For which pairs of natural numbers m and n does there exist a ring homomorphism $\Bbb Z/n\Bbb Z$→ $\Bbb Z/m\Bbb Z$?

Question

(a) Determine the set of units $(\Bbb Z/n\Bbb Z)^\times$ for $n = 4, 9, 15, 16$.

(b) For which pairs of natural numbers $m$ and $n$ does there exist a ring homomorphism $\Bbb Z/n\Bbb Z$→ $\Bbb Z/m\Bbb Z$? Where $\Bbb Z$ = set of integers.

My attempts:

a) I know that $a$ in $\Bbb Z$ is a unit if there exists $b$ in $\Bbb Z$ such that $a.b = b.a = 1$ where $b$ is the inverse of $a$.

So for $n=4$, $(\Bbb Z/4\Bbb Z)=\{0,1,2,3\}$ and $(\Bbb Z/4\Bbb Z)^\times = \{1,3\}$. And for $n=9$, $(\Bbb Z/9\Bbb Z)=\{0,1,2,3,4,5,6,7,8\}$ and $(Z/9Z)^\times = \{1, 2, 4, 5, 7, 8\}$. I'll do the same procedure for $n= 15$ and $n=16$.

Concerning part b), we didn't talk about it in the lecture, but I know that $\Bbb Z/n\Bbb Z$ →$\Bbb Z/m\Bbb Z$ is a ring homomorphism iff $m\mid n$ and so the pairs for $n=4$ are $(m, n)=(4,4), (4,8),(4,16),(4,20)$, etc... And for $n=9$ the pairs are $(m, n)=(9,9),(9,18),(9,27),(9,36)$, etc... Is part b) correct? If not can you show me what is its answer? Thank you.

Answer 1

As I said in my comments, you swapped $m$ and $n$ in your answer for part b). To have a homomorphism from $\mathbf Z/n\mathbf Z\to\mathbf Z/m\mathbf Z$ the condition is that $m$ divides $n$. Thus

• for $n=4$, you have the pairs $(m,n)=(1,4), (2,4)$ and $(4,4)$
• for $n=9$, similarly, you obtain the pairs $(1,9), (3,9)$ and $(9,9)$.

Note that, in the formulation of the question, there was some sort of trap: the modulus of the target of the morphism is enunciated before the modulus of its source.

Answer 2

Hint: Consider the kernel of such a homomorphism. It must be an additive subgroup (ideal of course) of the integers modulo $n$. The order of such a subgroup must be a divisor of $n$.