Let $f\colon\mathbb Z\to R$ be a surjective homomorphism. Show that $R$ is isomorphic either to $\mathbb Z$ or to the ring $\mathbb{Z/nZ}$ for suitable $n \geq 1$
I have no clue on this question, please help me, thanks
Let $f\colon\mathbb Z\to R$ be a surjective homomorphism. Show that $R$ is isomorphic either to $\mathbb Z$ or to the ring $\mathbb{Z/nZ}$ for suitable $n \geq 1$
I have no clue on this question, please help me, thanks
The first isomorphism theorem for rings says, among other things, that if $R,S$ are rings and $f: R \rightarrow S$ is a surjective homomorphism then $S \cong R / \ker f$. In the context of your question then it remains to determine what the possibilities are for $\ker f$. In particular $\ker f$ is an ideal of $\mathbb Z$. So you need to show that the ideals of $\mathbb Z$ are $n\mathbb Z$ and $\{0\}$.
Because $\mathbb Z$ is generated additively by $1$, $R$ is generated additively by $f(1)$. If $f(1)$ has infinite additive order, then you can show that $R\cong \mathbb Z$. If $f(1)$ has additive order $n$, then you can show that $R\cong \mathbb Z/n\mathbb Z$.
Hint: The only subgroups of $\mathbb{Z}$ are $\{0\}$ and $n\mathbb{Z}, \ n \in \mathbb{N}$. Thus the only possibilities of $\mathbb{Z}/\text{Ker} f$ are $\mathbb{Z}$ and $\mathbb{Z}_n , \ n \in \mathbb{N}\ldots$