Three different notions for "bigness" of a module (length, rank and "mass").

Question

Let $M$ be an $R-$module where $R$ is an integral domain. I'm trying to understand the relations between these three notions of size:

1. Let $S \subset M$ be a generating set of minimal cardinality. Define $|S| = mass(M)$
   (I don't know any formal term for this so i'll use this one. Sorry.).
2. Let $T \subset M$ be a maximal linearly independent set (A set $T$ is linearly independent iff the natural homomorphism from $R^{\oplus T}$ to $M$ is injective). Define $|T| = {rk}_R(M)$.
3. By Jordan–Hölder for modules, all composition series of a module have the same length. Define $length(M)$ as that length.

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• For vector spaces (i .e. if $R$ is a field) all the notions coincide. (please correct me if i'm wrong).

• Suppose $M$ is free and of finite length. We know $M$ is free iff it admits a basis. So $M \cong R^{\oplus B}$ where $B \subset M$ is a linearly independent subset. By splitting the composition series we find that $M$ is isomorphic to a sum of simple modules. That is: $$M \cong \bigoplus_{j \in J} N_j \cong R^{\oplus B}$$ Can we now deduce that $length(M)=mass(M)={rk}_R(M)$?

• Suppose $M$ is projective and of finite length. We have again, by splitting, that $M \cong \bigoplus_{j \in J} N_j $. Do we have $length(M)=mass(M)$? What about ${rk}_R(M)$?

• Suppose $M$ is noetherian. Clearly $mass(M)$ is finite. Must $length(M)$ be finite? What if we add that $M$ is projective (/free)?

I'd like to have a more organized picture of how these notions interact with each other on different modules. Comprehensive answers would be greatly appreciated (and rewarded ;)).

Answer 1

The relationship between the three numbers can be summed up in two inequalities: $$ rk_R(M)\leq mass(M) \leq length(M) . $$ There isn't much more that can be said, see below for an idea why.

First, let us prove the inequalities. Assume that $mass(M)=m<\infty$, and let $\{x_1, \ldots, x_m\}$ be a minimal generating set for $M$. Then we have a strictly increasing sequence of submodules $$Rx_1 \subset Rx_1 + Rx_2 \subset \ldots \subset Rx_1+\ldots +Rx_n=R. $$ This sequence can be refined into a composition series; hence $length(M)\geq mass(M)$. (Note that if $mass(M)=\infty$, then the same argument shows that if $M$ has a composition series, then it is of infinite length, so we still have equality).

Next, assume that $rk_R(M)=r$, and that $mass(M)=m$. Then there is an injective morphism $R^{\oplus r}\to M$ and a surjective morphism $R^{\oplus m}\to M$. Since $R$ is projective, we deduce the existence of an injective morphism $R^{\oplus r}\to R^{\oplus m}$; this implies that $r\leq m$ (since commutative rings have the invariant basis number property). This finishes the proof of the inequalities.

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Some comments:

• The difference between mass and rank can be arbitrarily big: over $\mathbb{Z}$, $M=\mathbb{Z}^{\oplus a}\oplus (\mathbb{Z}/2\mathbb{Z})^{\oplus b}$ has rank $a$ and mass $a+b$.

• The same holds for the difference between mass and length: if $p$ is prime, then $\mathbb{Z}/p^n\mathbb{Z}$ has mass $1$ and length $n$ over $\mathbb{Z}$.

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Finally, regarding, the special cases mentioned in the original question:

• As remarked in the original question, if $R$ is a field, then all three notions coincide.

• If $M$ is a free $R$-module, then $rk_R(M)=mass(M)$, and $length(M)=rk_R(M)\cdot length(R)$.

• If $M$ is projective, I don't see more what can be said.

• If $M$ is Noetherian, then $length(M)$ is only finite if $M$ is Artinian as well.