If there exists a vertex of $ \Gamma_{2}(R)\setminus J(R) $ which is adjacent to every other vertex then $ R \cong \mathbb{Z}_{2}\times F$

Question

I am reading the research paper Comaximal Graph of Commutative Rings by H.R. Maimani, M. Salimki, A. Sattari, S. Yassemi.

In this paper, $ R $ denotes a commutative ring with the identity element. $ \Gamma(R) $ is a graph with vertices as elements of $ R $, where two distinct vertices $ a $ and $ b $ are adjacent if and only if $ Ra + Rb = R $. $ \Gamma_{2}(R)$ denotes the subgraph of $ \Gamma(R) $ which consists of non-unit elements. In addition, $ J(R) $ is the Jacobson radical of $ R $ and $ Max(R) $ is the set of all maximal ideals of $ R $.

I try to understand the proof of Proposition 2.4, part (b). Part (b) states

If there exists a vertex of $ \Gamma_{2}(R)\smallsetminus J(R) $ which is adjacent to every other vertex then $ R \cong \mathbb{Z}_{2}\times F $, where $ F $ is a field.

Unfortunately, I can't understand the second line of the proof (part (b)). Can anyone please explain me how to show if $ x $ is comaximal with each non unit outside the Jacobson radical then $ x $ is idempotent, $ J(R)=(0) $ and $ \mathfrak{m}=\{0,x\} $ is a maximal ideal ?

If it is true, i can show that for each non unit $ s \in R\setminus \mathfrak{m} $, having $ xR+sR=R $ impies $ (1-x)sR=(1-x)R $.

Then how to show $ (1-x)R=F $ is a field and hence $ R \cong \mathbb{Z}_{2}\times F $ ?

Any hints/ideas are much appreciated.

Thanks in advance for any replies.

Answer 1

If $x^2\ne x$ then $xR+x^2R=R$, so $x$ is invertible, a contradiction.

Let $y\in J(R)$. If $y\ne 0$ then $x+y\ne x$. Since $xR\ne R$ there is a maximal ideal $M$ such that $x\in M$. Then $x+y\in M$, so $x+y$ is not invertible. Also $x+y\notin J(R)$, otherwise $x\in J(R)$. Thus $xR+(x+y)R=R$; but $xR,(x+y)R\subseteq M$, a contradiction.

If $2x\ne 0$, then $2x\in R-U(R)\cup J(R)$, so $xR+2xR=R$, false. So $2x=0$. Similarly we get $rx=0$ or $rx=x$ for any $r\in R$. This shows that $m=\{0,x\}$ is an ideal. If it is not maximal there is a maximal ideal $M$ strictly containing it, and for $y\in M-m$ we must have $xR+yR=R$, a contradiction.

Since $x$ is idempotent we have $R\simeq xR\times (1-x)R$. But $xR=\{0,x\}\simeq\mathbb Z_2$. The unity of $(1-x)R$ is $1-x$. A nonzero element of $(1-x)R$ writes $(1-x)s$ with $s\notin xR$ and from $(1-x)sR=(1-x)R$ we get $1-x=(1-x)st$ for some $t\in R$, so $(1-x)s$ is invertible.