Homomorphic images of $\mathbb{Z}$

Question

Suppose there is a homomorphism $f:\mathbb{Z}\to R$ where $R$ is a non-zero ring. By the definition of homomorphism $f(n)=f(1+1+1+1.....+1)=f(1)+f(1)+f(1)+....+f(1)$ Now suppose $f(1)=x$ then $f(n)=nx$ by definition. By varying x we can say that homomorphic images of $\mathbb{Z}$ is $0,\mathbb{Z},\mathbb{2Z},\mathbb{3Z}$...etc.

Am I right? Please provide some deep intuitions if any.

Answer 1

Take $R=\mathbf{Z}/n\mathbf{Z}$ and define $f\colon \mathbf{Z}\to \mathbf{Z}/n\mathbf{Z}$ to be the homomorphism $f(k) = k\pmod n$. The image is a ring with finitely many elements and meets your requirement.

Answer 2

Hint : if $\phi:\mathbb{Z} \to R $ is a homorphism then $ \mathbb{Z}/ker(\phi) \cong \phi(\mathbb{Z}) $

now, ker($\phi$) is an ideal of $\mathbb{Z}$ and any ideal of $\mathbb{Z}$ is of the form n$\mathbb{Z}$