Let $R$ by a commutative ring with $1$, and $I \subset R$ a non-zero integral ideal in $R$. When $R$ has finite quotients, and $I = P$ is prime in $R$, the group of units $(R/P)^{\times}$ of the finite ring $R/P$ is cyclic as $R/P$ is a finite field. Do there exist known sufficient and necessary conditions on $R$ and $I$ in general or for certain classes of unital rings for cyclicity of $(R/I)^{\times}$ ? In particular, do there exist more general analogues of the primitive root theorem, which answers this question for $R = \mathbb{Z}$ in terms of number-theoretic criteria on the positive generators of the principal ideals $I = (n)$?



Do there exist general conditions for cyclicity of unit groups of quotient rings (generalizations of the primitive root theorem)?
Throughout, let $R$ be a Noetherian ring and $I \subseteq R$ an ideal such that $R/I$ is finite. Then $R/I$ is Artinian, so we may write $I = I\_1 \cdots I\_r$ with $I\_i = \mathfrak m\_i^{n\_i}$ where $\mathfrak m\_1, \ldots, \mathfrak m\_r \subseteq R$ are pairwise distinct prime ideals and $\mathfrak m\_i^{n\_i} \subsetneq \mathfrak m\_i^{n\_i-1}$. Let $k\_i = R/\mathfrak m\_i$ and $p\_i = \operatorname{char} k\_i$, and write $q\_i$ for the order of $k\_i$ and $a\_i$ for the order of $(R/I\_i)^\times$.



> 
> **Theorem.** *Let $R, I, I\_i, \mathfrak m\_i, n\_i, k\_i, p\_i, q\_i, a\_i$ as above. Then $(R/I)^\times$ is cyclic if and only if all of the following hold:*
> 
> 
> 1. *The $a\_i$ are pairwise coprime;*
> 2. *If $n\_i > 1$, then $k\_i$ is the prime field $\mathbf F\_{p\_i}$ and $\mathfrak m\_i/\mathfrak m\_i^2$ has dimension $1$ (over $\mathbf F\_{p\_i} = k\_i$);*
> 3. *When $p\_i \neq 2$, either $n\_i \leq 2$ or $R/I\_i \cong \mathbf Z/p^{n\_i}$;*
> 4. *When $p\_i = 2$, we have $n\_i \leq 3$ and $R/I\_i \not\cong \mathbf Z/8$.*
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> 
> *If these hold, then $a\_i = q\_i - 1$ if $n\_i = 1$ and $a\_i = \phi(p\_i^{n\_i}) = p\_i^{n\_i-1}(p\_i - 1)$ if $n\_i > 1$, and the order of $(R/I)^\times$ is the product of the $a\_i$.*
> 
> 
> 


For $I = (m) \subseteq \mathbf Z$, the criteria are satisfied if and only if $m \mid 4$ or there exists an odd prime $p$ such that $m = p^n$ or $m = 2p^n$.


Indeed, criterion 2 is automatically satisfied. Criteria 3 and 4 give $n\_i \leq 2$ if $p\_i = 2$. We have $a\_i = \phi(p\_i^{n\_i})$, which is even as soon as $p\_i > 2$ or $n\_i \geq 2$. This gives the result for $R = \mathbf Z$.




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**Notation.** Given an Artinian local ring $(R,\mathfrak m)$ with $\mathfrak m^n = 0$ and $\mathfrak m^{n-1} \neq 0$, write $v \colon R \to \{0,\ldots,n\}$ for the function such that $(r) = \mathfrak m^{v(r)}$ for all $r \in R$. If $p = \operatorname{char}(R/\mathfrak m)$, write $e = v(p)$ (if $R = \mathcal O\_K/\mathfrak m^n$ for a finite extension $\mathbf Z\_p \to \mathcal O\_K$, then $e$ is the minimum of $n$ and the ramification index of $\mathbf Z\_p \to \mathcal O\_K$).


