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

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    $\begingroup$ It should work, because $-1$ is not a square modulo $7$. Why don't you think it works? $\endgroup$
    – user43208
    Sep 17, 2013 at 22:52
  • $\begingroup$ @user43208 Sorry I meant it works. I didn't type correctly $\endgroup$
    – user43758
    Sep 17, 2013 at 22:54
  • $\begingroup$ You seem to be using \mbox and \text solely for the purpose of putting spaces on both sides of the vertical slash in $\{a+bi\mid a,b\in\mathbb F_3\}$. There is a standard way to do that, not involving \mbox or \text. Here it is: \{a+bi\mid a,b\in\mathbb F_3\} I changed it. ${}\qquad{}$ $\endgroup$ Sep 17, 2013 at 22:57
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    $\begingroup$ Here $i$ presumably means the square root of $-1$. In $\mathbb F_5$, $-1 = 4$ has a square root $2$ in the field and so the set $\{a+bi\mid a, b \in \mathbb F_5\}$ is just $\mathbb F_5$. On the other hand, in $\mathbb F_7$, $-1 = 6$ does not have a square root in the field, and so the method works. $\endgroup$ Sep 17, 2013 at 23:01
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    $\begingroup$ @DilipSarwate It would be better to say that this ring $\mathbb{F}_5[x]/(x^2 + 1)$, which we may identify with the set $\{a + bi: a, b \in \mathbb{F}_5\}$ with the appropriate operations, is isomorphic to a cartesian product $\mathbb{F}_5 \times \mathbb{F}_5$. Not just $\mathbb{F}_5$. $\endgroup$
    – user43208
    Sep 17, 2013 at 23:06

1 Answer 1


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 $7$?


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