1

I know this's been asked here many times, and I now know a way to prove it using the first iso theorem on $\phi:R\to M$.

But my first approach was the other way around and something was wrong: I tried to show that $f:M\to R/I$ is iso.

More specifically, I defined $f(m)=f(\sum r_n m_n)=\sum r_n f(m_n):=\sum r_n+I$ where $m_n$'s are in the generating set of $M$.

Injectivity: since $\ker f\leq M$ and $M$ simple, $\ker f=\{0\}$ or $\ker f=M$.

If $\ker f = M$, then $f(M)=I\implies R=I$ but since $I$ is maximal, it is proper: $I<R$.

Surjectivity: $\forall r+I\in R/I,\, \exists m\in M$ s.t. $m=rx$ so that $f(m)=f(rx)=rf(x)=r+I$, where $x$ belongs to the generating set of $M$.

The thing is, that I never used the maximality of $I$. I only used its proper-ness. I don't think the maximality would be used to show the other axioms, like linearity or distributivity etc. What am I missing here?

user642721
  • 893

1 Answers1

1

The problem lies in showing that your map $f$ is well defined. For an arbitrary ideal $I$, you may have $\sum r_n m_n=0$ in $M$ without $\sum r_n\in I$ being satisfied.

A note: Since $M$ is simple, any non-zero element $x\in M$ will generate $M$ (since $Rx\subset M$ is a non-zero submodule). Setting $I=\{r\in R\mid rx=0\}$, you may take $\phi: M\to R/I,\: rx\mapsto r$, and show that it is a well-defined isomorphism. (But it is easier to use the first isomorphism theorem.)

Erik D
  • 921