# Nontrivial subring with identity of a ring without identity [duplicate]

I'm looking for an example a ring and a subring with $R \subset S$ such that $R$ has 1 but $S$ does not. Its easy to choose R to be the trivial ring with $0=1$, but are there any more exotic examples of this phenomenon?

## marked as duplicate by user26857 abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 15 '16 at 8:50

Let $R = \left\{\left(\begin{smallmatrix} a & 0 \\ 0 & 0 \end{smallmatrix}\right) : a \in K \right\}$, and let $S = \left\{\left(\begin{smallmatrix} a & b \\ 0 & 0 \end{smallmatrix}\right) : a,b \in K \right\}$. Note that, S is a rng under the standard operations in $M_2(K)$ whereas R is a ring with identity $\left(\begin{smallmatrix} 1 & 0 \\ 0 & 0 \end{smallmatrix}\right)$.
Let $R$ be your favorite ring with $1$, let $T$ be your favorite ring without $1$, and let $S=R\times T$ (identifying $R$ with $R\times\{0\}\subset S$). Your trivial example is just the special case of this when $R$ is the trivial ring.