Given a ring $R$, $R=(1)$ is a principal ideal

Question

An ideal generated by the element $a$ is defined to be the intersection of all ideals containing $a$. My book says $R=(1)$ is a principal ideal, and I know how to convince myself of this using different approaches, but I can't reconcile how $R=(1)$ and $(1)$ is the intersection of all ideals containing $1$ for the following reason: suppose I have elements $a$ and $b$, with $(a)<(b)$. They must both contain $1$, but $(1)$ can only be as large as $(a)$ which is not necessarily $R$.

Answer 1

If an ideal $I$ of $R$ contains $1$, then given any $r \in R$, $r\cdot 1 \in I$. Hence $R$ is contained in $I$, and $I$ is necessarily contained in $R$, so $I=R$. So any ideal containing $1$ is in fact all of $R$, and the intersection of all such ideals is $R$. This shows $R=(1)$. Now this ideal is principal because it is generated by a single element.