Linear independence and rank

#module_theory

The fundamental structure theorem for modules over a PID is a direct sum decomposition of each module into a free part and a torsion part. To understand both parts, it helps to look more closely at the ideas of linear (in)dependence and rank.

Definition of linear dependence

Let $M$ be a left $R$ -module. A set of elements ${m_{1}, \dots, m_{k}} \subseteq M$ is $R$ -linearly dependent if there exist $r_{1}, \dots, r_{k} \in R$ (not all zero) such that

r_{1} m_{1} + \dots + r_{k} m_{k} = 0_{M} .

Rank bounds the number of linearly independent elements

Let $R$ be an integral domain and $M$ be a free $R$ -module of finite rank $n$ . Then any set of more than $n$ elements in $M$ is $R$ -linearly dependent.

Let $S \subseteq M$ be any subset of more than $n$ elements. Let $F = Frac (R)$ be the field of fractions of $R$ . Since we have an $R$ -module isomorphism $M ≃ ⨁_{i = 1}^{n} R ≃ R^{n}$ and an injective $R$ -module morphism $R ↪ F$ (since $R$ is an integral domain), we also have an injective $R$ -module morphism $M ↪ ⨁_{i = 1}^{n} F ≃ F^{n}$ . Then (the image of ) $S$ is a set of more than $n$ elements in the $n$ -dimensional $F$ -vector space $F^{n}$ and hence must be $F$ -linearly dependent. For any nontrivial $F$ -linear dependence relation among the elements in $S$ , clearing denominators yields a nontrivial $R$ -linear dependence relation among the elements in $S$ . Thus, the set $S$ is $R$ -linearly dependent.

Definition of rank of a module

Let $R$ be an integral domain and $M$ be an $R$ -module. The rank of $M$ is the maximum^[1] number of $R$ -linearly independent elements in $M$ .

When $M$ is a free $R$ -module, this notion of rank agrees with our previous notion. When $R = F$ is a field, this notion of rank matches the dimension of an $F$ -module $M$ as a vector space.

For a general integral domain, however, an $R$ -module of rank $n$ need not have a "basis." Even if $M$ is torsion free, it still might not be a free $R$ -module. For example, you can show that when a ring $R$ is considered as an $R$ -module, an ideal $I \subseteq R$ is a free $R$ -module exactly when it is principal. So for example, in the ring $Z [x]$ the ideal $I = (2, x)$ is not principal and hence not a free $Z [x]$ -module. The ring $Z [x]$ is an integral domain, though, so $I$ is torsion free.

The structure of free modules over a PID

Let $R$ be a principal ideal domain and $M$ be a free $R$ -module of finite rank $k$ . For every submodule $N$ of $M$ :

$N$ is free of rank $l \leq k$ ; and
there exists a basis ${m_{1}, \dots, m_{k}}$ for $M$ and nonzero elements $a_{1}, \dots, a_{l} \in R$ such that ${a_{1} m_{1}, \dots, a_{l} m_{l}}$ is a basis for $N$ and^[2]

a_{1} ∣ a_{2} ∣ \dots ∣ a_{l} .

Let's walk through the proof of this one. If $N$ is the trivial submodule then its rank is 0 and its basis is the empty set, so there's nothing to prove. Now suppose $N$ is nontrivial. We'll break this long proof until manageable subsections.

The general idea

The general idea of the proof is to create a direct sum decomposition $M = (m_{1}) \oplus \dots \oplus (m_{k})$ that also induces a direct sum decomposition $N = (a_{1} m_{1}) \oplus \dots \oplus (a_{l} m_{l})$ with the prescribed properties. One of those properties (the divisibility condition on the $a_{i}$ ) tells us that $a_{1}$ should be the "smallest" element among the $a_{i}$ ; i.e., correspond to the largest ideal $(a_{1}) \subseteq R$ among the ideals $(a_{i}) \subseteq R$ . So that's where we begin: by looking for a projection from $M \to R$ for which the image of $N$ is the largest possible ideal of $R$ (obtainable from $N$ ).

