If $\pi:\mathbb{Z}\to \mathbb{F}$ so that $\pi(0)=0$ and $\pi(n+1)=\pi(n)+1$ why is the image a subfield? Suppose $\mathbb{F}$ is a field. Define $\pi:\mathbb{Z}\to \mathbb{F}$ so that $\pi(0)=0$ and $\pi(n+1)=\pi(n)+1$ for all $n \geq 0$. If $n<0$ then $\pi(n)=-\pi(-n)$. So $\pi$ is a homomorphism.
Suppose $\pi$ is not $1-1$. Then I have shown the smallest $n$ such that $\pi(n)=0$ is prime and $\operatorname{char}\mathbb{F}=n$
Can anyone help me prove that $\operatorname{im}(\pi)$ is a subfield of $\mathbb{F}$. It's easy to show it's an additive subgroup, but how do you show that it contains inverses for all it's elements?
Also how is this the smallest subfield and what is $|\operatorname{im}(\pi)|$ as it must be finite.
 A: Hint: Let $0 <a<n$. Then $a$ and $n$ are relatively prime, so there exist integers $x$ and $y$ such that $ax+ny=1$. Now see what the mapping does to the various numbers. You should be able to show that the object $x$ is mapped to is the inverse of $a$.  Please note that $x$ or $y$ need not be positive.
A: HINT $\ $ The image is a finite integral domain $ \Rightarrow$ field. Namely, it is finite since every element is congruent to some integer mod $\:n\:;\:$ further, being a subring of a field, it is necessarily a domain. 
Follow the above link for various simple proofs of this implication (which is best comprehended in this slightly more abstract context). Hint:  By definition, in a domain, $\rm\:c \ne 0\:$ $\Rightarrow$ $\rm\: c\:$ is cancellable, so $\rm\:x\mapsto c\:x\:$ is $1$-$1$ so onto (by finitenesss); hence $\rm\:c\:x = 1\:$ for some $\rm\:x\:,\:$ i.e.  elements $\ne 0$  are invertible. This is a prototypical algebraic application of the Pigeonhole Principle.
A: Well, all that $\pi$ seems to be doing is taking the integer $n$ to the element $n\cdot 1_\mathbb{F}=\underbrace{1_\mathbb{F}+1_\mathbb{F}+\cdots + 1_\mathbb{F}}_{n\textrm{ times}}$ (that is, if $n>0$...if $n<0$, then we're adding $-1_\mathbb{F}$ to itself, etc).  
So, if $char(F)=p>0$, then the image of $\pi$ is $\{0,1,\cdots,p-1\}$ (dropping the $\mathbb{F}$ in the subscripts), and you should have no trouble showing that this is isomorphic to $\mathbb{Z}_p$.  
A: If you have shown that $\pi$ is multiplicative, i.e. $\pi(mk)=\pi(m)\pi(k)$, then try to find, for each $m\in\mathbb{Z}$, a $k$ so that $mk=ln+1$ for some $k\in\mathbb{Z}$.  Then, by your assumption
$$
\pi(mk)=\pi(ln+1)=\pi(ln)+1=\pi(l)\pi(n)+1=1.
$$
