Let $ R$ be a commutative ring with $1$ and $S$ be a subring of $R$ containing $1$.

Question

Let $ R$ be a commutative ring with $1$ and $S$ be a subring of $R$ containing $1$.

Can we say $R=S$ always ?

Answer 1

If $S$ is a subring of $R$ with $1$ we can't say $R=S$ in general. Plenty of examples have been given already, for instance $1\in \mathbb{Z}$ and $\mathbb{Z\subseteq Q}$ but $\mathbb{Z\neq Q}$.

The statement would be true if $S$ was an ideal which is a subring satisfying the following property, if $r\in R$ and $s\in S$ then $rs\in S$. Then, if $S$ is an ideal containing $1$ we can always say $S = R$.

Answer 2

Certainly not. For example, $1\in\mathbb Z\subseteq\mathbb R$.

What's more, if $Q$ is a commutative ring with $1$, then $R=Q\times Q$ (with component-wise operations) is a commutative ring with $1$, that is, the one of $R$ is $(1,1)$, and $S=\{(q,q)|q\in Q\}$ is a aubring of $R$ containing $1$, but $S\neq R$.