Why is $\operatorname{Hom}_\Bbb{Z}(A,A)$ a ring

Question

I am trying to prove and understand why the homomorphisms from an abelian group A to itself $\operatorname{Hom}_\Bbb{Z}(A,A)$ is a ring. Is my reasoning correct in the following proof?

Show the homomorphisms from an abelian group A to itself $\text{Hom}_{\mathbb{Z}}(A,A)$ form a ring.

For two homomorphisms $\delta_1\ \colon V\to V$ and $\delta_2\ \colon V\to V$ $\in \text{Hom}_{\mathbb{Z}}(A,A)$ define their additive and multiplicative binary operataions as $[\delta_1+\delta_2](x)=\delta_1(x)+\delta_2(x)$
and $\delta_1 * \delta_2=\delta_1 \circ \delta_2$ respectively.

Additive Identity This would be the zero map so fix $\delta_1\in \text{Hom}_\mathbb{Z}(A,A)$ then $[\delta_1 + 0]=\delta_1$ $[0+\delta_1]=\delta_1$

Additive Associativity Addition of functions is associative.

Inverse Suppose we have a homomorphism $\delta$ in $\text{Hom}_{\mathbb{Z}}(A,A)$. Define another homomorphism from A to itself as $\delta^{-1}(v):=(\delta(v))^{-1}$ which exists because the image of A under $\delta$ is a subgroup of A (well known theorem of Homomorphisms).

Now we must show that the composition of these functions yields the zero function which is the identity of the Abelian group.

Let $v$ be an arbitrary element in $A$. Then $\delta(v)+\delta^{-1}(v)=\delta(v)+\delta(v^{-1})=\delta(v+v^{-1})=\delta(0)=0$ with the last equlaity following from the fact that $\delta$ is a group homomorphism.

Multiplicative Associativity Composition of functions is associative.

Multiplicative Identity This would be the identity homomorphism so fix $\delta_1 \in \text{Hom}_{\mathbb{Z}}(A,A)$ then $(\delta_1 + 1)=(\delta_1 \circ 1)=\delta_1$. Similarly, $(1 + \delta_1)=(1 \circ \delta_1)=\delta_1$

Multiplicative Distributive Laws Fix three homomorphisms $\delta_1, \delta_2, \delta_3$. $\delta_1 * (\delta_2 + \delta_3) (x)= \delta_1 \circ (\delta_2 + \delta_3) (x) =\delta_1((\delta_2 + \delta_3)(x))=\delta_1(\delta_2(x) + \delta_3(x))=
\delta_1(\delta_2(x)) + \delta_1(\delta_3(x))=(\delta_1*\delta_2 + \delta_1*\delta_3)$

Using the fact that $\delta_1$ is a homomorphism.

Answer 1

Just some feedback:

• Additive identity: If $0_A \in A$ is the identity element of $A$ ($a+0_A=a$ for any $a \in A$) define $0 : A \to A$ by $(\forall a \in A) \ 0(a) = 0_A$. Then it is easy to check that $0 \in \operatorname{Hom}_\mathbb Z(A,A)$ and $f+0=f$ for any $f \in \operatorname{Hom}_\mathbb Z(A,A)$ since $(f+0)(a) = f(a)+0(a) = f(a) + 0_A = f(a)$ for any $a \in A$.
• Additive inverse: instead of writting $a^{-1}$ for the inverse of $a$, write $-a$. So, the additive inverse of $f \in \operatorname{Hom}_\mathbb Z(A,A)$ is the map $-f : A \to A$ given by $(\forall a \in A) \ (-f)(a) = -f(a)$ (as above, check that $(f+(-f))(a)=0_A$ for any $a \in A$, which proves that $f+(-f)=0$).
• Multiplicative identity: Consider the identity map $\operatorname{id}_A : A \to A$ given by $(\forall a \in A) \ \operatorname{id}_A(a) = a$. Then it is easy to check that $\operatorname{id}_A \in \operatorname{Hom}_\mathbb Z(A,A)$ and that $\operatorname{id}_A \circ\, f = f \circ \operatorname{id}_A = f$ for any $f \in \operatorname{Hom}_\mathbb Z(A,A)$.
• Distributive laws: you also need to prove that $(\delta_1+\delta_2) \circ \delta_3 = \delta_1 \circ \delta_3 + \delta_2 \circ \delta_3$, which follows from the definition of the sum in $\operatorname{Hom}_\mathbb Z(A,A)$.