Jacobson radical $J$ is the largest ideal consisting of quasiregular elements

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Assume that $R$ is a commutative ring with unity and $J$ denotes the Jacobson radical of $R$. I have seen that $1-x$ is a unit for every $x \in J$. I will also see that the Jacobson radical $J$ is the largest ideal of $R$ such that $1-x$ is a unit for each $x \in J$.

In order to show that I need to consider an ideal $I$ of $R$ such that $1-x$ is a unit for each $x \in I$ and show that $I \subseteq J$.

Do you have any idea?

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Suppose $T$ is a maximal right ideal of $R$, and that $I$ isn't contained in it.

Then $I+T=R$, so that $x+t=1$ for some $x\in I$, $t\in T$.

Can you see the contradiction this presents immediately?

What's that tell you about $I$'s relationship to the maximal right ideals?