Question:
Find the number of non trivial ring homomorphisms from $\mathbb Z_{12}$ to $\mathbb Z_{28}$. ($f$ is not necessarily unitary, i.e., $f(1)$ need not be $1$.)
Suppose $f$ is a ring homomorphism from $\mathbb Z_{12}$ to $\mathbb Z_{28}$.
Consider $f$ as a additive group homomorphism. Let $k= |\ker f|$ and
$ t = |\operatorname{im}(f)|$. Then $k\mid 12$ and $t\mid 24$ and $kt=12$, by first isomorphism
theorem of groups.
There are two possibilities $k=3$, $t=4$ and $k=6$, $t=2$.
For the first case $f$ should map $1$ to an element of the subgroup generated by $7$ as there is a unique subgroup of $\mathbb Z_{28}$ of order $4$ generated by $7$. For the second case $1$ has to map to $14$, for the same reasoning.
So there are at most two ring homomorphisms from $\mathbb Z_{12}$ to $\mathbb Z_{28}$.
Question is how to check the possible maps which are ring homomorphisms.
Thanks.
Best Answer
Number of ring homomorphism from $\mathbb Z_m$ into $\mathbb Z_n$ is $2^{[w(n)-w(n/\gcd(m,n))]}$ , where $w(n)$ denotes the numbers of prime divisors of positive integer n. From this formula we get number of ring homomorphism from $\mathbb Z_{12}$ to $\mathbb Z_{28}$ is $2$.