Proving two finite fields are isomorphic So I'm asked to prove that $\mathbb{F}_9$, defined as $\{ a+bi$ | $a,b \in \mathbb{Z}_3,$ $i^2 = 2 \}$, is isomorphic to the field $F_1$, defined as $\mathbb{Z}_3[x]/ \langle x^2+2x+2 \rangle$, where $x^2 = x+1$. I know I need to define a function between the two in order to prove this (we have yet to be taught that since they both have 9 elements, they are isomorphic). I tried to define the function as $\phi(a+bi)=a+bx$, but then it isn't even homomorphic, because $\phi((a+bi)(c+di)) \neq \phi (a+bi)\phi(c+di)$. 
I am getting
$\phi((a+bi)(c+di)) = \phi (ac+adi+bci+bdi^2)$
$= \phi (ac+2bd+(ad+bc)i)$
$=ac+2bd+(ad+bc)x$
This is where I get stuck, because I know this should equal $ac+adx+bcx+bdx^2=ac+adx+bcx+bdx+bd$. 
Am I just completely wrong?
 A: Solved it. This is the isomorphism :
$$\Phi(a+bi) = 2b(x+1)+a + \langle x^2 + 2x + 2 \rangle$$
Seems to work fine.
A: Notice the following two facts.

  
*
  
*$\mathbb{F}_9 = \mathbb{Z}_3[X]/\langle X^2+1 \rangle$, where $i = X + \langle X^2+1 \rangle$
  
*The given polynomial $X^2+2X+2$ can be written as $(X+1)^2+1$ (quite similar to $X^2+1$, right?) 
  

Let's try to develop these ideas. Define $p_1,\ p_2 \in \mathbb{Z}_3[X] $ as follows:
\begin{align*}
p_1 &= X^2 + 2X + 2 \\
p_2 &= X^2 + 1
\end{align*}
Since $p_1=(X+1)^2+1$, we can consider the isomorphism $\phi:\mathbb{Z}_3[X] \longrightarrow \mathbb{Z}_3[X]$ determined by $\phi(X)=X+1$, so that $\phi(p_2)=p_1$.
Now we define the ideals
\begin{align*}
I_1 &= \langle\, p_1\rangle = \langle X^2 + 2X + 2 \rangle\\
I_2 &= \langle\, p_2\rangle = \langle X^2 + 1 \rangle
\end{align*}
(which satisfy $\phi(I_2)=I_1$). These two ideals are maximal, so we can define the fields
\begin{align*}
F_1 &= \mathbb{Z}_3[X]/I_1\\
F_2 &= \mathbb{Z}_3[X]/I_2
\end{align*}
It is quite easy to check that $F_1\simeq F_2$ with the isomorphism induced by $\phi$, more precisely $\widetilde{\phi}:F_2 \longrightarrow F_1$ is defined by:
$$\widetilde{\phi} (r + I_2) = \phi(r) + I_1$$
Since $F_2 = \mathbb{F}_9$, we are done.
Notation.
Note that the following is only useful if you need to make computations with the isomorphism.
It is common to define $x=X+I_1 \in F_1$, so that we can write $x^2 = x+1$. This is the same trick as defining $i = X+I_2 \in F_2$.
Now the isomorphism reads:
\begin{align*}
\widetilde{\phi} (a+bi)
&= \widetilde{\phi} (a+bX + I_2) \\
&= \phi(a + bX) + I_1\\
&= a + b(X+1) + I_1\\
&= (a + b) + b(X + I_1) \\
&= (a + b) + bx
\end{align*}
We can give the inverse of this isomorphism:
$$(\widetilde{\phi})^{-1} (s + I_1) = \phi^{-1}(s) + I_2$$
where $\phi^{-1}$ is determined by $\phi^{-1}(X)=X-1=X+2$, and with the new notation:
\begin{align*}
(\widetilde{\phi})^{-1} (a+bx)
&= (\widetilde{\phi})^{-1} (a+bX + I_1) \\
&= \phi^{-1}(a + bX) + I_2\\
&= a + b(X+2) + I_2\\
&= (a + 2b) + b(X + I_2) \\
&= (a + 2b) + bi
\end{align*}
Putting everything together, we have the isomorphism:
\begin{matrix}
\mathbb{F}_9 & \overset{1:1}{\longleftrightarrow} & F_1\\
a + bi & \longmapsto & (a+b) + bx\\
(a+2b) + bi & \longleftarrow \!\shortmid & a+ bx
\end{matrix}
A: First in $\mathbb Z_3$, I prefer think at $-1$ instead of $2$ (since $(-1)^2 = 1$ is more natural that $2^2=1$)
If you're searching an explicit isomorphism, you don't want $\phi(a+bi)= a+bx$, but $\phi(a+bi) = \alpha + \beta x$ where $\alpha$ and $\beta$ are unknown.
Since you want $\phi$ to be a homomorphism, $\phi(a+bi) = a\phi(1) + b\phi(i)$. Thus $\phi$ depends uniquely of $\phi(1)$ and $\phi(i)$.
Now:


*

*show that $\phi(1)$ is either 1 or -1  (hint: look at $\phi(3)$) (hint: look at $\phi(1)^3$ and remember about the properties of roots of a polynomial defined over a field). For simplicity, let's choose $\phi(1)=1$ (try also with $\phi(1)=-1$, it doesn't change a lot the following)

*since $\phi$ is a ring homomorphism, $\phi(1)=1$.

*notice that $\phi(i)^2 = \phi(i^2) = \phi(-1)$ which gives you simple conditions on $\alpha$ and $\beta$.


You now have a candidate, you only have to show that this function actually is an isomorphism.
