Number of non trivial ring homomorphisms from $\mathbb Z_{12}$ to $\mathbb Z_{28}$

Question

One homomorphism is $ 1 \mapsto 1$.

Other homomorphisms are:

We know that if $f$ is a homomorphism from $R$ to $S$ and $f(1_R) \neq 1_S$, then $f(1_R)$ is a zero diviasor in $S$. So zero divisor of $Z_{28}$ are 2,4,7 and also we know that if $f(1_R) = z$, then $z^2 = z$.

Hence there is only one Homomorphism

Please check my Solution and tell me right or not

Thank you for giving me your valuable time to check my Solution.

Answer 1

I'll assume that you don't require $f(1) = 1$ in the following: if you do require this condition, the ideas in this answer should help you figure out the answer in that case as well. Both underlying abelian groups are cyclic, so a ring homomorphism $\Bbb Z_{12}\to\Bbb Z_{28}$ must induce a group homomorphism $\Bbb Z_{12}\to\Bbb Z_{28}$ on the additive groups. Since the groups are cyclic, a homomorphism is completely determined by the image of $1$ (and you can verify that any such group homomorphism will extend to a ring homomorphism). Now, $1$ must be sent to an element of order dividing $12$ (why?), so this will determine the possible ring homomorphisms $\Bbb Z_{12}\to\Bbb Z_{28}$ (note that in particular $1\mapsto 1$ is not a well-defined homomorphism of rings).

Answer 2

You have $0=f(12\cdot1)=12f(1)$ and conclude that $4f(1)=28f(1)-24f(1)=0$, so that $f(1)=0,7$, $14$, or $21$. You can move further from this.