Show that, given $f:M→N$, if M is an artinian module so is N

Question

What it says on the tin. Note that $f:M→N$ is a surjective $R-module$ homomorphism.

This is what I have: Let ${I_n}$ denote a chain of submodules in $M$ such that ${I_0 \supseteq I_1 \supseteq ...}$. There is $k$ in $ \mathbb N$ such that $I_k = I_{k+1}...$. Take $f(I_n)$. By definition, the inclusions are preserved and so is the submodule structure.

Now, I think that I have to check the descending chain condition for the $f(I_n)$s, is that correct? How would I go about doing that? Also, I know that the surjectivity plays a part somewhere. Is it just the fact that $f$ maps every element of $N$ to a $f(I_n)$ thus guaranteeing that all of $N$ is artinian?

Answer 1

Here is a quick proof. While you have the correct idea, you need to start with a descending chain of submodules of $N$.

Let $\mathfrak{a}_0 \supseteq \mathfrak{a}_1 \supseteq \cdots$ be a descending chain of submodules of $N$ and let $M$ be an Artinian module with $f:M \to N$ an epimorphism of $R$-modules. For each $n \in \mathbb{N}$, define the submodule $\mathfrak{b}_n$ of $M$ by $$ \mathfrak{b}_n = \lbrace m \in M \; | \; f(m) \in \mathfrak{a}_n \rbrace. $$ That this is an Abelian group is a straightforward check, so let us just verify that it is a submodule of $M$. Pick an $m \in \mathfrak{b}_n$ and let $r \in R$ be arbitrary. Then $f(rm) = rf(m)$ because $f$ is $R$-linear and since $\mathfrak{a}_n$ is an $R$-module, $rf(m) \in \mathfrak{a}_n$ as well. Thus $f(rm) \in \mathfrak{a}_n$ so $rm \in \mathfrak{b}_n$ and hence $\mathfrak{b}_n$ is a submodule of $M$.

We now show that the $\mathfrak{b}_n$ form a descending chain. For this it suffices to show that for any $n \in \mathbb{N}$, $\mathfrak{b}_n \supseteq \mathfrak{b}_{n+1}$. To this end pick an arbitrary $m \in \mathfrak{b}_{n+1}$. Then $f(m) \in \mathfrak{a}_{n+1}$ and since $\mathfrak{a}_{n+1} \subseteq \mathfrak{a}_n$, we also have that $f(m) \in \mathfrak{a}_n$. Thus $m \in \mathfrak{b}_n$ so indeed we have $\mathfrak{b}_{n} \supseteq \mathfrak{b}_{n+1}$.

We now invoke that $N$ is Artinian to note that the descending chain $$ \mathfrak{b}_0 \supseteq \mathfrak{b}_1 \supseteq \cdots $$ stabilizes at some $k \in \mathbb{N}$. Now applying $f$ to the chain and noting as you did that this preserves the descent with $f(\mathfrak{b}_n) = \mathfrak{a}_n,$ we have that the chain $$ \mathfrak{a}_0 \supseteq \mathfrak{a}_1 \supseteq \cdots $$ stabilizes at $k \in \mathbb{N}$ as well. This gives that $N$ is Artinian.

Answer 2

To show that $N$ is artinian, you are supposed to consider descending chains of $R$-submodules of $N$ rather than descending chain of $R$-submodules of $M$.

Let such chain be $N_1\supseteq N_2\supseteq\cdots$. Then we can obtain a chain $f^{-1}(N_1)\supseteq f^{-1}(N_2)\supseteq\cdots$ of $R$-submodules of $M$. Since $M$ is artinian, the second chain collapses.

That is, there exists $K\in\mathbb{N}$ such that $f^{-1}(N_k)=f^{-1}(N_K)$ for all $k\geq K$.

Since $f$ is surjective (this is essential), it follows that $N_k=N_K$ for all $k\geq K$. That is, the original chain collapses as well.

Hence $N$ is artinian.