Determine all non-trivial finite rings with am unit element in which the sum of all elements is invertible. Mihai Opincariu
Problem
Source: Romanian NMO 2021 grade 12 P2
Tags: superior algebra, Rings
16.04.2023 21:16
I should be preparing my CS final and here my brain is telling me to do random algebra problem. Romanian NMO 2021 Grade 12/2 wrote: Determine all non-trivial finite rings with an unit element in which the sum of all elements is invertible. Mihai Opincariu The answer is $\mathbb{Z}/2\mathbb{Z}$. It is straightforward to check that this work since the sum of all elements in $\mathbb{Z}/2\mathbb{Z}$ is $1$, which is invertible. Now, we'll prove that there are no other non-trivial finite rings that work. For convenience, we name such a ring $R$ and let \[ y = \sum_{x \in R} x \]From the problem's condition, $y$ is invertible. Hence, \[ 2y = \sum_{x \in R} x + \sum_{x \in R} (-x) = 0 \implies 2 = 0y^{-1} = 0 \]Therefore, $R$ is a ring of characteristic $2$. Since $R$ is finite and nontrivial, by Cauchy's Theorem we conclude that $|R| = 2^n$ for some $n \ge 1$. Now, note that all the $2^n$ elements in $R$ can be grouped into $2^{n - 1}$ pairs of $(x,x+1)$ -- pair any $x \in R$ with $x + 1 \not= x$. The pairs are well-defined since $(x + 1) + 1 = x$. Furthermore, the sum of the two elements in any random pair is $1$. Therefore, we have \[ y = \sum_{x \in R} x = 2^{n - 1} \]is invertible, which forces $n = 1$. Thus, $R$ must have $2$ element. However, this ring has $0 \not= 1$. Thus, $R = \{ 0, 1 \} \cong \mathbb{Z}/2\mathbb{Z}$, as desired.
23.07.2024 22:04
Let $R$ be the ring and let $n\geq 2$ be its characteristic. Then $\mathbb{Z}_n$ is a subgroup of $(R,\,+)$. For $a\in R$, the sum of the elements in $a+\mathbb{Z}_n$ is equal to $na+0+1+\cdots+(n-1)=\left(\frac{n(n-1)}{2}\right)_R$. Summing this over all the cosets of $\mathbb{Z}_n$, we get $\sum_{a\in R}=\left(\frac{|R|}{n}\cdot\frac{n(n-1)}{2}\right)_R$. In order for this to be a unit, it must be relatively prime with $n$. This occurs only when $n=2$. Then $|R|$ is a power of $2$ so $|R|=2$. Thus the only ring is $\boxed{R\cong\mathbb{F}_2}$. $\square$