb{F}_9$, the (only) two possibilities are:

\begin{array}{c|c|c|c|c|c|c|c|c|} x & 0 & 1 & \alpha & \alpha^2 & \alpha^3 & \alpha^4 & \alpha^5 & \alpha^6 & \alpha^7 \\ \hline (x+1)_1 & 1 & \alpha^4 & \alpha^7 & \alpha^3 & \alpha^5 & 0 & \alpha^2 & \alpha & \alpha^6 \\ \hline (x+1)_2 & 1 & \alpha^4 & \alpha^2 & \alpha^7 & \alpha^6 & 0 & \alpha^3 & \alpha^5 & \alpha \\ \hline \end{array}

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  • $\begingroup$ In fact, if you pick some other power, then $\varphi(x+y) = \varphi(x) + \varphi(y)$ should fail, right? $\endgroup$ Commented Jan 29, 2021 at 13:36
  • $\begingroup$ Right, thanks! But then, is there any concrete way to find a valid isomorphism statisfying the condition (i.e. picking a suitable generator $\beta'\in F_2^\times$)? $\endgroup$
    – log_math
    Commented Jan 29, 2021 at 13:39
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    $\begingroup$ If $\alpha$ is a generator of the multiplicative group then the other generators are $\alpha^j$, $j\in\{3,5,7\}$. Of theses $\alpha\mapsto \alpha$ and $\alpha\mapsto \alpha^3$ both work and give an isomorphism. $\alpha\mapsto \alpha^5$ and $\alpha\mapsto \alpha^7$ won't work. Just what metamorphy said (+1). $\endgroup$ Commented Jan 29, 2021 at 14:14
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    $\begingroup$ @metamorphy Would be great, yes. :) But in general, it would be sufficient to show that $\varphi(\alpha^i+1)=\varphi(\alpha^i)+1$ for all $i$ in order to get an isomorphism $\varphi$, right? $\endgroup$
    – log_math
    Commented Jan 29, 2021 at 14:31
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    $\begingroup$ @st.math: I've edited the answer to include all of the above. $\endgroup$
    – metamorphy
    Commented Jan 30, 2021 at 20:26

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