Let $R$ be a finite ring (with unity) of order $p^2$, where $p$ is a prime. I know that there are exactly 11 rings of this order. My question is the following: Is there a classification of the group of units in every ring of order $p^2$? If the answer is yes, please recommend a reference. Thanks.
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Only 4 of the 11 rings of order $p^2$ have a unity.
$$ \mathbb{Z}/(p^2) $$
$$ \mathbb{F}_{p^2} $$
$$ \mathbb{F}_{p}\times \mathbb{F}_{p} $$
$$ \mathbb{F}_{p}[X]/(X^2) $$
See answers to the following question for an explanation: Classifying Unital Commutative Rings of Order $p^2$
For case 1: the group of units is $C_{p^2-p}$
Source: https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n#Cyclic_case
For case 2: the group of units is $C_{p^2-1}$
Source: Finite subgroups of the multiplicative group of a field are cyclic
For case 3: the group of units is $C_{p-1}\times C_{p-1}$
Source: Group of units of direct sum of rings is isomorphic to direct sum of the groups of units
For case 4: an element $a+bX+(X^2)$ in this ring is a unit iff $a\neq 0$, and the group is again cyclic $C_{p^2-p}$.
Sources:
Units in polynomial quotient ring
halrankard
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+1 Excellent referencing! – rschwieb Jun 29 '20 at 21:25
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If $\{0\}$ is maximal then its a field of $p^2$ elements and it's known that the units are the cyclic group of order $p^2-1$.
If $\{0\}$ is not maximal, then there is a maximal ideal of order $p$, call it $M$. If $M$ is unique, then $R\setminus M$ is the set of units, and as a group of order $p^2-p$. This case breaks further down into the case where the characteristic is $p$ or the characteristic is $p^2$.
If $M$ is not unique, then there's another maximal ideal $M'$ such that $M\cap M'=\{0\}$ (arguing by orders.). In this case $R\cong F_p\times F_p$, and the group of units is obviously $C_{p-1}\times C_{p-1}$.
rschwieb
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