Can every ring be embedded in a ring with identity of arbitrary characteristics?

Question

There is a theorem which says that every ring $R$ may be embedded in a $S$ ring with identity. In Hungerford's "Abstract Algebra" it says that the ring $S$ may be chosen to be either of characteristics zero or to have the same characteristics as $R$. I don't understand the second part of the theorem. Why is it so?

Proof of the first part is straightforward. One can show that $S=R\times \mathbb{Z}$ is a ring with addition and multiplication defined as follows: $$(r,n)+(s,m)=(r+s,n+m)$$ $$(r,n)\cdot(s,m)=(rs+mr+ns,nm)$$ The identity is the element $(0,1)$. It is obvious that the characteristics of this ring is zero. Similarly, we can define $S=R\times \mathbb{Z}_k$ which is also a ring with respect to $$(r,\bar{n})+(s,\bar{m})=(r+s,\bar{n}+\bar{m})$$ $$(r,\bar{n})\cdot(s,\bar{m})=(rs+mr+ns,\bar{n}\bar{m}),$$ where $\bar{n}$ and $\bar{m}$ denote images of $n$ and $m$ under the canonical map and operations on second coordinate are both modulo $k$. Now $S$ is a ring with unity $(0,1)$ but of characteristics $k$. It seems to me that the characteristics of the ring $S$ in no way depends on the characteristics of $R$ and can be an arbitrary $k\in\mathbb{N}_0$, not just $0$ or $char R$.

Answer 1

Note: I changed your $n$ in $\mathbb{Z}_n$ to $k$ (i.e. $\mathbb{Z}_k$) to avoid two conflicting uses of "$n$".

It does seem like changing from $\mathbb{Z}$ to $\mathbb{Z}_k$ doesn't effect the construction at all. But you're overlooking a subtle problem in the definition of multiplication... $$(r,n)\cdot(s,m)=(rs+mr+ns,nm)$$

When you use $\mathbb{Z}$, $mr$ and $ns$ are just $r$ and $s$ raised to additive exponents $m$ and $n$ (which is perfectly fine). But if $m,n \in \mathbb{Z}_k$, then $m$ and $n$ aren't integers anymore. They are representatives of equivalence classes mod $k$. So we need $mr$ and $ns$ to be uneffected if $m,n$ are swapped out for equivalent representatives. In other words, you need $mr=(m+k\ell)r$ for all $\ell \in \mathbb{Z}$ and $ns=(n+k\ell)s$ for all $\ell \in \mathbb{Z}$.

This means that you need $k\ell s =0$ and $k\ell r=0$ for all $\ell \in \mathbb{Z}$. In particular, you need $ks=0$ and $kr=0$. So unless the additive order of every element in $R$ is divides $k$, the proposed "definition" of multiplication isn't really well defined.

This means that for that multiplication to be well defined, you need the characteristic of $R$ to divide the proposed $k$ (new characteristic).