Prove that if $I$ contains a unit, then $I = R$.

Question

Let $R$ be a ring with identity, $I$ an ideal of $R$. Prove that if $I$ contains a unit, then $I = R$.

I am doing a double containment proof and am stuck with this ($\subseteq$) direction.

Proof : Let $R$ be a ring with identity and $I$ an ideal of $R$.

($\supseteq$) Suppose $U\in I$ such that $U$ is a unit. Then $U^{-1}\in R$. So $U\cdot U^{-1} = 1$. Since $U\in I$, $U\cdot U^{-1}\in I$. Thus $1 \in I$. Suppose $a \in R$. Then $a\cdot 1 \in I$. Thus, for every $a \in R$, $I = R$.

($\subseteq$) Suppose

Answer 1

An ideal is (by definition) a subset of a ring.

You only need to show "$\supseteq$" inclusion which you have done correctly.