# 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.

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This question pops up here almost every week or so ... what about using the search function. – Martin Brandenburg May 25 '13 at 12:54

## 3 Answers

1. $R = \mathbb Z$ and $S = 2\mathbb Z$
2. $R_1 = R_2 = \mathbb Z$ 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.

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1. 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.

2. 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.

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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.

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