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${Hom}_{Z} (Z / m, Z / n) ≅ Z / \gcd (m, n)$ for $n \geq 1$

Example

Let $m \in N$ and let $n \geq 1$ . Then, as $Z$ -modules,

${Hom}_{Z} (Z / m, Z / n) ≅ Z / \gcd (m, n) .$

The hypothesis $n \geq 1$ cannot be dropped. At $n = 0$ the module $Z / 0$ is a copy of $Z$ , and for $m \geq 2$ the left-hand side is the zero module while $\gcd (m, 0) = m$ makes the right-hand side $Z / m$ , which is not zero.

Both modules here are finitely generated over $Z$ , so Over a Noetherian ring the homomorphism module between two finitely generated modules is finitely generated predicts that the homomorphism module is finitely generated; the computation below identifies it outright.

Facts & Assumptions

Given: Natural numbers $m$ and $n$ with $n \geq 1$ , and the classes $[a]_{m} \in Z / m$ and $[b]_{n} \in Z / n$ .

[L1]

The integers modulo $n$ are the congruence classes $[a]_{n} = {b \in Z : b \equiv a (m o d n)}$ , and $[a]_{n} = [b]_{n}$ exactly when $a \equiv b (m o d n)$ ; at $n = 0$ each class is a singleton (The congruence class $[a]_{n}$ and the quotient set $Z / n$ ).

[L2]

Addition and multiplication of congruence classes are given by $[a]_{n} + [b]_{n} = [a + b]_{n}$ and $[a]_{n} [b]_{n} = [a b]_{n}$ (Addition and multiplication on $Z / n$ by $[a]_{n} + [b]_{n} = [a + b]_{n}$ and $[a]_{n} [b]_{n} = [a b]_{n}$ ).

[L3]

For every $n \in N$ , $(Z / n, +, [0]_{n})$ is an abelian group with $- [a]_{n} = [- a]_{n}$ (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold).

[L4]

Every abelian group carries a unique $Z$ -module structure whose scalar action is integer multiplication, and abelian groups and $Z$ -modules have the same objects and morphisms (Abelian groups and $Z$ -modules have the same objects and morphisms).

[L5]

A function $f ⁣ : M \to N$ between $R$ -modules is an $R$ -module homomorphism when $f (m + m^{'}) = f (m) + f (m^{'})$ and $f (r m) = r f (m)$ (Module homomorphism and isomorphism, kernel, image and cokernel).

[L6]

${Hom}_{R} (M, N)$ is an abelian group under pointwise addition (The abelian group ${Hom}_{R} (M, N)$ and maps induced by pre- and postcomposition).

[L7]

Over a commutative ring $R$ the group ${Hom}_{R} (M, N)$ is an $R$ -module under $(r f) (x) = r f (x)$ , with the published addition unchanged (Over a commutative ring the homomorphism group ${Hom}_{R} (M, N)$ is an $R$ -module).

[L8]

For $a, b \in Z$ , $a Z + b Z = \gcd (a, b) Z$ ( $a Z + b Z = \gcd (a, b) Z$ and $a Z \cap b Z = lcm (a, b) Z$ ; equivalently, in $(Z, +)$ the subgroup generated by ${a, b}$ is $⟨ \gcd (a, b) ⟩$ and $⟨ a ⟩ \cap ⟨ b ⟩ = ⟨ lcm (a, b) ⟩$ ).

[L9]

$\gcd (a, b)$ is the greatest common divisor of $a$ and $b$ , with $\gcd (0, 0) : = 0$ ; it satisfies $\gcd (a, b) \geq 0$ always, and $\gcd (a, b) \geq 1$ unless $a = b = 0$ (Common divisor, and the greatest common divisor $\gcd (a, b)$ , with the convention $\gcd (0, 0) : = 0$ ).

[L10]

Over a Noetherian commutative ring the homomorphism module between two finitely generated modules is finitely generated (Over a Noetherian ring the homomorphism module between two finitely generated modules is finitely generated).

[L11]

$Z$ is a Noetherian ring (Fields and $Z$ are Noetherian, and so are their polynomial rings in finitely many variables).

Verification

technique · direct

1.1L1L2L3L4L5L6L7L9given

Write $d : = \gcd (m, n)$ , so $d \geq 1$ because $n \geq 1$ , and $d$ divides both $m$ and $n$ . Both $Z / m$ and $Z / n$ are abelian groups and hence $Z$ -modules with the integer-multiplication action, and a $Z$ -module homomorphism between them is exactly an additive map; ${Hom}_{Z} (Z / m, Z / n)$ is a $Z$ -module under pointwise addition and the integer action. Each $Z / m$ is generated by $[1]_{m}$ , since $[a]_{m} = a [1]_{m}$ .

