3

Let $R$ be a ring. I want to prove that every quotient of a cyclic $R$-module is cyclic.

Can I assume that the question is about the quotient module or it could be just quotient?

Because in the first case if $M$ is a cyclic $R$-module then $M= \langle m\rangle$ for some $m$, and a quotient $N$ of $M$ would be given by a surjective $R$-module homomorphism $H:M\to N$, but this means that $N=\langle H(m)\rangle$, which is to say $N$ is cyclic.

Is this proof right?

user26857
  • 52,094
Cure
  • 4,051
  • 3
    I don't understand: what is the assumption you're trying to make? It seems to me like the problem is rather trivial: if $M=Rm$ for some $m\in M$, then certainly $M/N=R(m+N)$ – tomasz Nov 04 '14 at 11:40
  • @tomasz Could you explain further? – Cure Nov 04 '14 at 13:08

1 Answers1

3

Yes the proof is correct. (If unsure, you should explain in detail why $N = \langle H(m) \rangle$.)

If $M$ is a module, then a quotient of $M$ is a module $Q$ with a surjective homomorphism $M \to Q$. If $U$ is the kernel, it follows $Q \cong M/U$. Conversely, $M/U$ is always a quotient of $M$. So up to isomorphism there is no difference.

Martin Brandenburg
  • 163,620