*Proof of Theorem.* The Chinese remainder theorem gives
$$R/I \cong \prod\_{i=1}^r R/I\_i,$$
so we get the same statement for the unit groups. Since a product $\prod\_i A\_i$ of finite abelian groups is cyclic if and only the $A\_i$ are cyclic of pairwise coprime degrees (again by the Chinese remainder theorem!), we get criterion 1 and reduce to the case $r = 1$. We will drop all subscripts $i$ and write $I = \mathfrak m^n$ with residue field $k$ of characteristic $p > 0$. Replacing $R$ by $R/I$ we may assume that $\mathfrak m^n = 0$ and $\mathfrak m^{n-1} \neq 0$, and we use the notation of Notation above. For $0 \leq i \leq j \leq n$ we get a short exact sequence
$$1 \to \frac{1+\mathfrak m^i}{1+\mathfrak m^j} \to \big(R/\mathfrak m^j\big)^\times \to \big(R/\mathfrak m^i\big)^\times \to 1.\tag{1}\label{1}$$
Moreover, for $1 \leq i \leq j \leq 2i$ we have an isomorphism
\begin{align\*}
\psi \colon \frac{\mathfrak m^i}{\mathfrak m^j} &\stackrel\sim\to \frac{1+\mathfrak m^i}{1+\mathfrak m^j}\tag{2}\label{2}\\
x &\mapsto 1+x.
\end{align\*}
Indeed, it is clearly a bijection, and the formula
$$(1+x)(1+y) = 1 + x + y + xy \equiv 1 + x + y \pmod{\mathfrak m^{2i}}$$
shows that $\psi$ is a homomorphism.


**Criteria 2, 3, 4 are necessary**
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We will first show that criteria 2, 3, and 4 are necessary. We will use repeatedly that subquotients of cyclic groups are cyclic. For criterion 2, if $n > 1$ then the sequence (\ref{1}) and the isomorphism (\ref{2}) for $(i,j) = (1,2)$ show that
$$\mathfrak m/\mathfrak m^2 \hookrightarrow \big(R/\mathfrak m^2\big)^\times.$$
Thus if $R^\times$ is cyclic, so are $(R/\mathfrak m^2)^\times$ and hence $\mathfrak m/\mathfrak m^2$, so
$$\dim\_{\mathbf F\_p} \mathfrak m/\mathfrak m^2 = 1.$$
This also forces $k = \mathbf F\_p$ since $\mathfrak m/\mathfrak m^2$ is actually a $k$-vector space, proving criterion 2. This also implies that $\mathfrak m^i/\mathfrak m^{i+1} \cong \mathbf F\_p = \mathbf Z/p$ for $i < n$, which together with the sequence (\ref{1}) and the isomorphism (\ref{2}) proves the formula
$$a = \begin{cases}q-1, & n = 1, \\ p^{n-1}(p-1), & n > 1. \end{cases}$$
For criterion 3, assume $p>2$. If $e = 1$, then $(p) = \mathfrak m$, so the unique map $\mathbf Z/p^n \to R$ is surjective (see e.g. Tag [00DV](https://stacks.math.columbia.edu/tag/00DV)(11)), hence an isomorphism by length considerations. Thus it suffices to show that if $e > 1$ and $n \geq 3$, then $(R/\mathfrak m^3)^\times$ is not cyclic. The truncated exponential
\begin{align\*}
\exp \colon \mathfrak m/\mathfrak m^3 &\to \big(R/\mathfrak m^3\big)^\times\\
x &\mapsto 1 + x + \tfrac{x^2}{2}
\end{align\*}
is an injective group homomorphism (here we use $p > 2$). Since $e \geq 2$, every element in $\mathfrak m/\mathfrak m^3$ is killed by $p$, so we conclude that $(R/\mathfrak m^3)^\times$ contains $\mathfrak m/\mathfrak m^3 \cong \mathbf Z/p \oplus \mathbf Z/p$, hence cannot be cyclic. This shows criterion 3.


