corner of a ring

Does there exist a subset $S$ of a ring $R$ which is a ring with a multiplicative identity, but not a subring of $R$?

Let $R$ be a ring without the assumption that $R$ has a multiplicative identity. Further, assume that $e$ is an idempotent of $R$. Then the subset of the form $eRe$ is called a corner of the ring $R$.

It’s not hard to see that $eRe$ is a ring with $e$ as its multiplicative identity:

1. $eae+ebe=e(a+b)e\in eRe$,

2. $0=e0e\in eRe$,

3. $e(-a)e$ is the additive inverse of $eae$ in $eRe$,

4. $(eae)(ebe)=e(aeb)e\in eRe$, and

5. $e=ee=eee\in eRe$, with $e(eae)=eae=(eae)e$, for any $eae\in eRe$.

If $R$ has no multiplicative identity, then any corner of $R$ is a proper subset of $R$ which is a ring and not a subring of $R$. If $R$ has 1 as its multiplicative identity and if $e\ne 1$ is an idempotent, then the $eRe$ is not a subring of $R$ as they don’t share the same multiplicative identity. In this case, the corner $eRe$ is said to be proper. If we set $f=1-e$, then $fRf$ is also a proper corner of $R$.

Remark. If $R$ has 1 with $e\ne 1$ an idempotent. Then corners $S=eRe$ and $T=fRf$, where $f=1-e$, are direct summands (as modules over $\mathbb{Z}$) of $R$ via a Peirce decomposition.