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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?

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2 Answers 2

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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.

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  • $\begingroup$ Fact 2: every non-unit element is included in some maximal ideal. Is this true? $\endgroup$ Commented Sep 23, 2020 at 17:30
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    $\begingroup$ 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 $a\in\mathfrak{m}$. $\endgroup$ Commented Sep 23, 2020 at 17:32
  • $\begingroup$ Thank you so much $\endgroup$ Commented Sep 23, 2020 at 17:33
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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 $1-yx \in R^\times$

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  • $\begingroup$ So we can conclude that $R^{\times}=R\backslash \cup m_{maximal}$? Am I right? $\endgroup$ Commented Sep 23, 2020 at 17:39
  • $\begingroup$ Yes, this is actually a usual fact. I just said it differently $\endgroup$
    – user598294
    Commented Sep 23, 2020 at 17:45

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