Subrings and homomorphisms of unitary rings Let $(R,+,\cdot)$ be a ring (in the definition i use multiplication is associative operation and it's not assumed that there is unity in the ring).
I've seen two definitons of subring. 
1) non-empty subset $S \subset R$ is called subring of ring $(R,+,\cdot)$ iff $(S,+,\cdot)$ is a ring 
2) Let $(R,+,\cdot)$ be unitary ring with unity $e$. Non-empty subset $S \subset R$ is called subring of ring $(R,+,\cdot)$ iff $(S,+,\cdot)$ is a ring and $e \in S$.
I'm looking for an example of such ring R and its subset S that $(S,+,\cdot)$ is a ring but not a unitary ring.
I will be also very grateful for an example of such UNITARY rings $(R_1,+_1,\cdot_1)$, $(R_2,+_2,\cdot_2)$ and function $f: ~~ R_1 \longrightarrow R_2$ that 
$(1) \forall a,b \in R_1 ~~~ f(a+_1 b) = f(a) +_2 f(b) $ 
$(2) \forall a,b \in R_1 ~~~ f(a\cdot_1 b) = f(a) \cdot_2 f(b)$
$(3) f(e_1) \neq e_2$ 
where $e_1$ is unity in $R_1$; $e_2$ - in $R_2$.
Thanks in advance.
 A: *

*$R = \mathbb Z$ and $S = 2\mathbb Z$

*$R_1 = R_2 = \mathbb Z$ and $f(a) = 0$.


More generally let $R$ be a unitary ring. Any proper ideal $S$ provides an example for 1. For arbitrary unitary rings $R_1$ and $R_2\neq\{0\}$, the zero map is always an example for 2.
A: *

*Take any $S$ without $1$ and join to it extra unity $1$, so you get the unitary ring $R=\{a+n\cdot 1|a\in S, n\in \mathbb{Z}\}$ in which $S$ is a subring.

*Let $M$ be a monoid with an idempotent $e\ne 1$, $\mathbb{Z}M $ its semigroup ring. Then the embedding $\mathbb{Z}e\to \mathbb{Z}M$ is what you want.
A: Possibly the simplest example of the second kind is provided by the rings of matrices
$$
R_1 = \left\{ \begin{bmatrix}a&0\\0&0\end{bmatrix} : a \in \Bbb{Z}
\right\},
\qquad
R_2 = \left\{ \begin{bmatrix}a&0\\0&b\end{bmatrix} : a, b \in \Bbb{Z}
\right\},
$$
with
$$
f\left(\begin{bmatrix}a&0\\0&0\end{bmatrix}\right) = \begin{bmatrix}a&0\\0&0\end{bmatrix}.
$$
The identity of $R_1$ is
$$
e_1 = \begin{bmatrix}1&0\\0&0\end{bmatrix},
$$
that of $R_2$ is 
$$
e_2 = \begin{bmatrix}1&0\\0&1\end{bmatrix},
$$
and $f(e_1) = e_1 \ne e_2$.
The same example can also be seen via $f : \Bbb{Z} \to \Bbb{Z} \times \Bbb{Z}$ given by $f(a) = (a, 0)$, but using matrices makes it slightly more elementary.
Also, $R_1$ is a subring of $R_2$, both rings are unitary, but the identity elements are different.