We begin by fixing a temporary basis ${m_{1}^{'}, \dots, m_{k}^{'}}$ for $M$ . This is equivalent to fixing an $R$ -module isomorphism $M ≃ F ({x_{1}, \dots, x_{k}}) ≃ R \oplus \dots \oplus R ≃ R^{k}$ . This also allows us to define the $R$ -module projection morphisms $π_{i} : M \to R$ . Using these projections, each element $m \in M$ can be written uniquely as

m = \sum_{i = 1}^{k} π_{i} (m) m_{i}^{'} .

We will use this later in the proof.

Finding the element $a_{1}$

Note that for every $R$ -module morphism $ϕ : M \to R$ the image $ϕ (N)$ of $N$ is a submodule of $R$ , i.e., an ideal of $R$ . Since $R$ is a PID this ideal is principal, say $ϕ (N) = (a_{ϕ})$ for some $a_{ϕ} \in R$ . Now consider the collection $S$ of all such principal ideals in $R$ that are also nontrivial:

S = {(a_{ϕ}) ∣ ϕ \in {Hom}_{R} (M, R), a_{ϕ} \neq 0} .

We first note that this collection is nonempty: since $N$ is not the trivial submodule, for at least one of the projection morphisms $π_{i} : M \to R$ the image $π_{i} (N)$ must be nontrivial, otherwise we would have for all $n \in N$

n = \sum_{i = 1}^{k} π_{i} (n) m_{i}^{'} = \sum_{i = 1}^{k} 0_{R} m_{i}^{'} = 0.

Since $R$ is Noetherian the collection $S$ has at least one maximal element. In other words there is an $R$ -module morphism $ν : M \to R$ so that the principal ideal $ν (N) = (a_{ν})$ is not properly contained in any other element of $S$ .^[3] Let $a_{1} = a_{ν}$ and $n_{1} \in N$ be any element with $ν (n_{1}) = a_{1}$ . Note that $a_{1} \neq 0$ by the definition of $S$ .

Constructing the element $m_{1}$

Our next goal is to construct an element $m_{1} \in M$ so that $ν (m_{1}) = 1_{R}$ . Intuitively, we already have $ν (n_{1}) = a_{1}$ and so it would be nice to simply take $m_{1} = a_{1}^{- 1} n_{1}$ . We would then have $ν (m_{1}) = ν (a_{1}^{- 1} n_{1}) = a_{1}^{- 1} ν (n_{1}) = a_{1}^{- 1} a_{1} = 1_{R}$ . However, there is no guarantee that the element $a_{1} \in R$ is actually invertible. We only know that it is nonzero and that $R$ is a PID (and not necessarily a field). So we need to be a little bit tricky.

We first show $a_{1}$ divides $ϕ (n_{1})$ for every $R$ -module morphism $ϕ : M \to R$ . To see this, fix some $R$ -module morphism $ϕ : M \to R$ and let $I = (a_{1}, ϕ (n_{1}))$ be the ideal generated by $a_{1}$ and $ϕ (n_{1})$ . Since $R$ is a PID this ideal is principal, so $I = (d)$ for some $d \in R$ . We can then write $d = r_{1} a_{1} + r_{2} ϕ (n_{1})$ for some $r_{1}, r_{2} \in R$ . But now consider the $R$ -module morphism $ψ : M \to R$ defined by $ψ = r_{1} ν + r_{2} ϕ$ . By construction we have $ψ (n_{1}) = r_{1} ν (n_{1}) + r_{2} ϕ (n_{1}) = r_{1} a_{1} + r_{2} ϕ (n_{1}) = d$ , so that $d \in ψ (N)$ and hence $(d) \subseteq ψ (N)$ . But we also have $(a_{1}) \subseteq (d) \subseteq ψ (N)$ so by the maximality of $(a_{1})$ we must have equality: $(a_{1}) = (d) = ψ (N)$ . This prove $(a_{1}) = (d)$ and hence $ϕ (n) \in (a_{1})$ ; i.e., $a_{1}$ divides $ϕ (n_{1})$ .

We now apply the above property to the projection morphism $π_{i} : M \to R$ , and so we see that $a_{1}$ divides $π_{i} (n)$ for each $i = 1, \dots, k$ . Write $π_{i} (n_{1}) = a_{1} b_{i}$ for some $b_{i} \in R$ and define

m_{1} = \sum_{i = 1}^{k} b_{i} m_{i}^{'} .

By construction we have

a_{1} m_{1} = \sum_{i = 1}^{k} a_{1} b_{i} m_{i}^{'} = \sum_{i = 1}^{k} π_{i} (n_{1}) m_{i}^{'} = n_{1} .

