Exercise A0.III.7.05

Lemma L3.5

A homomorphic image of a Noetherian module is itself Noetherian.

Proof

Let $M$ be a Noetherian module and let $M/K$ be a homomorphic image of $M.$ Let $\pi : M \to M/K$ be the quotient map.

Fix an arbitrary submodule $X$ of $M/K$ and an inclusion $i_X : X \to M/K.$ Let $j : R^{\oplus A} \to X$ be a surjection with $A$ an arbitrary generating set.

Since free modules are projective, we have a morphism $s : R^{\oplus A} \to M$ satisfying $i_X \circ j = \pi \circ s.$

Using the canonical decomposition we can factor $s$ as $i_{\im s} \circ \hat{s},$ with $\hat{s} : R^{\oplus A} \to \im s$ an epimorphism and $i_{\im s} : \im s \to M$ an inclusion.

Note that $\im s$ is a subobject of $M,$ and is thus finitely generated. Fix a surjection $k : R^{\oplus n} \to \im s.$

Since free modules are projective this gives us a map $u : R^{\oplus A} \to R^{\oplus n}$ with $\hat{s} = k \circ u.$ Similarly, we have a map $u' : R^{\oplus n} \to R^{\oplus A}$ with $k = \hat{s} \circ u'.$

$$\begin{align*} i_X \circ j \circ u' \circ u & = \pi \circ s \circ u' \circ u \\ & = \pi \circ i_{\im s} \circ \hat{s} \circ u' \circ u \\ & = \pi \circ i_{\im s} \circ k \circ u \\ & = \pi \circ i_{\im s} \circ \hat{s} \\ & = \pi \circ s \\ & = i_X \circ j \\ \end{align*}$$

Since $i_X$ is a monomorphism, $j \circ u' \circ u = j$ and since $j$ is an epimorphism, $j \circ u'$ is epic as well. Hence, $X$ is finitely generated.

$\blacksquare$

Proposition

Let

$$\cdots \longrightarrow L \longrightarrow M \longrightarrow N \longrightarrow \cdots$$

be an exact complex, and let $L$ and $N$ be Noetherian. Then $M$ is also Noetherian.

Proof

Let $f : L \to M$ and $g : M \to N$ be the boundary maps. Using canonical decomposition we factor $f = i_{\im f} \circ \hat{f}$ where $\hat{f} : L \to \im f$ is an epimorphism, and $i_{\im f} : \im f \to M$ is an inclusion.

Similarly, we factor $g = i_{\im g} \circ \hat{g}$ where $\hat{g} : M \to \im g$ is an epimorphism, and $i_{\im g} : \im g \to N$ is an inclusion.

Let $X$ be an arbitrary subobject of $M$ with an inclusion $i_X : X \to M.$ Let $h : X \to \im g$ be the composition $\hat{g} \circ i_X.$

We can factor $h$ as $h = i_{\im h} \circ \hat{h}$ with $\hat{h} : X \to \im h$ an epimorphism, and $i_{\im h} : \im h \to \im g$ an inclusion. Since $\im h$ is a subobject of $\im g,$ which itself is a subobject of a Noetherian module $N.$ By transitivity, $\im h$ is a subobject of $N$ and is finitely generated.

Let $i_{\ker h} : \ker h \to X$ be an inclusion.

The kernel of $h$ is killed by $g,$ since

$$g \circ i_X \circ i_{\ker h} = i_{\im g} \circ h \circ i_{\ker h} = i_{\im g} \circ 0 = 0.$$

By the universal property of $\ker g$ we have an inclusion $j : \ker h \to \ker g.$ But $\ker g = \im f,$ hence $\ker h$ is a submodule of a homomorphic image of $L.$ By L3.5 we know that homomorphic images of Noetherian modules are themselves Noetherian.

Thus, we have a short exact sequence

$$0 \longrightarrow \ker h \longrightarrow X \longrightarrow \im h \cong X/\ker h \longrightarrow 0$$

with both $\ker h$ and $X/\ker h$ finitely generated. By 6.18 this establishes that $X$ is finitely generated. Thus, $M$ is Noetherian.

$\blacksquare$