Proving two finite fields are isomorphic

Question

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?

Answer 1

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.

Answer 2

Solved it. This is the isomorphism : $$\Phi(a+bi) = 2b(x+1)+a + \langle x^2 + 2x + 2 \rangle$$

Seems to work fine.

Answer 3

Notice the following two facts.

1. $\mathbb{F}_9 = \mathbb{Z}_3[X]/\langle X^2+1 \rangle$, where $i = X + \langle X^2+1 \rangle$
2. 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}