# Generator of the multiplicative groups of units in $\mathbb{Z_5[x]}/(x^{2}+3x+3)$

I would like to find a generator of the multiplicative groups of units in $\mathbb{Z_5[x]}/(x^{2}+3x+3)$. This is a field since $x^{2}+3x+3$ is irreducible, so every coset with $bx+a\not=0$ as a representative should be a unit...

I do not understand how to go from here though... Since $b\not=0$ we have $4$ diffrent choices for $b$ and $5$ different options for coefficient $a$ hence $24$ elements in the multiplicative group of units.

Should any $bx+a$ with order $24$ be a generator? How do I go from here?

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What is $\mathbb{Z}_5$? For some, this is the field with five elements; for many, it is the ring of $5$-adic integers. To avoid ambiguity, you might use $\mathbb{Z}/(5)$ or some similar notation for the finite field. – Lubin Aug 12 '12 at 20:16
There are some general methods. An amount of searching is involved. But if you, for example, stumble upon elements of order $8$ and $3$ while looking for a generator, then their product would necessarily have order $24$ (think inside a cyclic group). Here $\phi(24)=8$, so a random choice will work with probability $1/3$. High enough to suggest that trial-and-error will work reasonably well. See my answer for a proof that the first guess works. There is scope for being smart, I guess, but 24 is a small enough number anyhow... – Jyrki Lahtonen Aug 12 '12 at 20:24
As J. Lubin notes, something like $\mathbb F_5$ would be much better than $\mathbb Z_5$, if not $\mathbb Z/5$, especially as one goes out into the big world... – paul garrett Aug 12 '12 at 22:04

## 1 Answer

It suffices to rule out the maximal factors of $24$, i.e. $8=24/3$ and $12=24/2$ as possible orders. The (coset of) $x$ fits the bill, because $$x^4=(-x^2)^2\equiv (3x+3)^2=4(x^2+2x+1)=-x^2+3x+4\equiv6x+7=x+2,$$ and therefore neither $$x^8=(x^4)^2\equiv(x+2)^2=x^2+4x+4\equiv x+1$$ not $$x^{12}=x^8\cdot x^4\equiv(x+2)(x+1)=x^2+3x+2\equiv-1$$ are equal to $1$.

The point is that other possible orders ($2,3,4,6$) that are less than $24$ are factors of either $8$or $12$.

Now that we settled that $x$ is a generator, the other generators can be found by reducing the powers $x^j$, $0<j<24, \gcd(j,24)=1$ modulo the generator $x^2+3x+3$.

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