## In a ring with 1, an ideal which contains 1 is the entire ring

Let $R$ be a ring with 1 and let $I \subseteq R$ be an ideal. Show that if $1 \in I$, then $I = R$.

Recall that if $a \in R$ and $b \in I$, then $ab \in I$. Letting $a$ be arbitrary and $b = 1$, then, we have $R \subseteq I$. Hence $R = I$.

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