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Given the following set $Z_3[i]=\{a+bi | a,b \in Z_3\}$, which is a ring, prove that it is also a field.

To be a field, every nonzero element of the ring has to be invertible for the multiplication. So:

$(a+bi)^{-1} = \frac{1}{a+bi} = \frac{a-bi}{(a+bi)(a-bi)} = \frac{a}{a^2+b^2}-\frac{bi}{a^2+b^2}$.

We can consider the case in which $a=b=1$, so we have the elements $(1+i)$ and $(\frac{1}{2}-\frac{1}{2}i)$. Both elements are the inverse of each other, but, in the second element, $a=\frac{1}{2} \notin Z_3$.

We can conclude that $Z_3[i]$ is not a field, which contradict what I watch in a video in which the youtuber says that this set is a field (she didn't show the proof).

What is the correct result and why?

Sergio Durán Vega
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  • The issue here seems to be that you are trying to perform standard multiplication, but this does not work. Attempting to "divide" in the normal sense would assume a priori that this is a field, which you have not done. Think about what multiplication is in this ring. – JJ Hoo May 08 '22 at 15:54
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    $(\Bbb Z/3)[i]$ is isomorphic to the quotient ring $\Bbb Z[i]/(3)$, and $(3)$ is a maximal ideal. So the quotient is a field - see here. – Dietrich Burde May 08 '22 at 16:02
  • My recommendation: write out the multiplication table for your set — it’s only nine by nine, so won’t take long (and it’s an instructive exercise, too). Then notice that every nonzero element has its multiplicative inverse. – Lubin May 08 '22 at 18:13

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