Question about extension theorems in ring theory I know there is a theorem which ensures that every ring $R$ can be embedded into a ring with unity $S$. Moreover, you can always choose S in a way that $\operatorname{char}(S)=0$, or, if $\operatorname{char}(R)=n$ then you can also construct $S$ in order to $\operatorname{char}(S)=\operatorname{char}(R)=n$.
On the other hand, there are examples where a ring $R$ with unity $1_R$ can have a subring $S$ with unity $1_S$ such that $1_S \neq 1_R$.
One example is in $\mathbb{Z}_{18}$, the subring $\langle \bar2 \rangle$ has unity $\overline{10}$.
This sort of cases make me wonder if it is possible to somehow choose, maybe with some restrictions, the characteristic of the bigger ring $S$ which contains an isomorphic copy of the original one $R$.
 A: You can always extend $R$ with $\mathbb Z$, because $R$ is always a $\mathbb Z$ module. So every rng (with or without identity) can be embedded in a characteristic $0$ ring.
You can extend $R$ with $\mathbb Z_n$ as long as $R$ is a $\mathbb Z_n$ module, otherwise things aren't compatible. Being a $\mathbb Z_n$ module entails that $nr=0$ for every $r\in R$, so the characteristic of $R$ would have to divide $n$. So if $R$ has finite characteristic $k$, you can extend it by $\mathbb Z_n$ for any $n$ divisible by $k$.
Notice that you can't get from $R=(2)\subset \mathbb Z_{18}$ back to $\mathbb Z_{18}$ using an extension like this. $R$ has 9 elements, so you'd need to extend by $\mathbb Z_2$, but that isn't possible: if you adjoined an identity with additive order $2$, then $0=2(2)=4\in R$ is a contradiction. (Remember, $R$ has to be a $\mathbb Z_n$ module for some $n$ divisible by $9$ to avoid this.)
But, of course, just taking the ordinary direct product with $\mathbb Z_2$, you get a new ring with identity since $R$ had an identity, and the identity has order $18$. Clearly this doesn't help if $R$ doesn't have an identity.
So, using ordinary ring products, you can take any finite ring of order $n$ with identity and embed it into a ring with characteristic $gcd(n, k)$ for whatever $\mathbb Z_k$ you would like. That's a little bit of freedom choosing the characteristic, I suppose...
A: Characteristic $0$ means that there exists no positive integer $k$ such that $kx=0$ for every $x\in R$. Obviously a ring extension of $R$ must have characteristic $0$.
Suppose conversely that the characteristic of $R$ is $n>0$. This means that $n$ is the minimum positive integer such that $nx=0$ for all $x\in R$ or, which is the same, it is the unique positive generator of the ideal of $\mathbb{Z}$ consisting of all integers $k$ such that $kR=\{0\}$, let's call it $c(R)$.
Suppose $S$ is a ring extension of $S$ having characteristic $m>0$. Then, easily, $m\mathbb{Z}=c(S)\subseteq c(R)=n\mathbb{Z}$, which means that $m$ is a multiple of $n$.
Now take $R$ of characteristic $n>0$ and $m$ any multiple of $n$. Then
$$
S=R\times(\mathbb{Z}/n\mathbb{Z})
$$
can be endowed with a ring structure. Addition is the obvious one:
$$
(x,\bar{a})+(y,\bar{b})=(x+y,\bar{a}+\bar{b})
$$
Multiplication is defined by
$$
(x,\bar{a})(y,\bar{b})=
(xy+\bar{a}y+\bar{b}x,\bar{a}\bar{b})=
(xy+ay+bx,\bar{a}\bar{b})
$$
and it's just a matter of verifying that the ring axioms are satisfied. Note that the fact that $m$ is a multiple of $n$ ensures $\bar{a}x=ax$, for $\bar{a}\in\mathbb{Z}/m\mathbb{Z}$ and $x\in R$ is well defined.
Clearly, $(0,\bar{1})$ is the identity element in $S$ which has characteristic $m$ because it contains a subring isomorphic to $\mathbb{Z}/m\mathbb{Z}$.
