
Lemma 10.108.11. Let $R$ be a ring. Suppose we have a module $M = \bigcup _{e \in E} M_ e$ where the $M_ e$ are submodules well-ordered by inclusion. Assume the quotients $M_ e/\bigcup \nolimits _{e' < e} M_{e'}$ have projective dimension $\leq n$. Then $M$ has projective dimension $\leq n$.

Proof. We will prove this by induction on $n$.

Base case: $n = 0$. Then $P_ e = M_ e/\bigcup _{e' < e} M_{e'}$ is projective. Thus we may choose a section $P_ e \to M_ e$ of the projection $M_ e \to P_ e$. We claim that the induced map $\psi : \bigoplus _{e \in E} P_ e \to M$ is an isomorphism. Namely, if $x = \sum x_ e \in \bigoplus P_ e$ is nonzero, then we let $e_{max}$ be maximal such that $x_{e_{max}}$ is nonzero and we conclude that $y = \psi (x) = \psi (\sum x_ e)$ is nonzero because $y \in M_{e_{max}}$ has nonzero image $x_{e_{max}}$ in $P_{e_{max}}$. On the other hand, let $y \in M$. Then $y \in M_ e$ for some $e$. We show that $y \in \mathop{\mathrm{Im}}(\psi )$ by transfinite induction on $e$. Let $x_ e \in P_ e$ be the image of $y$. Then $y - \psi (x_ e) \in \bigcup _{e' < e} M_{e'}$. By induction hypothesis we conclude that $y - \psi (x_ e) \in \mathop{\mathrm{Im}}(\psi )$ hence $y \in \mathop{\mathrm{Im}}(\psi )$. Thus the claim is true and $\psi$ is an isomorphism. We conclude that $M$ is projective as a direct sum of projectives, see Lemma 10.76.3.

If $n > 0$, then for $e \in E$ we denote $F_ e$ the free $R$-module on the set of elements of $M_ e$. Then we have a system of short exact sequences

$0 \to K_ e \to F_ e \to M_ e \to 0$

over the well-ordered set $E$. Note that the transition maps $F_{e'} \to F_ e$ and $K_{e'} \to K_ e$ are injective too. Set $F = \bigcup F_ e$ and $K = \bigcup K_ e$. Then

$0 \to K_ e/\bigcup \nolimits _{e' < e} K_{e'} \to F_ e/\bigcup \nolimits _{e' < e} F_{e'} \to M_ e/\bigcup \nolimits _{e' < e} M_{e'} \to 0$

is a short exact sequence of $R$-modules too and $F_ e/\bigcup _{e' < e} F_{e'}$ is the free $R$-module on the set of elements in $M_ e$ which are not contained in $\bigcup _{e' < e} M_{e'}$. Hence by Lemma 10.108.9 we see that the projective dimension of $K_ e/\bigcup _{e' < e} K_{e'}$ is at most $n - 1$. By induction we conclude that $K$ has projective dimension at most $n - 1$. Whence $M$ has projective dimension at most $n$ and we win. $\square$

Comment #2977 by Dario Weißmann on

Maybe it should be mentioned in the lemma that we assume $M_{e'}\subset M_{e}$ for $e\leq e$?

Comment #3101 by on

OK, I reformulated the lemma. Can you take a look here to see if you think it is better? Maybe it is worse, but I think it is more precise without being longer. Thanks in any case.

Comment #3131 by Dario Weißmann on

Yes, I think it's better. Although I really would like the term 'well-ordered by inclusion' in there somewhere. Maybe something like:

Let $R$ be a ring. Suppose we have a module $M = \bigcup_{e \in E} M_e$ where the $M_e$ are submodules well-ordered by inclusion. Further assume that the quotients $M_e/\bigcup\nolimits_{e' < e} M_{e'}$ have projective dimension $\leq n$. Then $M$ has projective dimension $\leq n$.

It's a bit longer though...

Comment #3135 by on

@#3131 This is good and when I remove the word "Further" it is the same length as what I had. Thanks! Changes here.

Comment #3157 by Dario Weißmann on

Shouldn't it be 'submoduleS' though? Sorry for bringing this up again.

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