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I proved part (i) of the following:

Proposition 6.3. Let $0 \to M' \xrightarrow{\alpha} M \xrightarrow{\beta} M'' \to 0$ be an exact sequence of $A$-modules. Then

i) $M$ is Noetherian $\iff$ $M'$ and $M'"$ are Noetherian;

ii) $M$ is Artinian $\iff$ $M'$ and $M''$ are Artinian

Proof. We shall prove i); the proof of ii) is similar.

$\implies$: An ascending chain of submodules of $M'$ (or $M''$) gives rise to a chain in $M$, hence is stationary.

Can you tell me if my proof of $(i)\Longleftarrow$ is correct? Here goes:

Let $M^\prime$ and $M^{\prime \prime}$ be Noetherian. Let $L_n$ be an ascending chain of submodules in $M$. Then $\alpha^{-1}(L_n)$ is an ascending chain of submodules in $M^\prime$. Hence $\alpha^{-1}(L_n)$ is stationary, that is, $\alpha^{-1}(L_n) = \alpha^{-1}(L_{n+1})$ for $n$ large enough. Hence $L_n = L_{n+1}$ for $n$ large enough since $\alpha$ is injective hence $L_n$ is stationary.

The proof given in Atiyah-Macdonald is the following but I don't understand why they need both maps, $\alpha$ and $\beta$:

$\Longleftarrow$: Let $(L_n)_{n \ge 1}$ be an ascending chain of submodules of $M$; then $(\alpha^{-1}(L_n))$ is a chain in $M'$, and $(\beta(L_n))$ is a chain in $M''$. For large enough $n$ both these chains are stationary, and it follows that the chain $(L_n)$ is stationary. $\Box$

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    $\begingroup$ I don't think a proof-verification can be a duplicate of a newer question. Unless someone posts the exact same proof with the exact same mistakes. $\endgroup$ – Rudy the Reindeer Jan 30 '16 at 3:36
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If this proof worked then we would be able to conclude that a module is Noetherian if it has a Noetherian submodule. This is certainly not the case — as a silly example, the trivial submodule is always Noetherian. If we forget about $\beta$ then it is true that $(\alpha^{-1}(L_n))_n$ will stabilize, but it's not as if $\alpha(\alpha^{-1}(L_n)) = L_n$, since $\alpha$ need not be surjective.

To use Atiyah and MacDonald's proof, you want to show that if $N_1 \subset N_2$ are submodules of $M$ such that $\alpha^{-1}(N_1) = \alpha^{-1}(N_2)$ and $\beta(N_1) = \beta(N_2)$, then $N_1 = N_2$. For this, take $x \in N_2$. By assumption there is a $y \in N_1$ such that $\beta(y) = \beta(x)$. By exactness, then, there is a $z \in M'$ such that $\alpha(z) = x - y \in N_2$. Thus $z \in \alpha^{-1}(N_2) = \alpha^{-1}(N_1)$. Do you see how to conclude that $x \in N_1$?

[We do need that $N_1 \subset N_2$, by the way. Think about $M = \mathbb R^2$ with $M' = \text{the $x$-axis}$ and the family of lines $y = mx$ for $m \neq 0$.]

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    $\begingroup$ From $\alpha^{-1}(x-y) \in \alpha^{-1}(N_2)$ it follows that $x-y \in \alpha(\alpha^{-1}(N_2)) \subset N_2$ and since by assumption, $y \in N_2$, we may conclude that $x \in N_2$. Is that correct? $\endgroup$ – Rudy the Reindeer Jul 21 '12 at 12:02
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    $\begingroup$ @Clark Looks good to me! $\endgroup$ – Dylan Moreland Jul 21 '12 at 16:35
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    $\begingroup$ Cool, thank you very much! $\endgroup$ – Rudy the Reindeer Jul 21 '12 at 16:43

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