Proving that $M$ is a simple $R-$module.

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Here is the question I want to answer:

A module is simple if it is not the zero module and it has no proper nonzero submodule.

$(a)$ Let $M$ be an $R-$module. Show that the following conditions are equivalent.

  1. $M$ is a simple $R-$module.

  2. There exists a maximal ideal $\mathfrak{m} \in R$ such that $M \cong_R R/ \mathfrak{m}.$

  3. for any nonzero $x,y \in M$ there exists $r \in R$ such that $rx = y$ (i.e. the action of $R$ on $M$ is transitive).

My trial is:

$1 \implies 2$

Let $M$ be a simple $R-$module. We want to show that there exists a maximal ideal $\mathfrak{m} \in R$ such that $M \cong_R R/ \mathfrak{m}.$

If $M$ is simple, then $0 \neq M$, so we can pick $x \in M, x \neq 0$. Then $Rx$ is a non-zero submodule of $M.$ Since $M$ is simple, we have $Rx=M$ as a simple module has no proper nonzero submodule .

If we define $f: R \to M$ as $f(r)=rx$ then clearly it is surjective (is this correct in general or just because my $M$ is cyclic in that case)and it is an $R-$module homomorphism because:

Let $r,s,r' \in R.$

$1- f(r + s) = (r + s)x = rx + sx = f(r) + f(s).$

$2- f(r'r) = (r'r)x = r'(rx) = r'f(r).$

Now, by the first isomorphism theorem, we have $R/\operatorname{Ker}(f) \cong \operatorname{im}(f)= M$.

Since we assumed that $R-$ module means left and right $R-$module i.e. $R$ is commutative, then we can prove the maximality of the ideal $\mathfrak{m} = \operatorname{Ker}(f)$ as follows:

Since $R/\mathfrak{m} \cong M$ then it is simple, hence the only submodules (ideals) of $R/ \mathfrak{m}$ is $0$ and itself. This show that $R/ \mathfrak{m}$ is a field, so $\mathfrak{m}$ is maximal by a theorem.

$2 \implies 3$

Assume that there exists a maximal ideal $\mathfrak{m} \in R$ such that $M \cong_R R/ \mathfrak{m}$ and we want to show that for any nonzero $x,y \in M$ there exists $r \in R$ such that $rx = y$ (i.e. the action of $R$ on $M$ is transitive)

Take the action of $R$ on $M$ as $f_x(r) = rx$ in the previous proof we said that $f_x$ is surjective and hence for any nonzero $x,y \in M$ there exists $r \in R$ such that $rx = y.$

My questions are:

1- what is the precise reason for surjectivity of $f$?

2- Is my proof of $2 \implies 3$ correct?

3- how can I prove $3 \implies 1$?

4- Is there is a better way of proving that question?

Could anyone help me in answering those questions please?

Best Answer

  1. If $f$ is not surjective, then $Rx$ is a proper nonzero submodule of $M$, contrary to our assumption that $M$ is simple.

  2. It could be better - it seems like you're using $1$ here instead of $2$. I think I'd go with the following: let $f:R\to R/\mathfrak{m}\cong M$ be the natural projection, and consider $Rx\subset M$ for some nonzero $x\in M$. Then $I=f^{-1}(Rx)$ is an $R$-submodule of $R$ (or, an ideal) which contains $\mathfrak{m}$. As $x\neq 0$, we must have that $I\neq \mathfrak{m}$: if $I=\mathfrak{m}$, then $f(I)=0$, but $x\neq 0$ is in $f(I)$. So $I=R$, and $f^{-1}(Rx)=R$, therefore $Rx=f(R)=M$ and there is some $r\in R$ with $rx=y$ for any $0\neq y\in M$.

  3. Suppose $S\subset M$ is a nonzero submodule. We need to show that it's all of $M$. If it's not, then there's an $m\in M\setminus S$. Do you see where to go from here?

Pick $s\in S$ and $r\in R$ so that $rs=m$. Thus $m\in S$, so $S=M$.

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