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Pete L. Clark
Pete L. Clark
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$\newcommand{\F}{\mathbb{F}} \newcommand{\Z}{\mathbb{Z}}$ [Throughout this answer all rings will be commutative (and unital!).]

It seems that van Dobben de Bruyn has essentially rediscovered a theorem of Gilmer:

Gilmer, Robert W., Jr. Finite rings having a cyclic multiplicative group of units. Amer. J. Math. 85 (1963), 447-452.

A couple of preliminary comments: (i) In van Dobben de Bruyn's answer, we may as well take $I = (0)$: that is, he is giving necessary and sufficient conditions on a finite commutative ring to have cyclic unit group. (ii) A finite ring $R$ is indeed Artinian, hence a finite product $\prod_{i=1}^r R_i$ of local rings $R_i$, each of which must have prime power order. As seen in his answer, we quickly find that $R^{\times}$ is cyclic iff each $R_i^{\times}$ is cyclic and $\# R_1^{\times},\ldots, \# R_r^{\times}$ are pairwise coprime. Thus the critical case is the classification of finite local rings with cyclic unit group. Here is Gilmer's result:

Theorem Let $R$ be a finite local ring. Then $R^{\times}$ is cyclic iff $R$ is isomorphic to one of the following rings:
(A) A finite field $\F$.
(B) $\Z/p^a \Z$ for an odd prime number $p$ and $a \in \Z^+$.
(C) $\Z/4\Z$.
(D) $\Z/p\Z[t]/(t^2)$ for a prime number $p$.
(E) $\Z/2\Z[t]/(t^3)$.
(F) $\Z[t]/\langle 2t,t^2-2 \rangle$, a $\Z/4\Z$-algebra of order $8$.

To compare Gilmer's classification to van Dobben de Bruyn's it is helpful to observe that the local rings of order $p^2$ are $\F_{p^2}$, $\Z/p^2\Z$ and $\Z/p\Z[t]/(t^2)$ and to know the fivesix local rings of order $8$.

By the way, Gilmer's Theorem appears as Theorem 5.14 in this expository note of mine, where it used to derive Theorem 5.15, a 2013 result of Hirano-Matsuoka that explicitly determines the product over all elements of the unit group of a finite ring. (Thus it is a generalization of Wilson's Theorem that $(p-1)! \equiv -1 \pmod{p}$. It seems weird that it is so recent.) I wanted to include the proof of Gilmer's Theorem in the note, but it is rather long and computational. The proof of van Dobben de Bruyn looks a bit shorter!

A final comment leading to a question: It turns out that all the rings in Gilmer's classification are principal, i.e., every ideal is principal. (This is obvious except for (F), in which case you can see my paper if you don't want to do the calculation yourself.) In other words, for a finite ring $R$ the property that the unit group be cyclic forces every $R$-submodule of $R$ to be cyclic. Is that just a coincidence, or can it be proved directly?

$\newcommand{\F}{\mathbb{F}} \newcommand{\Z}{\mathbb{Z}}$ [Throughout this answer all rings will be commutative (and unital!).]

It seems that van Dobben de Bruyn has essentially rediscovered a theorem of Gilmer:

Gilmer, Robert W., Jr. Finite rings having a cyclic multiplicative group of units. Amer. J. Math. 85 (1963), 447-452.

A couple of preliminary comments: (i) In van Dobben de Bruyn's answer, we may as well take $I = (0)$: that is, he is giving necessary and sufficient conditions on a finite commutative ring to have cyclic unit group. (ii) A finite ring $R$ is indeed Artinian, hence a finite product $\prod_{i=1}^r R_i$ of local rings $R_i$, each of which must have prime power order. As seen in his answer, we quickly find that $R^{\times}$ is cyclic iff each $R_i^{\times}$ is cyclic and $\# R_1^{\times},\ldots, \# R_r^{\times}$ are pairwise coprime. Thus the critical case is the classification of finite local rings with cyclic unit group. Here is Gilmer's result:

Theorem Let $R$ be a finite local ring. Then $R^{\times}$ is cyclic iff $R$ is isomorphic to one of the following rings:
(A) A finite field $\F$.
(B) $\Z/p^a \Z$ for an odd prime number $p$ and $a \in \Z^+$.
(C) $\Z/4\Z$.
(D) $\Z/p\Z[t]/(t^2)$ for a prime number $p$.
(E) $\Z/2\Z[t]/(t^3)$.
(F) $\Z[t]/\langle 2t,t^2-2 \rangle$, a $\Z/4\Z$-algebra of order $8$.

To compare Gilmer's classification to van Dobben de Bruyn's it is helpful to observe that the local rings of order $p^2$ are $\F_{p^2}$, $\Z/p^2\Z$ and $\Z/p\Z[t]/(t^2)$ and to know the five local rings of order $8$.

By the way, Gilmer's Theorem appears as Theorem 5.14 in this expository note of mine, where it used to derive Theorem 5.15, a 2013 result of Hirano-Matsuoka that explicitly determines the product over all elements of the unit group of a finite ring. (Thus it is a generalization of Wilson's Theorem that $(p-1)! \equiv -1 \pmod{p}$. It seems weird that it is so recent.) I wanted to include the proof of Gilmer's Theorem in the note, but it is rather long and computational. The proof of van Dobben de Bruyn looks a bit shorter!

A final comment leading to a question: It turns out that all the rings in Gilmer's classification are principal, i.e., every ideal is principal. (This is obvious except for (F), in which case you can see my paper if you don't want to do the calculation yourself.) In other words, for a finite ring $R$ the property that the unit group be cyclic forces every $R$-submodule of $R$ to be cyclic. Is that just a coincidence, or can it be proved directly?

