# The number of ring homomorphisms from $\mathbb{Z}_m$ to $\mathbb{Z}_n$

I face the problem of finding how many non-trivial ring or group homomorphisms there are from $\mathbb{Z}_m$ to $\mathbb{Z}_n$, where $m<n$. Is there any general formula?

At the moment, I want to know how many ring homomorphisms there are when $m=12,n=28$.

• A ring homomorphism requires sending the identity to the identity, so this question isn't very interesting. Do you mean module homomorphisms instead? That is a better question, in my opinion. Commented Dec 21, 2012 at 6:05
• @Potato: There are some who disagree with including that requirement in the definition of ring homomorphism (e.g. Herstein). I imagine it mainly arises because, if rings are not required to have multiplicative identities in the first place, one should make a distinction between a ring homomorphism between two rings each of which happens to have a unity, vs. a ring-with-unity homomorphism. Commented Dec 21, 2012 at 6:12
• @ZevChonoles I see. I was not aware -- I learned from Jacobson, who requires sending the identity to the identity. Commented Dec 21, 2012 at 6:14
• For groups, see math.stackexchange.com/questions/273169 Commented Jan 19, 2017 at 20:53

A ring homomorphism $f:\mathbb Z_m\to \mathbb Z_n$ is uniquely determined by the conditions: $mf(1)=0$ and $f(1)^2=f(1)$.
In order to find out how many ring homomorphisms there are we have to count the number of elements of the set $\{e\in\mathbb Z_n:e^2=e,me=0\}$. It is not hard to see that this equals $2^{\omega(n)-\omega(n/(m,n))}$, where $\omega(a)$ is the number of distinct prime factors of $a$. (See Gallian and Van Buskirk, The Number of Homomorphisms from $\mathbb Z_m$ to $\mathbb Z_n$, AMM, 91(1984), 196-197.)

For particular cases we don't need the above result. For instance, if $m=12$ and $n=28$ we get $0=me=12e$ in $\Bbb Z_{28}$ iff $28\mid 12e$ iff $\,7\mid e,\,$ so $f(1)\in\{0,7,14,21\}$ and only $0$ and $21$ are idempotent in $\Bbb Z_{28},$ so there are two ring homomorphisms from $\mathbb Z_{12}$ to $\mathbb Z_{28}$.

Edit. If one asks $f(1)=1$ then the problem becomes trivial: there is a unitary ring homomorphism iff $n\mid m$.

Hint: regardless of whether you're considering ring or group homomorphisms, where you send $1$ determines where you must send everything else, because for any $n$, $$f(n)=f(\underbrace{1+\cdots+1}_{n\text{ times}})=\underbrace{f(1)+\cdots+f(1)}_{n\text{ times}}$$

But where can you send $1$? Remember, the resulting function must be a homomorphism: $$f(a+b)=f(a)+f(b)\qquad f(ab)=f(a)f(b)$$ Try to figure out what conditions this imposes on your choice of $f(1)$. See user26857's answer if you are stuck.

Note that the answer will depend on whether you require that a ring homomorphism $f:R\to S$ must preserve multiplicative identities, i.e. $f(1_R)=1_S$.

• @Kuttus: How many generating elements are there in $\mathbb Z_{12}$? just one element? Commented Dec 21, 2012 at 6:05
• @Kuttus: Even if you don't require ring homomorphisms to preserve multiplicative identities, you cannot send 1 to any element of $Z_{28}$. Homomorphisms send 0 to 0, and must preserve addition... Commented Dec 21, 2012 at 6:05
• @Kuttus: there is at most one ring homomorphism. Is the actual number one or zero? Commented Dec 21, 2012 at 8:02
• @user26857: The idea of a hint is to help the OP figure an answer out for themselves. Additionally (and as I pointed out in my answer), it's quite common to require $f(1)=1$ for a ring homomorphism, and under that definition your answer is incorrect. That seems like an important thing to have missed. Commented Dec 7, 2015 at 17:44
• I'd have taken this as a good reason, but unfortunately the whole discussion into the comments shows that the OP has no idea what's going on and you didn't help him clarify things. Commented Dec 7, 2015 at 17:56