Inverse of a complex number in finite field

Question

I am working with a finite field $\mathbb{F}_p(i)$ for $p = 431$. The elements of this field are of the form $u + vi$ where $u, v \in \mathbb{F}_p$. I have a confusion with the inverse operation in this field. For example, if some element $z = 324 + 171i$, then normally for complex numbers, i.e. not in a finite field, I understand that this should be:

$$ z^{-1} = \frac{\bar{z}}{z\bar{z}} = \frac{324 - 171i}{(324 + 171i)(324 - 171i)} = \frac{324 - 171i}{134217} $$ Obviously this would give me decimal numbers. I have tried another way to do it. I have generated all possible numbers of the form $u + vi$ where $u, v \in \mathbb{F}_p$ and multiplied it with $z$ to see if it gives me the multiplicative identity $ 1 + 0i$. But I can not find any. I think I do not know the proper way to find an inverse in this case. Any help would be greatly appreciated.

Answer 1

The element $i$ is not a complex number. It represents a solution to the equation $x^2+1=0$ relative to the field $\mathbb{F}_p$ (assuming that $p\equiv 3\pmod{4}$ so that the polynomial $x^2+1$ is irreducible). In particular, $(p-1)i+i=0$, so it cannot possibly be the complex number $i$.

Two methods:

1. When you see a fraction $\frac{a}{b}$, that really means “multiply $a$ by the multiplicative inverse of $b$”. So $\frac{324-171i}{134217}$ means “multiply $324-171i$ by the multiplicative inverse of $134217$“. So find the solution to $1324217x\equiv 1\pmod{p}$, and that is the value of $x$ you should multiply $324-171i$ by to get the value of the “fraction”.

2. Taking $324+171i$, consider the polynomials $p(x)=324+171x$, and $q(x)=x^2+1$. Since $\mathbb{F}_p(i) = \mathbb{F}_p[x]/(x^2+1)$, find polynomials $r(x)$ and $s(x)$ in $\mathbb{F}_p[x]$ such that $r(x)p(x) + s(x)q(x) = 1$, which should be possible because $x^2+1$ is irreducible and therefore relatively prime to any polynomial of degree $1$. Evaluating at $i$ you get $r(i)p(i) + s(i)q(i)=1$. But $q(i)=0$, so this gives $r(i)p(i)=1$. Since $p(i) = 324+171i$, then $r(i)$ is the multiplicative inverse of $324+171i$.

Answer 2

$p=431$ is prime $\equiv 3\bmod 4$ so $\Bbb{F}_p(i)$ is a field. $324+171i$ is inversible iff its conjugate $324-171i$ is iff $324^2+171^2$ is inversible iff $p\nmid 324^2+171^2$. In that case $$\frac1{324+171i} = \frac{324-171i}{324^2+171^2}$$ The division by $324^2+171^2$ means multiplication by some $n$ such that $p| n(324^2+171^2)-1$.

$n$ is found from the Euclidean algorithm applied to $\gcd(p,324^2+171^2)$.