Ring homomorphisms $\mathbb{Z}\to\mathbb{C}$ Let $f\colon\mathbb{Z}\to\mathbb{C}$  be a homomorphism of rings? Prove that the kernel of $f$ can not be equal to $12\mathbb{Z}$.
I also  wondering if the kernel can be equal to $13\mathbb{Z}$?  
 A: Edit: I should suppose I mention that I use the notion of a rng homomorphism in the following argument, because the question is trivial if $f(1)=1$.
Let $f: \mathbb Z \rightarrow \mathbb C$ be a ring homomorphism. Then $\ker f = \mathbb Z$ or $ \{0\}$, that is $f$ is injective or trivial. This follows from observing that since $\mathbb C$ has characteristic $0$ and is a domain so we know that $f(n)=n*f(1)$
is equal to zero if and only if $n$ or $f(1)$ is zero. In particular if $f(1)$ is not zero, then for any nonzero $n$ we have $f(n)$ is nonzero. If $f(1)=0$ then $\ker f= \mathbb Z$.
Now what about group homomorphisms from $\mathbb Z$ to $\mathbb C$. Well if we map $\mathbb Z$ to the additive group of $\mathbb C$ then we have the same result. But if we map $\mathbb Z$ into $\mathbb C^\times=\mathbb C \setminus \{0\}$ things are a little bit more interesting. For every $n \in \mathbb Z$ we can find a homomorphism $f: \mathbb Z \rightarrow \mathbb C^\times$ such that $\ker f=n\mathbb Z$. We can do this by sending $1$ to a primitive $n^{th}$ root of unity. For instance the mapping $f_n: \mathbb Z \rightarrow \mathbb C$ given by
$$f_n(k)=e^{2\pi ik/n}$$
is a mapping with kernel $n\mathbb Z$. We can also find an injective map into $\mathbb C^\times$ by picking an irrational number, say $\sqrt{2}$ and creating the mapping $g: \mathbb Z \rightarrow \mathbb C$ as
$$g(n)=e^{ni\pi\sqrt{2}}.$$
A: I assume $\mathbb Z$ and $\mathbb C$ denote the integers and complex numbers respectively. If $f:\mathbb Z\to \mathbb C$ is a ring homomorphism then $\mathbb Z/\ker(f)$ is isomorphic to a subring of $\mathbb C$ since $\mathbb C$ is a field  $\mathbb Z/\ker(f)$ is an integral domain. Thus $\ker(f)\ne 12\mathbb Z$.
A: If we can write an integer $n$ as $n=pq$ where $p,q>1$ then we can't find a homomorphism $f$ between $\mathbb Z$ and $\mathbb C$ such that $\ker f=pq\mathbb Z$. Indeed, $f(p)\neq 0$ and $f(q)\neq 0$ so we should have $f(n)=f(pq)\neq 0$ ($\mathbb C$is a field) which is a contradiction. 
