# An isomorphism between two finite fields

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?

• 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 $f$. Commented Jan 29, 2021 at 14:30
• 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.) Commented Jan 29, 2021 at 23:01

$$\varphi(\alpha):=\beta$$ (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 $$0\leqslant k. In particular, for $$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:

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

• In fact, if you pick some other power, then $\varphi(x+y) = \varphi(x) + \varphi(y)$ should fail, right? Commented Jan 29, 2021 at 13:36
• Right, thanks! But then, is there any concrete way to find a valid isomorphism statisfying the condition (i.e. picking a suitable generator $\beta'\in F_2^\times$)? Commented Jan 29, 2021 at 13:39
• 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). Commented Jan 29, 2021 at 14:14
• @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? Commented Jan 29, 2021 at 14:31
• @st.math: I've edited the answer to include all of the above. Commented Jan 30, 2021 at 20:26