# Dimension of a module over a left-Ore domain

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:

Do you happen to know any reference for that?

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 $$rv\in(S)$$.

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 $$C$$.

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 $$c$$.$$\square$$

Theorem. All basis of $$M$$ have the same cardinality, written $$\dim_R M$$.

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 $$\dim_R M=\dim_R M/N+\dim_R N.$$

(3) Let $$f:M\rightarrow N$$ be a morphism of $$R$$-module, then $$\dim_R M=\dim_R\ker f+\dim_R {\rm Im} f.$$

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 $$N$$.$$\square$$

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 $$\dim_R K+\dim_R I=\dim_R M$$.