TITLE: Homomorphic images of $\mathbb{Z}$
QUESTION [1 upvotes]: 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.

REPLY [4 votes]: 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$\mathbb{Z}$
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