Definition 15.23.1. Let $R$ be a domain with fraction field $K$. The rank of an $R$-module $M$ is $\dim _ K(M \otimes _ R K)$.
15.23 Ranks of modules
Here is our definition.
This definition does not conflict with the notion of a locally free module of rank $r$ in Algebra, Definition 10.78.1 when both definitions apply.
Lemma 15.23.2. Let $R$ be a domain. If $M \to M'$ is a map of $R$-modules whose kernel and cokernel are torsion, then the rank of $M$ equals the rank of $M'$.
Proof. Omitted. Hint: the induced map $M \otimes _ R K \to M' \otimes _ R K$ is an isomorphism if $K$ is the fraction field of $R$. $\square$
Lemma 15.23.3. Let $R$ be a domain. Let $0 \to M \to M' \to M'' \to 0$ be a short exact sequence of $R$-modules. Then the rank of $M'$ is the sum of the ranks of $M$ and $M''$.
Proof. Omitted. $\square$
Lemma 15.23.4. Let $R$ be a domain. Let $M$ and $N$ be $R$-modules.
The rank of $M \otimes _ R N$ is the product of the ranks of $M$ and $N$.
If $M$ is a finitely presented $R$-module, then the rank of $\mathop{\mathrm{Hom}}\nolimits _ R(M, N)$ is the product of the ranks of $M$ and $N$.
Proof. Part (1) follows from the same fact for vector spaces. Assume $M$ is finitely presented as an $R$-module. Then $\mathop{\mathrm{Hom}}\nolimits _ R(M, N) \otimes _ R K$ is isomorphic to $\mathop{\mathrm{Hom}}\nolimits _ K(M \otimes _ R K, N \otimes _ R K)$ as a $K$-vector space by Algebra, Lemma 10.10.2. Since also $M \otimes _ R K$ is finite dimensional, we conclude from the corresponding fact for vector spaces. $\square$
Lemma 15.23.5. Let $R \subset R'$ be an extension of domains. If $M$ is an $R$-module, then the rank of $M$ over $R$ is equal to the rank of $M \otimes _ R R'$ over $R'$.
Proof. This is true because the dimension of a vector space is invariant under extension of ground field. $\square$
Lemma 15.23.6. Let $R$ be a domain. Let $M$ be a finite $R$-module. Then
$M$ has finite rank $r \geq 0$,
there is a map $M \to R^{\oplus r}$ whose kernel and cokernel are torsion modules,
there is an $f \in R$, $f \not= 0$ such that $M_ f$ is free of rank $r$, and
there is an injective map $R^{\oplus r} \to M$ whose cokernel is a torsion module.
Proof. Since $M$ is finite, $M \otimes _ R K$ is a finite dimensional vector space over the fraction field $K$ of $R$. Thus the rank $r$ is finite.
Choose generators $x_1, \ldots , x_ n$ of $M$. Choose a basis $e_1, \ldots , e_ r$ for $M \otimes _ R K$. We may write $x_ i \otimes 1 = \sum (a_{ij}/b_{ij}) e_ j$ for some $a_{ij} , b_{ij} \in R$ with $b_{ij} \not= 0$. Setting $b = \prod b_{ij}$ and replacing $a_{ij}$ by $a_{ij}b/b_{ij}$ we may assume that $x_ i \otimes 1 = \sum (a_{ij}/b) e_ j$. Note that if we have a relation $\sum r_ i x_ i = 0$ in $M$ then $\sum r_ ia_{ij} = 0$ in $R$ for all $j = 1, \ldots , r$.
Proof of (2). The last comment above tells us the map $M \to \bigoplus Re_ j$ sending $x_ i$ to $\sum a_{ij} e_ j$. is well defined. The fact that $e_1, \ldots , e_ r$ forms a basis of $M \otimes _ R K$ tells us that this map becomes an isomorphism upon tensoring by $K$. Hence this is a map as in (2).
Proof of (3). The cokernel of the map $M \to R^{\oplus r}$ constructed in (2) is finite and torsion. Hence we can find a $g \in R$, $g \not= 0$ which annihilates this cokernel. In other words, such that the map $M_ g \to R_ g^{\oplus r}$ is surjective. Then we can choose a splitting $M_ g = N \oplus R_ g^{\oplus r}$. Then $N$ is a finite $R_ g$-module with $N \otimes _{R_ g} K = 0$. In other words, $N$ is torsion and finite over $R_ g$. Thus we can find a $g' \in R$ which annihilates $N$. Setting $f= gg'$ we see that (3) holds.
To prove (4) choose a basis $y_1, \ldots , y_ r$ of $M_ f$ where $f$ is as in (3). Then write $y_ j = m_ j/f^{t_ j}$ for some $m_ j \in M$ and $t_ j \geq 0$. The map $R^{\oplus r} \to M$ sending the $j$th basis vector to $m_ j$ will be as in (4). Details omitted. $\square$
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