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Let $R$ be a ring and let $J$ be an ideal of $R$. Assume $J$ contains a unit of $R$. Prove $J=R$, rigorously.

Again I feel here as if the statement is intuitive so I am not able to make it rigorous. If you could show all steps in the proof that would be nice.

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1 Answer 1

up vote 5 down vote accepted

If $J$ contains a unit of $R$, say $a$ then we must have $a^{-1} \cdot a = 1$ being in $J$. However if $J$ contains $1$, recall that an ideal must "absorb" multiplication from the outside. In our case, this means in particular that for all $r\in R$, we have

$$r \cdot 1 = r \in J$$

so that $R \subseteq J$. Since a priori $J \subseteq R$ we have $J = R$.

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