Ring homomorphism: $0 < \mathrm{char}(f(R)) \leq \mathrm{char}(R)$ I'm learning ring theory and need help to understand the solution to the following exercise : 
Let $f : R \to R'$ be a ring homomorphism where $\mathrm{char}(R) > 0$. Show that 
$$0 < \mathrm{char}(f(R)) \leq \mathrm{char}(R).$$
Here's the solution: 
Let $\mathrm{char}(R) = n > 0$. Then, for all $r \in R$
$$n.f(r) = \underbrace{f(r) + \ldots + f(r)}_{n -\text{times}} = f(\underbrace{r + \ldots + r}_{n-\text{times}}) = f(n.r) = f(0) = 0.$$
(?) Therefore, $0 < \mathrm{char}(f(R)) \leq \mathrm{char}(R)$.
I understand that the second equality holds because $f$ is a homomorphism. The only part which I don't understand is the conclusion. 
Since we proved that $n.f(r) = 0$ for every $r \in R$, by the definition of the characteristic of a ring, I would have conclude that $\mathrm{char}(f(R)) = \mathrm{char}(R) = n$. What am I missing here? 
 A: First at all, you can't conclude that $\mathrm{char}(f(R)) = \mathrm{char}(R) = n$, because you have to show that $n$ it's the minimum with that property; now you have showed that $n\cdot f(r)=0$ for all $r\in f(R)$. You can see the conclusion as follows, let $S=\{m\in\mathbb{N} \mid m\cdot f(r)=0 \quad \forall r\in f(R) \}$, as $n\in S$ and $S\subseteq \mathbb{N}$, it follows for the well ordering principle that there exists $\mathrm{min}(S)=m$, and of course $m\leq n$, by the definition of the characteristic of a ring we have that $\mathrm{char}(f(R))=m$ (here we have $m>0$), another way it's to show that $\mathrm{char}(f(R)) \mid \mathrm{char}(R)$, your proof shows $\mathrm{char}(f(R))\neq 0$, now let $\mathrm{char}(f(R))=m$, by the division algorithm there exists unique $q,r\in \mathbb{Z}$ with $0\leq r<m$ such that $n=mq+r$, but:
\begin{align}
0&=nf(x)=(mq+r)f(x)=mqf(x)+rf(x)=qmf(x)+rf(x)=q\cdot 0+rf(x)\\
&=rf(x) \quad \forall x\in f(R)
\end{align}
Therefore, for the minimality of $m$ it follows that $r=0$, then $n=mq$, as we want to show.
