Embedding of a ring into a ring with unity

Question

I was reading the theorem on Embedding of a ring into a ring with unity which is as follows:

Let R be ring and $R\times \mathbb Z=\{(r,n)|r\in R,n\in \mathbb Z\}$. This is a ring with addition defined as $(r,n)+(s,m)=(r+s,n+m).$ and multiplication defined as :$(r,n).(s,m)=(rs+ns+mr,nm)$ .this ring has unity as $(0,1)$.

Now we can easily show that
If we define an homomorphism from $R\to R_1$ as $f(r)=(r,0)$ $\forall r\in R$ then $R \cong f(R)\subseteq R_1$. Hence $R$ is embedable in $R_1$ which has unity $(0,1)$.

I can't understand why in the ring $R\times \mathbb Z=\{(r,n)|r\in R,n\in \mathbb Z\}$ we had to define multiplication as $(r,n).(s,m)=(rs+ns+mr,nm)$, why can't we define it as $(r,n).(s,m)=(rs,nm)$ ?

Answer 1

With your multiplication, what is your proposed unity?

Answer 2

it may be helpful to look on the ring with unity as $$ eR \oplus \mathbb{Z} $$ where $e$ is an idempotent which commutes with all other elements and satisfies $e^2=e$

then every element is of the form $er +m$ and we have:

$$ (er+m)(es+n) = e^2rs+ern+mes +mn = e(rs+nr+ms) + mn $$ the unity is the element $1$

Answer 3

Legend has it that one may construct $\mathbb Z$ from $\mathbb W:=\mathbb N\cup\{0\}$ by $\mathbb N\times \mathbb N / \sim$, where $(a,b)\sim (x,y)$ is defined by a+y=b+x (meaning $b-a=y-x$ intuitively).

Next I saw similar definitions of addition and multiplication, namely

$$(a,b)(x,y) = (ay+bx, ax+by).$$

I looks a little different right? Because its background thought is

$$ (ax+by)-(ay+bx) = (b-a)(y-x).$$

So in the rIng-rng embedding case, I change the definition of multiplication by the thought (Yes, just unreasonably do it, for an idea! )

$$(m-r)(n-s)= mn - (rn+sm-rs),$$

i.e. $(r,m)(s,n):=(rn+sm-rs, mn)$. Then the mul-asso, distribution law also hold, and so does the identity and embedding property.