For criterion 4, it is clear that $(\mathbf Z/8)^\times$ is not cyclic. Similar to the above, we see that $e = 1$ iff $R = \mathbf Z/2^n$, so it suffices to show that $(R/\mathfrak m^4)^\times$ is not cyclic if $n \geq 4$ and $e \geq 2$. For $x \in \mathfrak m^2$, we get
$$(1+x)^2 = 1 + 2x + x^2 \in 1 + \mathfrak m^4$$
since $2 \in \mathfrak m^2$. Thus, all $4$ elements of $(1+\mathfrak m^2)/(1+\mathfrak m^4)$ have order $2$, so $(R/\mathfrak m^4)^\times$ is not cyclic. This shows criterion 4.


**Criteria 2, 3, 4 are sufficient.**
------------------------------------


Conversely, given a finite Artinian local ring $(R,\mathfrak m)$ satisfying criteria 2, 3, and 4 (where $n = v(0)$ is the smallest integer such that $\mathfrak m^n = 0$), we have to show that $R^\times$ is cyclic. Clearly the case $n = 1$ is good, since $\mathbf F\_q^\times$ is cyclic of order $q-1$. The case $n = 2$ is also good, by the sequence (\ref{1}) and the isomorphism (\ref{2}): by assumption $\mathfrak m/\mathfrak m^2$ is cyclic of order $p$, and $k^\times$ is cyclic of order $p-1$. Then the sequence (\ref{1}) for $(i,j) = (1,2)$ splits and the middle term is cyclic by the Chinese remainder theorem.


For $p > 2$ we have to show that $(\mathbf Z/p^n)^\times$ is cyclic. This follows since the $p$-adic exponential
\begin{align\*}
\exp \colon p\mathbf Z\_p &\to 1 + p\mathbf Z\_p\\
x &\to \sum\_{i=0}^\infty \frac{x^i}{i!}
\end{align\*}
converges (in general, it converges when $v(x) > \tfrac{e}{p-1}$, so we're using that $e = 1$ and $p > 2$) and defines isomorphisms $p^i\mathbf Z\_p \cong 1 + p^i\mathbf Z\_p$ for all $i \geq 1$, hence an isomorphism
$$\frac{1 + p\mathbf Z\_p}{1+p^n\mathbf Z\_p} \cong \frac{p\mathbf Z\_p}{p^n\mathbf Z\_p} \cong \mathbf Z/p^{n-1}.$$
Then the sequence (\ref{1}) again splits (this time with $(i,j) = (1,n)$), and the Chinese remainder theorem shows that $(\mathbf Z/p^n)^\times$ is cyclic. (In fact the $p$-adic exponential gives $\mathbf Z\_p^\times \cong \mu\_{p-1} \times p\mathbf Z\_p$, where $\mu\_{p-1}$ are the $(p-1)^{\text{st}}$ roots of unity, so $\mathbf Z\_p^\times$ is procyclic with generator $(\zeta\_{p-1},p) = \zeta\_{p-1}\exp(p)$ for a primitive $(p-1)^{\text{st}}$ root of unity $\zeta\_{p-1}$.)


For $p = 2$ we have to show that $R^\times$ is cyclic if $n = 3$ and $e > 1$. We claim that $R^\times$ is generated by $1+\pi$ for any $\pi \in \mathfrak m \setminus \mathfrak m^2$. Indeed, $(1+\pi)^2 = 1 + 2\pi + \pi^2 \equiv 1 + \pi^2 \pmod{\mathfrak m^3}$ since $2 \in \mathfrak m^2$. Thus, $(1+\pi)^2 \neq 1$, so $1 + \pi$ has order $4$, hence generates.


This shows that the criteria are sufficient. We already saw that they are necessary and that the final statement holds. $\square$


