A ring with a subring that is a field I'm looking for an example of a ring $R$ such that $R$ has no multiplicative identity, but R has a subring $A$ which is a field
 A: A slightly different example would be the rng (= ring without 1) of 2x2 matrices over the reals of the form
$$
M(a,b)=\pmatrix{a&b\cr0&0\cr}.
$$
We have $M(a,b)M(c,d)=M(ac,ad)$, so no (2-sided) multiplicative identity element exists, and this is a rng with respect to the usual matrix operations.
OTOH the matrices of the form $M(a,0)$ form a subring isomorphic to $\mathbf{R}$.
A: I'll go against the temptation of posting a trivial answer as a comment. Polynomials in $X$ (or more indeterminates) over any given field are probably the best known example of the situation you describe, the subring being the constant polynomials. Any decent algebra course should give this example immediately after defining rings and fields.
Added This answer is wrong, I misread the question (because for me rings always have a multiplicative identity, which is apparently not intended here): I thought it was about non-field rings with a subring that is a field. Maybe the title should say "A rng with a subring that is a field"
A: Here a simple example:
take $\mathbb K$ a field and consider the ring $R=xK[x]$. The ring $\mathbb K \times R$ has no identity, but it has a sub ring $\mathbb K \times \{0\}$ is isomorphic to $\mathbb K$, so it's a field.
