$S^{-1}M$ is a Noetherian $S^{-1}R$ module

Question

Let $R$ be a ring, $S$ a multiplicative subset and $M$ a Noetherian module. Show that $S^{-1}M$ is a Noetherian $S^{-1}R$ module.

My attempt: Let $K$ be a submodule of $S^{-1}M$. Then $\varphi^{-1}_S K\subset M$ is finitely generated because $M$ is Noetherian. Let $m_1,...,m_k$ generate $\varphi^{-1}_S K$. Thus, $m_1/1,...,m_k/1$ generate $S^{-1}\varphi^{-1}_S K$ over $S^{-1}R$. But $K=S^{-1}\varphi^{-1}_S K$.

Therefore, $K$ is finitely generated over $S^{-1}R$ and so, $S^{-1}M$ is Noetherian over $S^{-1}R$.

My question: I am not sure, but why is it that $\varphi^{-1}_S K$ is a module; also, I wasn't able to show that $K=S^{-1}\varphi^{-1}_S K$. I appreciate any help on this. Thanks in advance!

Answer 1

First of all, you can define an $R$-module structure on $S^{-1}M$ by restriction of scalars. In other words, the scalar multiplication is $r.\frac{m}{s}=\frac{r}{1}\cdot\frac{m}{s}$. It is easy to check that any $S^{-1}R$-submodule of $S^{-1}M$ is also an $R$-submodule.

Now, we can define a map $f:M\to S^{-1}M$ by $f(m)=\frac{m}{1}$. Again, it is straightforward that this is a homomorphism of $R$-modules. Thus, if $K\subseteq S^{-1}M$ is a submodule over $S^{-1}R$ (and thus also a submodule over $R$) then $f^{-1}(K)$ is a submodule of $M$ as an inverse image of a submodule of $S^{-1}M$.

Now let $K\subseteq S^{-1}M$ be an $S^{-1}R$-submodule, and let $N=f^{-1}(K)$. We will show that indeed $K=S^{-1}N$. If $\frac{n}{s}\in S^{-1}N$ where $n\in N, s\in S$ then by definition $f(n)=\frac{n}{1}\in K$, and since this is an $S^{-1}R$-module we have $\frac{n}{s}=\frac{1}{s}\cdot\frac{n}{1}\in K$ as well. Conversely, suppose $\frac{n}{s}\in K$. Then $\frac{n}{1}=\frac{s}{1}\cdot\frac{n}{s}\in K$ as well, and so $n\in f^{-1}(K)=N$, which means $\frac{n}{s}\in S^{-1}N$. So by two sided inclusion indeed $S^{-1}N=K$.

Alright, now your exercise is really easy. Suppose $K_1\subseteq K_2\subseteq K_3\subseteq...$ is an increasing sequence of $S^{-1}R$-submodules of $S^{-1}M$. For each $n\in\mathbb{N}$ let $N_n=f^{-1}(K_n)$. Then $N_1\subseteq N_2\subseteq...$ is an increasing sequence of $R$-submodules of $M$. Since $M$ is a Noetherian $R$-module there is some $n\in\mathbb{N}$ such that $N_i=N_n$ for all $i\geq n$. As we have shown, this implies $K_i=S^{-1}N_i=S^{-1}N_n=K_n$ for all $i\geq n$.