If $R$ is a domain, and $M$ a (left) $R$-module, what are the different notions of dimension of $M$ and their respective assets, what do they measure?


I found out that if $\dim\_RM$ is the cardinal of any maximal independent subset of $M$, then it does not depend of maximal independent subset chosen if $R$ is a left-Ore domain, and the rank-nullity theorem holds for $\dim\_R$. This seems very natural, and I wonder whether this is known:



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> > Do you happen to know any reference for that?
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**Edit.** Let me still give a proof that would fit in a second year algebra course.


Let $R$ be a *left-Ore* domain, that is for any no-zero $a,b\in R$, one as $Ra\cap Rb\neq(0)$. Let $M$ be a left $R$-module. Call a family $\bar v\in M^n$ *dependent* if there is a non-zero $\bar r\in R^n$ s.t. $r\_1v\_1+\dots+r\_nv\_n=0$, or *independent* otherwise. It is a *basis* if independent and maximal such.


**Lemma (incomplete basis).** Any independent family extends to a (possibly empty) basis.


If $\bar b$ is a basis of $M$, for every $v\in M\setminus\bar b$, the set $\bar b\cup\{v\}$ is dependent, so there is a non-zero $r\in R$ such that $rv\in (\bar b)$. For any $S\subset M$ and $v\in M$, say that *$v$ is algebraic over $S$* if there is a nonzero $r\in R$ st $rv\in(S)$.



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> > **Lemma (transitivity of agebraicity).** Let $A,B,C$ be subsets of $M$. If $A$ is algebraic over $B$ and $B$ over $C$, then $A$ is algebraic over $C$.
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**Key idea of the 8 lines proof.** Suppose $ra=sb$ and $tb=uc$ with $r$ and $t$ nonzero. By assumption, there is a nonzero $r's=r''t\in Rs\cap Rt$, so $r'ra=r''uc$, hence $a$ is algebraic over $c$.$\square$



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> > **Theorem.** All basis of $M$ have the same cardinality, written $\dim\_R M$.
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**Proof.** Standard argument using transitivity and exchange property: Treat the particular case where $M$ has a finite basis $\bar b=(b\_1,\dots,b\_n)$. Let $(c\_1,\dots,c\_m,\dots)$ be another basis of $M$. By maximality of $\bar b$, one can write $\displaystyle rc\_1=\sum r\_ib\_i$ for some non-zero $r\in R$. As $c\_1$ is free, $r\_1$ say is nonzero. So $b\_1$ is algebraic over $(c\_1,b\_2,\dots,b\_n)$. As $M$ is algebraic over $\bar b$, by transitivity, $M$ is algebraic over $(c\_1,b\_2,\dots,b\_n)$. One concludes in a similar way that $M$ is algebraic over $(c\_1,c\_2,b\_3,\dots,b\_n)$, and iterating, one can add every $c\_i$. If $m>n$, we conclude that $c\_{m}$ is algebraic over its predecessors, a contradiction, so $m\leqslant n$, and all basis of $M$ are finite. By symmetry, one has $n=m$.$\square$


**Lemma.** Let $M$ be an $R$-module.


(1) Let $N$ be an $R$-module, then $$\dim\_R M\oplus N=\dim\_R M+\dim\_R N.$$


(2) Let $N\subset M$ be a submodule, then $$\dim\_R M=\dim\_R M/N+\dim\_R N.$$



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> > (3) Let $f:M\rightarrow N$ be a morphism of $R$-module, then $$\dim\_R M=\dim\_R\ker f+\dim\_R {\rm Im} f.$$
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**Proof (sketch).** (1) If $\bar b$ and $\bar c$ are basis of $M$ and $N$, there union is a basis of $M\oplus N$ (uses the Ore-condition again). (2) If $\bar b+N$ is a basis for $M/N$ and $\bar c$ for $N$, then $\bar b\cup \bar c$ is a basis for $M$. (3) Considering the induced bijection $M/\ker f\rightarrow {\rm Im}f$ and in view of (2), we may assume that $f$ is bijective. It is then straightforeward that $\bar b$ is independent in $M$ iff $f(\bar b)$ is independent in $N$.$\square$



Dimension of a module over a left-Ore domain
I do not know of an explicit reference, but the facts in your question follow from the exactness of Ore localization and the corresponding linear algebra facts. A possible reference for the exactness of Ore localization is Exercise 18 at the end of $\S$10 in Lam's "Lectures on Modules and Rings".


In more detail, write $S=R\setminus\{0\}$ and suppose $R$ is a left Ore domain with left fraction division ring $D=S^{-1}R$. Given a left $R$-module $M$, it is
easy to see using $\ker(M\to S^{-1}M)=\{m\in M\,:\,sm=0~\text{for some}~s\in S\}$ (op.cit.), that a collection of elements $\{m\_\lambda\}\_\lambda$ in $M$ is $R$-independent if and only if its image in $S^{-1}M$ is $D$-independent. Since every element $m\in S^{-1}M$ admits some $s\in S$ with $sm\in \mathrm{im}(M\to S^{-1}M)$, one can turn any $D$-independent collection in $S^{-1}M$ into an $R$-independent collection of the same cardinality in $M$. This means that $\dim\_R M$ defined in your question is just the $D$-dimension of the left $D$-vector space $S^{-1}M$. In particular, all maximal $R$-independent sets in $M$ have the same cardinality, namely $\dim\_D S^{-1}M$. 


The rank-nullity theorem can be derived from this observation together with the fact that the functor $M\mapsto S^{-1}M$ is exact. Indeed, if $f:M\to N$ is a $R$-module homomorphism with kernel $K$ and image $I$, then $0\to S^{-1}K\to S^{-1}M\to S^{-1}I\to 0$ is exact, and taking $D$-dimensions gives $\dim\_R K+\dim\_R I=\dim\_R M$.


