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?
Best Answer
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?