How can I calculate $\iota^{-1} \text{ in } \mathbb F_9$?

Question

How can I calculate $\iota^{-1} \text{ in } \mathbb F_9$? Can I do that the way I would do it in the complex numbers?

$\mathbb F_9[X]:= \mathbb F_3[X] \text{ mod } (x^2+1) \text{ and } \iota := [X]_{x^2+1} $

$\mathbb F_k := \mathbb Z \text{ mod } k$

Answer 1

If $\iota = [X]$, then you know $\iota^2 + 1 = 0$, so $\iota^2 = -1$. Multiplying by $-1$, you find $-\iota^2 = 1$, so $\iota(-\iota) = 1$ and hence $-\iota=\iota^{-1}$. Note that this doesn't use anything about $\Bbb F_3$. In fact, this calculation shows that if you have a ring $R$ and an element $r\in R$ such that $r^2 = -1$, then $r$ is a unit and $r^{-1} = -r$.

Answer 2

In general $\Bbb{F}_q$ is not the same as $\Bbb{Z}/q\Bbb{Z}$. This is only true if $q$ is a prime number. It is true that $\Bbb{F}_9\cong\Bbb{F}_3[X]/(X^2+1)$. The latter shows that the field $\Bbb{F}_9$ is very small; you can simply list the elements and multiply them by $X$ to see which one is $\iota^{-1}$.