Exercise A0.III.6.17

Proposition (H1)

Let $M$ be a [[Exercise A0.III.6.16#definition-cyclic-module|cyclic]] $R$-module, with $M \cong R/I$ for an ideal $I$ (6.16.C2), and let $N$ be another $R$-module. Let $Q$ be a submodule of $N$ given by

$$Q := \{n \in N \mid (\forall a \in I): an = 0\}.$$

Then

$$\Hom_{\RMod}(M,N) \cong Q.$$

Proof

Let $\pi : R \to R/I$ be the quotient map and let $a \in I.$ If $\varphi: M \to N$ is a homomorphism, then

$$a \varphi(\pi(1)) = \varphi(a\pi(1)) = \varphi(\pi(a)) = 0.$$

Now, consider the map $v : \Hom_{\RMod}(M,N) \to Q$ given by $\psi \mapsto \psi(\pi(1)).$ It is immediate that it is a module homomorphism. It is injective, since each element of $\Hom_{\RMod}(M,N)$ is determined uniquely by its value on $\pi(1)$ due to $M$ being cyclic.

It is surjective. By the universal property of a quotient, whenever a map $\tau : R \to N$ sends $I$ to $0$, it must factor uniquely through the quotient $M,$ giving a homomorphism $\varphi_n : M \to N, \pi(1) \mapsto n$ for every value $n \in N$ such that for all $a \in I,$ $an = 0.$

So $v$ is an isomorphism.

$\blacksquare$

NOTE

The same proof shape works without assuming commutativity of $R,$ giving an isomorphism of abelian groups $\Hom_{\RMod}(M,N) \cong Q.$

Proposition (H2)

For all $a,b \in \mathbb{Z}_{>0}$ we have $\Hom_\mathbf{Ab}(\mathbb{Z}/a\mathbb{Z}, \mathbb{Z}/b\mathbb{Z}) \cong \mathbb{Z}/\gcd(a,b)\mathbb{Z}.$

Proof

DANGER

This proof is trash but I don’t know any better way. Trying to use H1 leads to even more number chasing.

A homomorphism $f : \mathbb{Z}/a\mathbb{Z} \to \mathbb{Z}/b\mathbb{Z}$ is defined by its action on $[1]_a.$

Suppose the order of $[k]_b$ divides $a$ and $f([1]_a) = [k]_b$. Then the map $q : \mathbb{Z} \to \mathbb{Z}/b\mathbb{Z}$ given by $q(1) = [k]_b$ (well-defined by the universal property of $\mathbb{Z}$) has the property $q(an) = [ank]_b = [0]_b.$ So $a\mathbb{Z} \subseteq \ker(q)$ and hence it factors uniquely (via the universal property of quotients) through $\mathbb{Z}/a\mathbb{Z}.$

This shows that $\Hom_\mathbf{Ab}(\mathbb{Z}/a\mathbb{Z}, \mathbb{Z}/b\mathbb{Z})$ contains all homomorphisms sending $[1]_a$ to an element of order dividing $a,$ and only those homomorphisms (since for any homomorphism $g$, $|g(x)|$ divides $|x|$).

Recall that every subgroup of $\mathbb{Z}/b\mathbb{Z}$ is cyclic and has order that is a factor of $b.$ Recall also that for two subgroups $A, B \leq \mathbb{Z}/b\mathbb{Z}$ the subgroup $A+B$ is the smallest subgroup containing both $A$ and $B.$

Now take the sum of all subgroups of $\mathbb{Z}/b\mathbb{Z}$ whose order divides $a.$ The order of this sum would be the product of the multiset of prime factors of $b$ contained within the multiset of prime factors of $a.$ But this is, by definition, $\gcd(a,b).$

So there are exactly $\gcd(a,b)$ suitable elements of $\mathbb{Z}/b\mathbb{Z}.$ Let $[m]_b \in \mathbb{Z}/b\mathbb{Z}$ be an element of order $\gcd(a,b)$ (in $\mathbf{Ab}$, every factor of a finite group’s order corresponds to a subgroup). By definition, $[m]_b$ generates the cyclic group $\mathbb{Z}/\gcd(a,b)\mathbb{Z}.$

Take $f \in \Hom_\mathbf{Ab}(\mathbb{Z}/a\mathbb{Z}, \mathbb{Z}/b\mathbb{Z})$ defined by $f([1]_a) = [m]_b.$ Note that $nf$ sends $[1]_a$ to $n[m]_b.$ Take any element $[u]_b$ of $\mathbb{Z}/\gcd(a,b)\mathbb{Z}.$ Since $[m]_b$ generates $\mathbb{Z}/\gcd(a,b)\mathbb{Z},$ $[u]_b = k[m]_b$ for some $k \in \mathbb{Z}.$ So $kf$ gives the unique homomorphism sending $[1]_a$ to $[u]_b.$ Thus, $f$ generates the whole group.

Any two cyclic group of identical order are isomorphic.

$\blacksquare$