<s>[INST] For which prime $p$, is the group units $R^*$ cyclic? I'm studying for my qualifying exam and found the following question in the question bank of previous years paper:

Let $p$ be a prime and $\mathbb{F}_p$ the field with $p$ elements. Set $R=\mathbb{F}_p[x]/(x^3)$. For which prime $p$ is the group of units $R^*$ cyclic?

What I've got so far is that: $R$ is local ring and any element in $R$ with non-zero constant term will be a unit. I have proven it for $p=2$, but I don't know if it will be true for any other primes. Also, I have proven that if $r=a\overline{x}^2+b\overline{x}+c$ is the generator of the cyclic group $R^*$ then $b\not=0$. I tried to look at the powers of $r$ hoping that I might be able find some element in $R^*$ which can't be expressed as $r^n$, for some $n$, but it got complicated and now I'm completely lost.
 [/INST] First, note that if $ax^2 + bx + c$ has $c \neq 0,$ then $\gcd(x^3, ax^2 + bx + c) = 1$ in $\mathbb{F}_p[x]$ (since the only prime factor of $x^3$ is $x,$ which won't divide $ax^2+bx+c$ for $c\neq 0$) and so $ax^2+bx+c$ is invertible mod $x^3.$ It's easy to see that nothing else can be invertible mod $x^3$ from a similar gcd argument. If $p=2$ this ring $R$ thus consists only of $x^2+x+1, x^2 + 1,$ and $x+1.$ It is cyclic with generator $x+1,$ since $(x+1)^2 = x^2 + 1$ and $(x+1)^3 = (x^2+1)(x+1) = x^2 + x + 1$ in $R.$
Suppose that $R$ had cyclic group of units, and let it be generated by $ax^2 + bx + c.$
Note that, mod $x^3,$ we find
$$(ax^2 + bx + c)^n = ((ax^2+bx)+c)^n = c^n + nc^{n-1}(ax^2+bx) + \binom{n}{2}c^{n-2}(ax^2+bx)^2 = c^n + nc^{n-1}(ax^2 + bx) + \binom{n}{2}c^{n-2}b^2x^2.$$
This lets us immediately rule out $b \neq 0,$ since otherwise we'd never generate a polynomial like $x+1$ with nontrivial linear term.
Now, as $n$ varies, we need $(c^n, nc^{n-1}b, \binom{n}{2}c^{n-2}b^2 + nc^{n-1}a)$ to run through all possible triples $(x,y,z)$ in $\mathbb{F}_p$ where $x \neq 0.$
The value of $c^n$ depends only on $n$ mod $p-1,$ since $c$ is fixed. Similarly, $n, \binom{n}{2}$ only depend on the value of $n$ mod $p$ for $p > 2,$ so the entire triple $(c^n, nc^{n-1}b, \binom{n}{2}c^{n-2}b^2 + nc^{n-1}a)$ depends only on $n$ mod $p(p-1).$ Thus, there are only $p^2 - p$ distinct possibilities for this triple at most.
How many $(x,y,z)$ triples with $x \neq 0$ are there, though? Well, there's $p^2(p-1) > p(p-1).$ So, it's not cyclic if $p > 2$.
(This breaks for $p=2$ since, as indicated above, $\binom{n}{2}$ mod $2$ cannot be determined just by knowing $n$ mod $2,$ since i.e. $\binom{3}{2} = 3$ is odd but $\binom{1}{2} = 0$ is even. This is because $\binom{n}{2} = n(n-1)/2$ and division by $2$ only makes since modulo an odd prime.)
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