# Field in $\mathbb{F}_3$

Prove that the set of symbols $\{a+bi \mid a, b \in \mathbb{F}_3\}$ forms a field with nine elements, if the laws of composition are made to mimic addition and multiplication of complex numbers. Will the same method work for $\mathbb{F}_5$? For $\mathbb{F}_7$?

I was able to prove that $\{a+bi \mid a, b \in \mathbb{F}_3\}$ is a field and that the method does not work for $\mathbb{F}_5$. But could someone explain to me why it works in $\mathbb{F}_7$ ?

Thank you

#### Solutions Collecting From Web of "Field in $\mathbb{F}_3$"

Everything is completely straightforward over $\mathbb{F}_p$, for any prime $p$, with the possible exception of showing that every nonzero element has a multiplicative inverse.

If you remember how to express reciprocals $\frac1{a + bi}$ by “rationalizing the denominator” (multiply numerator and denominator by the conjugate $a – bi$), then the idea is that this should work here too, as long as $a^2 + b^2$ is guaranteed to be non-zero (as long as one of $a, b$ is nonzero modulo $7$). So you have to show that $a^2 +b^2 = 0$ has no solutions modulo $7$ besides $a = 0 = b$ (modulo $7$, of course). Can you show this is equivalent to $-1$ being a nonsquare modulo $7$?