# Prove that $x \in {\rm Jac}(R)$ if and only if $1-xy \in R^{\times}$ for all $y$.

I know the definition of Jacobson radical $$J(R)=\cap m$$ over maximal ideals. I want to prove the following property, but I can prove just one side of it.

Fact: $$x \in {\rm Jac}(R) \Longleftrightarrow \forall y, 1-xy \in R^{\times}$$

Proof: $$\Leftarrow$$: suppose on contrary that $$x \notin {\rm Jac}(R)$$, then there exists a maximal ideal $$m$$ such that $$x \notin m$$, so $$\langle x\rangle+m=R$$, so there exist $$y \in R$$ and $$m_0 \in m$$ such that $$xy+m_0=1$$. By assumption $$m_0=1-xy \in m$$ is a unit, and thus $$m=R$$, which contradicts the maximality of $$m$$.

$$\Longrightarrow$$: I don't have any idea for this side.

Edit: I think I did something, but I can not complete it!

$$x \in {\rm Jac}(R)$$, and let $$y$$ be arbitrary and let $$m$$ be an arbitrary maximal ideal. We claim that $$1-xy \notin m$$. suppose on contrary that $$1-xy \in m$$, then $$1=(1-xy)+xy \in m$$, so $$m=R$$ which contradicts the maximality of $$m$$.

so I proved that $$1-xy$$ is not included in no maximal ideal. but I can not go further.

Considering both of the answers, I guess that $$R^{\times}=R\setminus \cup m_{{\rm maximal}}$$, am I correct?

Suppose that $$x\in\text{Jac}(R)$$, but there exists $$y\in R$$ such that $$1-xy$$ is not invertible. Then, there exists a maximal ideal $$\mathfrak{m}$$ such that $$1-xy\in\mathfrak{m}$$. Set $$m=1-xy$$. By hypothesis, $$x\in\text{Jac}(R)\subseteq\mathfrak{m}$$, so $$x\in\mathfrak{m}$$. But $$\mathfrak{m}$$ being an ideal implies that $$xy\in\mathfrak{m}$$, and then $$m+xy=1-xy+xy=1\in\mathfrak{m},$$ which is absurd because $$\mathfrak{m}$$ is a proper ideal.
• yes this is true, if $a\in R^\times$ is not a unit, just take $I=(a)$ the principal ideal generated by $a$.Then $I$ is cointained in some maximal ideal $\mathfrak{m}$, and so $a\in\mathfrak{m}$. Commented Sep 23, 2020 at 17:32
Take $$x \in Jac(R)$$ and $$y \in R$$. Then see that $$1-yx$$ can't belong to any maximal ideal so the ideal generated by $$1-yx$$ is the whole ring $$R$$. You should be able to conclude that $$1-yx \in R^\times$$
• So we can conclude that $R^{\times}=R\backslash \cup m_{maximal}$? Am I right? Commented Sep 23, 2020 at 17:39