Concerning Ideals and invertible elements in a commutative ring

Question

Here is the problem that I have:

Let $R$ be a commutative ring with unity and let $I$ be an ideal in $R$. Prove that $I=R$ if and only if $I$ contains some invertible element of the ring $R$.

Here is what I did: Let $a\in I$, and $a$ is invertible. That will mean that $1\in I$, as $I$ includes $a^{-1}a$. And if $1\in I$, $I=R$, as for any $a\in R$, $a1=a$ and $1a=a$.

How can I prove the second part of the if and only if(i.e. prove that if $I=R$, then $I$ or $R$ includes invertible element)?

Answer 1

($\Rightarrow$) Suppose $I=R$, then $1\in R$ which is an invertible element (unit).

($\Leftarrow$) We need to show that $1\in I$, as if this happens, then by definition of ideal. $1.a\in R$ i.e. every $a\in R$ implies $a\in I$. Thus $I\subseteq R$

Case I: $1\in R$. By definition of ideals $I=R$

Case II: If $a$ is a unit in R and $a\in I$. Then $1=a^{-1}a\in I$. Again by above statement $I=R$

Answer 2

If $I = R$, then $R$ contains unity, which is its own inverse, so $I$ contains an invertible element.