2.1L1L2L5L6L7step 1.1

Evaluation at $[1]_{m}$ is a $Z$ -module isomorphism from ${Hom}_{Z} (Z / m, Z / n)$ onto $A : = {y \in Z / n : m y = 0}$ . It lands in $A$ , because $m [1]_{m} = [m]_{m} = [0]_{m}$ gives $m f ([1]_{m}) = f ([0]_{m}) = 0$ . It is injective, because $f ([a]_{m}) = a f ([1]_{m})$ determines $f$ . It is surjective: given $y \in A$ , the assignment $[a]_{m} \mapsto a y$ is well defined, since $[a]_{m} = [a^{'}]_{m}$ means $m$ divides $a - a^{'}$ and then $(a - a^{'}) y$ is a multiple of $m y = 0$ , and it is additive. Additivity and $Z$ -homogeneity of the evaluation map are immediate from the pointwise operations.

2.2L8L9step 1.1

${b \in Z : n divides m b} = (n / d) Z$ . For the inclusion from right to left, $b = (n / d) t$ gives $m b = (m / d) n t$ , a multiple of $n$ , using that $d$ divides $m$ . For the other inclusion, Bézout in the form $m Z + n Z = d Z$ supplies integers $u, v$ with $d = u m + v n$ ; if $n$ divides $m b$ then $n$ divides $b d = u (m b) + v (b n)$ , say $b d = n s$ , and dividing by $d \geq 1$ gives $b = (n / d) s$ .

3.1L1L2L3step 2.1step 2.2

$A$ is the cyclic submodule generated by $[n / d]_{n}$ , and $A ≅ Z / d$ . Indeed $y = [b]_{n}$ lies in $A$ exactly when $[m b]_{n} = [0]_{n}$ , that is when $n$ divides $m b$ , which by step 2.2 says $b \in (n / d) Z$ ; so $A = {[(n / d) t]_{n} : t \in Z}$ . The map $Z / d \to Z / n$ sending $[c]_{d}$ to $[c (n / d)]_{n}$ is well defined, since $d$ dividing $c - c^{'}$ makes $n = (n / d) d$ divide $(c - c^{'}) (n / d)$ ; it is additive; its image is $A$ ; and it is injective, since $[c (n / d)]_{n} = [0]_{n}$ means $c (n / d) = n t = (n / d) d t$ for some $t$ , whence $c = d t$ because $n / d \geq 1$ is nonzero, so $[c]_{d} = [0]_{d}$ .

4.1L1L9L10L11step 2.1step 3.1∎

Combining steps 2.1 and 3.1 gives ${Hom}_{Z} (Z / m, Z / n) ≅ Z / d$ with $d = \gcd (m, n)$ . This is consistent with the general finiteness statement: $Z$ is Noetherian and $Z / m$ , $Z / n$ are generated by one element each, so the homomorphism module had to be finitely generated, and here it is cyclic. The hypothesis $n \geq 1$ is used in step 1.1 to make $d \geq 1$ and in steps 2.2 and 3.1 to divide by $n / d$ ; at $n = 0$ the conclusion is false for $m \geq 2$ , since an additive $f ⁣ : Z / m \to Z / 0$ has $m f ([1]_{m}) = 0$ in a copy of $Z$ , forcing $f ([1]_{m}) = 0$ and $f = 0$ , while $\gcd (m, 0) = m$ makes the claimed answer $Z / m$ , which has more than one element.

Remarks

The two boundary values behave differently, and only one of them is admitted. At $m = 0$ the module $Z / 0$ is a copy of $Z$ and the formula reads ${Hom}_{Z} (Z, Z / n) ≅ Z / n$ , which is correct and is the case $d = \gcd (0, n) = n$ . At $n = 0$ it fails, as step 4.1 records. The asymmetry is the asymmetry between the source and the target of a homomorphism, not an artefact of the $\gcd$ convention.
The displayed isomorphism is canonical for these quotient presentations. The classes $[1]_{m}$ and $[n / d]_{n}$ are distinguished by the standard quotient maps, not chosen generators. Evaluation at $[1]_{m}$ , followed by the inverse of $[c]_{d} \mapsto [c (n / d)]_{n}$ , therefore gives the isomorphism without an auxiliary choice.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Cyclic group (Wikipedia), §Tensor product and Hom of cyclic groups (standard reference, not scraped)
A. Altman and S. Kleiman, A Term of Commutative Algebra, 13th ed., Exercise (16.20) (standard reference, not scraped)

Hom⁡Z(Z/m,Z/n)≅Z/gcd⁡(m,n)HomZ​(Z/m,Z/n)≅Z/gcd(m,n) for n≥1n≥1