# 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 $13\mathbb{Z}$?

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Are $Z$ and $C$ the integers and complex numbers? – Thomas Andrews Mar 29 '12 at 14:02
When asking questions about rings, it's very important to be clear about conventions. Does your definition of ring include a multiplicative identity? In particular, that means $f(1) = 1$ for a ring homomorphism, which is a very relevant constraint for the problem.... – Hurkyl Mar 29 '12 at 14:37
@Hurkyl: you can require that your rings have identities without requiring that your homomorphisms preserve them, although I suppose this would be a somewhat unusual convention. – Qiaochu Yuan Mar 29 '12 at 15:15
I'm not 100% sure about the following. So I will leave it as a comment. Proof by contradiction. Assume ${\rm ker}(f) = 12{\mathbb Z},$ then by the first isomorphism theorem, $\mathbb{Z}/{\rm ker}(f) \cong {\rm im}(f),$ where ${\rm im}(f)$ is a subring of $\mathbb C.$ But note that (1) ${\rm im}(f) = \mathbb{Z}_{12}$ is not an integral domain (2) Every subring of $\mathbb C$ is an integral domain. Hence ${\rm im}(f)$ can not be a subring of $\mathbb C.$ Contradiction. – user2468 Mar 29 '12 at 21:35
@J.D. : I'm happy to confirm that your argument is 100% correct. – Georges Elencwajg Mar 31 '12 at 20:36

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 $\ker(f)\ne 12\mathbb Z$.

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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

$$g(n)=e^{ni\pi\sqrt{2}}.$$

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I don't think you meant "f is surjective" in that second sentence. – Hurkyl Mar 29 '12 at 14:35
@Hurkyl, no I did not. Thank you. – JSchlather Mar 29 '12 at 14:42
Depending on how you define things, this could be circular. There is a unique ring homomorphism from $\mathbb{Z}$ to $\mathbb{C}$ (I include the condition that $1\mapsto 1$ for a ring homomorphism). The phrase "$\mathbb{C}$ has characteristic zero," by definition, means that this unique ring homomorphism is injective. So you can't really use the fact that $\mathbb{C}$ has characteristic zero to prove that the homomorphism is injective. – Keenan Kidwell Mar 31 '12 at 19:25
It's not true that $\rm\:n\!\:r = 0\:\Rightarrow\: n=0\ or\ r=0\:$ in a ring of characteristic $0$, e.g. consider $\rm\mathbb Z[x]/(n\!\:x).\qquad$ – Bill Dubuque Mar 31 '12 at 19:34
@KeenanKidwell, well I'm looking at a rng homomorphism so it preserves addition and multiplication, but not necessarily identities. Anyway I slightly modified my argument to include that C is a domain, which I think is sufficient for my conclusion. – JSchlather Apr 1 '12 at 0:23

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.

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