Prove that isomorphic rings have the same characteristic If $\phi: A \to B$ is a ring isomorphism, I have to proof that $\operatorname{char} A= \operatorname{char} B$.
I know that  an isomorphism $\phi$ is bijective and:
$$\phi(x+y)=\phi(x)+ \phi(y)$$
$$\phi(x*y)=\phi(x)*\phi(y)$$
$$\phi(1)=1$$
I have supposed that $\operatorname{char} A=n$, so: $n1=\underset{n\text{ times}}{\underbrace{1+1+\ldots+1}}=0$ in $A$.
As $\phi$ is surjective, $n1\in A$ exists, where $x=\phi(n1)$.
Now, $\phi(n1)=\phi(0)=\phi(1+1+\ldots+1)=\phi(1)+\ldots+\phi(1)=n*\phi(1)=n$.
I don't know if this is correct and if this is how to finish the proof. Could you help me please?
 A: Part 1: Take $1_B$ and show that if you add it up $n$ times you get $0_B$.
Part 2: Show that if you add up $1_B$ a total of $m$ times for $m < n$ then you don't get $0_B$.
For the first part, the proof could be something like:
Observe $1_B = \phi(1_A)$. Then 
$$1_B + \cdots + 1_B = \phi(1_A) + \cdots + \phi(1_A) = \phi(1_A + \cdots + 1_A) = \phi(0_A) = 0_B$$
where each sum has $n$ terms. Can you justify each of the equalities above?
A: Since $R\cong S$, there exists a ring isomorhism $\theta:R\longrightarrow S$.
Assume that $\operatorname{Char}(R)=n>0$. Let $b\in S$. Since $\theta$ is surjective, there exists some $a\in R$ such that $b=\theta(a)$. Then, $nb=n\theta(a)=\theta(na)=\theta(0_R)=0_S$. So, $0<\operatorname{Char}(S)\leq n$.
Suppose now for contradiction that $\operatorname{Char}(S)=m$ with $0<m<n$. Then, for all $a\in R$ observe that
$$m\theta(a)=0_S\iff m\theta(a)=0_S \iff \theta(ma)=0_S \iff \theta(ma)=\theta (0_R) \iff ma=0_R,$$
since $\theta$ is injective. This is a contradiction by the minimality of $n$. Hence, $m\geq n$. So, $\operatorname{Char}(S)\geq n$ and since we saw that $\operatorname{Char}(S)\leq n$ we eventually have $\operatorname{Char}(S)= n$.
Assume now that $\operatorname{Char}(R)=0$. If it was $\operatorname{Char}(S)=n>0$, then like before for all $a\in R$ we would have
$n\theta(a)=0_S$ iff $na=0_R$. Thus, $\operatorname{Char}(R)>0$, contradiction. So $\operatorname{Char}(S)=0$.

Note: This is a nice way to show that two rings are not isomorphic. For instance, $\mathbf{Z}_4$ and $\mathbf{Z}_2\times \mathbf{Z}_2$ are not isomorphic since $4=\operatorname{Char}(\mathbf{Z}_4)\neq \operatorname{Char}(\mathbf{Z}_2\times \mathbf{Z}_2)=2$.
Hope that helps.
A: The characteristic of a ring $\mathcal R $  is the nonnegative integer  $n $ such that  $\rm {ker}(h) =(n)  $, where $h $ is the unique homomorphism $\mathbb Z\longrightarrow\mathcal R $.
Then if $j:\mathbb Z\longrightarrow\mathcal S $ is the corresponding homomorphism for $S $, where $\mathcal R\stackrel i{\cong }\mathcal S $, we have a couple commutative triangles, and easily get $(n)\subset (s) $ and $ (s)\subset  (n) $ , so that $(n)=(s) $, where $s=\rm {char}\mathcal S $.
This follows from the universal property of initial objects, since $\mathbb Z $ is an initial object in the category of rings, $\bf {Ring} $.
