On the Rank and the Span Rank of Modules
We work over a commutative ring \(R\) with unity. The rank \(\operatorname{rank}_R M\) of a module \(R\)-module \(M\) is defined as the maximal cardinality of a \(R\)-linearly independent subset of \(M\). A homologist might write this as an exact sequence \[ 0 \longrightarrow R^{\oplus n} \longrightarrow M \] where \(n\) is as large as possible.
It’s also tempting to define the rank in other ways. For example, if \(R\) is an integral domain with field of fractions \(k\), one might define the rank of an \(R\)-module \(M\) as the dimension of the \(k\)-vector space \(k \otimes_R M\). Are these two definitions equivalent?
Proof. To see this, consider the following commutative diagram, where downward arrows are given by tensoring with \(k\):
It suffices to show that the top arrow is injective if and only if the bottom arrow is injective.
For the downward direction, note that \(k\) is a flat \(R\)-module, so tensoring with \(k\) preserves injections.
For the upward direction, suppose the bottom arrow is injective. Consider an element \((a_i)_{i=1}^n \in R^{\oplus n}\) that maps to zero in \(M\). Its image in \(k^{\oplus n}\) also maps to zero in \(k \otimes_R M\). Since the bottom arrow is injective, the image in \(k^{\oplus n}\) must be zero, which means each \(a_i\) is zero in \(k\). Since \(R\) is an integral domain, the localization \(R \to k\) is injective, hence each \(a_i\) is zero in \(R\), indicating that the the top arrow is injective.
Remark (Generalizing to localizations). Note that we have only used the flatness of \(k\) and that \(R\) is an integral domain. Since every localization of a ring is flat, the same proof shows that as long as \(R\) is an integral domain, localzation does not change the rank of a module.
[TODO] It’s interesting to ask if the integral domain condition is necessary. At least, the one side inequality \(\operatorname{rank}_R M \leq \dim_k (k \otimes_R M)\) is always true.
Another possible definition of rank is the minimal number of generators of a module. We call this the span rank \(\operatorname{spanrank}_R M\) of an \(R\)-module \(M\). A homologist might write this as an exact sequence \[ R^{\oplus m} \longrightarrow M \longrightarrow 0 \] where \(m\) is as small as possible.
Clearly, \(\operatorname{rank}\) and \(\operatorname{spanrank}\) are equal for free modules, since the definition of a basis requires both linear independence and spanning. For finitely generated modules over a PID, having the structure theorem in hand, the \(\operatorname{rank}\) is exactly the free rank, while the \(\operatorname{spanrank}\) is the number of the direct summands in the decomposition. We see that \(\operatorname{spanrank}M \geq \operatorname{rank}M\) in this case. This is in general true:
Proof. Consider the following commutative diagram with exact rows:
where \(R^n \to R^m\) is constructed by the projective nature of \(R^n\). \(R^n \to M\) is injective, hence so do \(R^n \to R^m \to M\) and in turn \(R^n \to R^m\). Thus, we have an injection \(R^n \to R^m\). Tensoring with any residue field of \(R\), we get an injection of vector spaces over that field, which implies \(n \leq m\).
Regarding Proposition 1, one might wonder if there is an analogous result for \(\operatorname{spanrank}\). Let’s say \(\mathfrak m\) is a maximal ideal of \(R\) and the residue field \(k := R/\mathfrak m\). Tensoring the exact sequence \[ R^{\oplus m} \longrightarrow M \longrightarrow 0 \] with \(k\), we get an exact sequence \[ k^{\oplus m} \longrightarrow k \otimes_R M \longrightarrow 0 \] So indeed \(\operatorname{spanrank}_R M \geq \dim_k (k \otimes_R M)\). Sadly the converse is not true in general. For example:
An arithmetic example: Let \(R = \mathbb Z\) and \(M = \mathbb Z/2\mathbb Z\). Then \(\operatorname{spanrank}_{\mathbb Z} M = 1\) but \(\mathbb Z/2\mathbb Z \otimes_{\mathbb Z} \mathbb Z/3\mathbb Z \cong 0\).
An geometric example: Let \(R = \mathbb R[x]\) and \(M = \mathbb R[x]\). Then \(\operatorname{spanrank}M = 1\). But \(\mathbb R[x] / (x) \cong \mathbb R\) while \(\mathbb R[x] / (x^2 + 1) \cong \mathbb C\). We see that different residue fields give different dimensions. Blame the failure of Hilbert’s Nullstellensatz over non-algebraically closed fields for this irregularity!
Luckily, it turns out that this irregularity is completely caused by the arbitrariness of residue fields. Once we restrict ourselves to a local ring, the world becomes much nicer:
Proof. Lifting the exact sequence \[ k^{\oplus m} \longrightarrow k \otimes_R M \longrightarrow 0 \] to \[ R^{\oplus m} \longrightarrow M \longrightarrow 0 \] is a direct consequence of Nakayama’s lemma. cf. [1, theorem 2.8] for example.
Remark (Necessity of finite generation). At most of the time, we work in Noetherian rings, hence the finitely generated condition is rather mild.
Nevertheless, it is indeed necessary. Consider the following example:
Let \(M = k[[x]]\) be the formal power series ring over a field \(k\), regarded as a module over \(R = k[x]\). Then as \(k[x]\)-modules, \[ k[x] / (x) \otimes_{k[x]} k[[x]] \cong k[[x]] / x k[[x]] \] has dimension \(1\) over \(k\), while \(\operatorname{spanrank}_{k[x]} k[[x]]\) is infinite.
Localizing \(R = k[x]\) at the maximal ideal \((x)\) does not change any statements above, providing an counterexample over a local ring.
Remark. [TODO] What can we say about the module \(M\) given that \(\operatorname{rank}_R M = \operatorname{spanrank}_R M\)?