If $S$ is a subring of $R$ then $\operatorname{char}(S)\leq\operatorname{char}(R)$, when $S$ and $R$ have finite characteristic $\newcommand{\ch}{\operatorname{char}}$
a) Let $S$  be a  subring of a ring $R$ and let $\ch (S)$  and $\ch (R)$ be finite. Then $\ch (S) \le \ch (R)$. Could someone give me a hint.
b) Prove that if $S$ and $R$ have the same unity then $\ch (R)$ and $\ch (S)$ are equal. 

$\newcommand{\ch}{\operatorname{ch}}$Let us start with the second part. If a ring has a unity, then its characteristic is the additive period of the unity. So if $\mathbf{r}$ and its subring $\mathbf{s}$ have the same unity, then they have the same characteristic.
In a ring that may or may not have a unity, the characteristic is the (possibly infinite) least common multiple (lcm) of the periods of its elements. If $\ch(\mathbf{r}) = n$ is finite, then $n x = 0$ for all elements of $x \in \mathbf{r}$, and thus for all elements $x \in \mathbf{s} \subseteq \mathbf{r}$. Thus the period of all elements of $\mathbf{s}$ divides $n$, and so the lcm $m = \ch(\mathbf{s})$ of all these periods divides $n = \ch(\mathbf{r})$.
