If $ f: \mathbb Z_{m} \to \mathbb Z_{n} $ is a ring homomorphism, then $f(0) = 0, mf(1) = 0, f(1)^2 = f(1)$. This is easy to show.

But, is the converse true?

Edit-1 :

In the below comments it is shown that the answer is a clear no. i.e, the converse does not hold. But, my doubt is, what is the justification for the answer given by user26857 in the following post to use the above identities to find the number of homomorphisms from $\mathbb Z_{m}$ to $ \mathbb Z_{n}$: The number of ring homomorphisms from $\mathbb{Z}_m$ to $\mathbb{Z}_n$ ?

  • 2
    $\begingroup$ Are you asking if a map of the underlying sets which satisfies those properties is necessarily a homomorphism? If so, then the answer is definitely no. Consider any function $\mathbb{Z_m}\rightarrow \mathbb{Z_m}$ which sends $0$ to $0$ and $1$ to $1$. Most of them aren't homomorphisms $\endgroup$ – Badam Baplan Sep 27 '18 at 6:25
  • $\begingroup$ For example, even a bijection on $\mathbb{Z_4}$ defined by $f(0) = 0, f(1) = 1, f(2) = 3, f(3) = 2$ is not a homomorphism. $\endgroup$ – Badam Baplan Sep 27 '18 at 6:28
  • $\begingroup$ @BadamBaplan Yes, I get it now. Thanks. But, can you answer the above-edited question? $\endgroup$ – Suhan Shetty Sep 27 '18 at 7:06

Expanding a bit on the comment of Badam Baplan:

Let $f:\mathbb{Z}_m\rightarrow \mathbb{Z}_n$ be a group homomorphism. Since $1$ generates the group $\mathbb{Z}_m$, the map $f$ is completely determined by $f(1)$. Now notice that $mf(1)=0$, hence the order $\text{ord}(f(1))$ of $f(1)$ divides $m$. On the other hand $nf(1)=0$ since $n$ is the order of $\mathbb{Z}_n$, it follows that the order $\text{ord}(f(1))$ divides $n$ as well. It follows that $\text{ord}(f(1))\mid \gcd(m,n)$.

Hence if $m,n$ are coprime, it follows that $\text{ord}(f(1))=1$ which in turn yields $f(1)=0$. It follows that $f=0$. Hence in general there are not to many group morphisms between $\mathbb{Z}_m$ and $\mathbb{Z}_n$, let alone ring morphisms.

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