We therefore have that $a_{1} = ν (n_{1}) = ν (a_{1} m_{1}) = a_{1} ν (m_{1})$ and hence (since $a_{1}$ is nonzero and $R$ is an integral domain)

ν (m_{1}) = 1_{R} .

Verifying $m_{1}$ can be part of a basis for $M$

We will now verify that $m_{1}$ can be taken as one element in a basis for $M$ and that $a_{1} m_{1}$ can be taken as one element in a basis for $N$ . First, let $m \in M$ be an arbitrary element and write

m = ν (m) m_{1} + (m - ν (m) m_{1}) .

Note that

ν (m - ν (m) m_{1}) = ν (m) - ν (m) ν (m_{1}) = ν (m) - ν (m) \cdot 1_{R} = 0_{R}

and so $m - ν (m) m_{1}$ is in the kernel of $ν : M \to R$ . This shows that we at least have $M = (m_{1}) + \ker (ν)$ . To see that this is a direct sum decomposition, suppose $r m_{1} \in \ker (ν)$ for some $r \in R$ . Then

0_{R} = ν (r m_{1}) = r ν (m_{1}) = r \cdot 1_{R} = r .

Thus, we do indeed have $(m_{1}) \cap \ker (ν) = (0)$ and hence we have a direct sum decomposition $M = (m_{1}) ⨁ \ker (ν)$ . This implies that $m_{1}$ can indeed be taken as one element in a basis for $M$ .

Verifying $a_{1} m_{1}$ can be part of a basis for $N$

Observe that for every $n^{'} \in N$ the element $ν (n^{'})$ is divisible by $a_{1}$ (since $a_{1}$ generates the ideal $ν (N)$ ). So given any $n^{'} \in N$ we can write $ν (n^{'}) = b a_{1}$ for some $b \in R$ . Then we can write

\begin{aligned} n^{'} & = ν (n^{'}) m_{1} + (n - ν (n^{'}) m_{1}) \\ = b a_{1} m_{1} + (n^{'} - b a_{1} m_{1}) . \end{aligned}

By the same computation as above, the second term in the above sum is an element of $N$ that is in the kernel of $ν$ :

ν (n^{'} - b a_{1} m_{1}) = ν (n^{'}) - b a_{1} ν (m_{1}) = b a_{1} - b a_{1} \cdot 1_{R} = b a_{1} - b a_{1} = 0_{R} .

We therefore have $N = (a_{1} m_{1}) + (N \cap \ker (ν))$ , and once again the trivial intersection between those two submodules proves this is a direct sum decomposition, $N = (a_{1} m_{1}) \oplus (N \cap \ker (ν))$ . This implies $a_{1} m_{1}$ can indeed be taken as one element in a basis for $N$ .

Proving $N$ is free of rank no more than $k$

We now prove $N$ is free of rank $l \leq k$ , by induction on $l$ . (Recall that the rank of $N$ is defined as the maximum number of linearly independent elements in $N$ .)

If $l = 0$ , then for every $n \in N$ the set ${n}$ is $R$ -linearly dependent; i.e., $r n = 0_{N}$ for some nonzero $r \in R$ . But $N$ is a submodule of the free $R$ -module $M$ , which is torsion free, so we must have $n = 0$ . This implies $N = (0)$ is the trivial submodule, a contradiction to our running assumption that $N$ is nontrivial.

Now assume that the rank of $N$ is $l > 0$ and that all submodules of $M$ of rank less than $l$ are free. In our direct sum decomposition $N = (a_{1} m_{1}) ⨁ (N \cap \ker (ν))$ , the submodule $N \cap \ker (ν)$ has rank $l - 1$ and hence by our induction hypothesis is free. By the direct sum decomposition, adjoining $a_{1} m_{1}$ to any basis for $N \cap \ker (ν)$ gives a basis for $N$ , so $N$ is also free (of rank $l$ ).

Proving property (2) of the proposition

Finally, we prove property (2) of the proposition by induction on the rank $k$ of $M$ . Applying property (1) to the submodule $\ker (ν)$ shows that this submodule is free, and because of the direct sum decomposition

The general idea

Finding the element a1

Constructing the element m1

Verifying m1 can be part of a basis for M