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The exercise states that, for the field $(\mathbb{Z}/3\mathbb{Z})[i]$, the group of units is cyclic. And it asks for the generator for this group.

My question is, since $(\mathbb{Z}/3\mathbb{Z})[i]$ is a field, all the elements in the field are units. Then it is asking for a generator for this entire field?...

Any help would be appreciated!

Eugene Zhang
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loveneighbor
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    All elements except $0$ are units. It is asking for a unit $x \in \mathbb Z/3\mathbb Z[i]$ such that every unit is of the form $x^n$ for some $n$. – Bruno Joyal Mar 19 '16 at 00:16

1 Answers1

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In this field we have nine elements: $a+bi$ where $a,b\in\{0,\pm1\}$.

Well, the multiplicative order of $-1$ is $2$, of $\pm i$ is $4$. We are looking for an element with order $8$.

Next number in this field coming to mind is $1+i$. Its square is $2i=-i$, whose order is $4$, so $1+i$ will be of order $8$.

By the way, its powers are: $1,\ 1+i,\ -i,\ 1-i,\ \ -1,\ -1-i,\ i,\ -1+i$.

Berci
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