$\newcommand{\F}{\mathbb{F}} \newcommand{\Z}{\mathbb{Z}}$ [Throughout this answer all rings will be commutative (and unital!).]

It seems that van Dobben de Bruyn has essentially rediscovered a theorem of Gilmer:

Gilmer, Robert W., Jr. Finite rings having a cyclic multiplicative group of units. Amer. J. Math. 85 (1963), 447-452.

A couple of preliminary comments: (i) In van Dobben de Bruyn's answer, we may as well take $I = (0)$: that is, he is giving necessary and sufficient conditions on a finite commutative ring to have cyclic unit group. (ii) A finite ring $R$ is indeed Artinian, hence a finite product $\prod_{i=1}^r R_i$ of local rings $R_i$, each of which must have prime power order. As seen in his answer, we quickly find that $R^{\times}$ is cyclic iff each $R_i^{\times}$ is cyclic and $\# R_1^{\times},\ldots, \# R_r^{\times}$ are pairwise coprime. Thus the critical case is the classification of finite local rings with cyclic unit group. Here is Gilmer's result:

Theorem Let $R$ be a finite local ring. Then $R^{\times}$ is cyclic iff $R$ is isomorphic to one of the following rings:
(A) A finite field $\F$.
(B) $\Z/p^a \Z$ for an odd prime number $p$ and $a \in \Z^+$.
(C) $\Z/4\Z$.
(D) $\Z/p\Z[t]/(t^2)$ for a prime number $p$.
(E) $\Z/2\Z[t]/(t^3)$.
(F) $\Z[t]/\langle 2t,t^2-2 \rangle$, a $\Z/4\Z$-algebra of order $8$.

To compare Gilmer's classification to van Dobben de Bruyn's it is helpful to observe that the local rings of order $p^2$ are $\F_{p^2}$, $\Z/p^2\Z$ and $\Z/p\Z[t]/(t^2)$ and to know the six local rings of order $8$.

By the way, Gilmer's Theorem appears as Theorem 5.14 in this expository note of mine, where it used to derive Theorem 5.15, a 2013 result of Hirano-Matsuoka that explicitly determines the product over all elements of the unit group of a finite ring. (Thus it is a generalization of Wilson's Theorem that $(p-1)! \equiv -1 \pmod{p}$. It seems weird that it is so recent.) I wanted to include the proof of Gilmer's Theorem in the note, but it is rather long and computational. The proof of van Dobben de Bruyn looks a bit shorter!

A final comment leading to a question: It turns out that all the rings in Gilmer's classification are principal, i.e., every ideal is principal. (This is obvious except for (F), in which case you can see my paper if you don't want to do the calculation yourself.) In other words, for a finite ring $R$ the property that the unit group be cyclic forces every $R$-submodule of $R$ to be cyclic. Is that just a coincidence, or can it be proved directly?

Source Link
Pete L. Clark
Pete L. Clark
  • 65.4k
  • 12
  • 241
  • 381

$\newcommand{\F}{\mathbb{F}} \newcommand{\Z}{\mathbb{Z}}$ [Throughout this answer all rings will be commutative (and unital!).]

It seems that van Dobben de Bruyn has essentially rediscovered a theorem of Gilmer:

Gilmer, Robert W., Jr. Finite rings having a cyclic multiplicative group of units. Amer. J. Math. 85 (1963), 447-452.

A couple of preliminary comments: (i) In van Dobben de Bruyn's answer, we may as well take $I = (0)$: that is, he is giving necessary and sufficient conditions on a finite commutative ring to have cyclic unit group. (ii) A finite ring $R$ is indeed Artinian, hence a finite product $\prod_{i=1}^r R_i$ of local rings $R_i$, each of which must have prime power order. As seen in his answer, we quickly find that $R^{\times}$ is cyclic iff each $R_i^{\times}$ is cyclic and $\# R_1^{\times},\ldots, \# R_r^{\times}$ are pairwise coprime. Thus the critical case is the classification of finite local rings with cyclic unit group. Here is Gilmer's result:

Theorem Let $R$ be a finite local ring. Then $R^{\times}$ is cyclic iff $R$ is isomorphic to one of the following rings:
(A) A finite field $\F$.
(B) $\Z/p^a \Z$ for an odd prime number $p$ and $a \in \Z^+$.
(C) $\Z/4\Z$.
(D) $\Z/p\Z[t]/(t^2)$ for a prime number $p$.
(E) $\Z/2\Z[t]/(t^3)$.
(F) $\Z[t]/\langle 2t,t^2-2 \rangle$, a $\Z/4\Z$-algebra of order $8$.

To compare Gilmer's classification to van Dobben de Bruyn's it is helpful to observe that the local rings of order $p^2$ are $\F_{p^2}$, $\Z/p^2\Z$ and $\Z/p\Z[t]/(t^2)$ and to know the five local rings of order $8$.

By the way, Gilmer's Theorem appears as Theorem 5.14 in this expository note of mine, where it used to derive Theorem 5.15, a 2013 result of Hirano-Matsuoka that explicitly determines the product over all elements of the unit group of a finite ring. (Thus it is a generalization of Wilson's Theorem that $(p-1)! \equiv -1 \pmod{p}$. It seems weird that it is so recent.) I wanted to include the proof of Gilmer's Theorem in the note, but it is rather long and computational. The proof of van Dobben de Bruyn looks a bit shorter!

A final comment leading to a question: It turns out that all the rings in Gilmer's classification are principal, i.e., every ideal is principal. (This is obvious except for (F), in which case you can see my paper if you don't want to do the calculation yourself.) In other words, for a finite ring $R$ the property that the unit group be cyclic forces every $R$-submodule of $R$ to be cyclic. Is that just a coincidence, or can it be proved directly?