Ideal of Unit is Whole Ring

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Theorem

Let $\struct {R, +, \circ}$ be a ring with unity.

Let $J$ be an ideal of $R$.

If $J$ contains a unit of $R$, then $J = R$.


Corollary

Let $\struct {R, +, \circ}$ be a ring with unity.


Let $J$ be an ideal of $R$.

If $J$ contains the unity of $R$, then $J = R$.


Proof

Let $u \in J$, where $u \in U_R$.

Also by definition, we have $u^{-1} \in U_R$.

Let $x \in R$.

\(\ds \) \(\) \(\ds x \in R\)
\(\ds \) \(\leadsto\) \(\ds x \circ u^{-1} \in R\) as $R$ is closed
\(\ds \) \(\leadsto\) \(\ds \paren {x \circ u^{-1} } \circ u \in J\) Definition of Ideal of Ring
\(\ds \) \(\leadsto\) \(\ds x \in J\) Ring properties: $u \circ u^{-1} = 1_R$


Thus $R \subseteq J$.

As $J \subseteq R$ by definition, it follows that $J = R$.

$\blacksquare$


Sources

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