# Rank of a module

The rank of a left module $M$ over a ring $R$ imbeddable in a skew-field $k$ is the dimension of the tensor product $k \otimes _ {R} M$, regarded as a vector space over $k$. If $R = \mathbf Z$, the ring of integers, the definition coincides with the usual definition of the rank of an Abelian group (cf. Rank of a group). If $k$ is a flat $R$- module (say, $k$ is the skew-field of fractions of $R$, cf. Flat module), then the ranks of the modules in an exact sequence

$$0 \rightarrow M ^ \prime \rightarrow M \rightarrow M ^ {\prime\prime} \rightarrow 0$$

satisfy the equality

$$\mathop{\rm rk} M = \mathop{\rm rk} M ^ \prime + \mathop{\rm rk} M ^ {\prime\prime} .$$

The rank of a free module $M$ over an arbitrary ring $R$( cf. Free module) is defined as the number of its free generators. For rings that can be imbedded into skew-fields this definition coincides with that in 1). In general, the rank of a free module is not uniquely defined. There are rings (called $n$- FI-rings) such that any free module over such a ring with at most $n$ free generators has a uniquely-defined rank, while for free modules with more than $n$ generators this property does not hold. A sufficient condition for the rank of a free module over a ring $R$ to be uniquely defined is the existence of a homomorphism $\phi : R \rightarrow k$ into a skew-field $k$. In this case the concept of the rank of a module can be extended to projective modules as follows. The homomorphism $\phi$ induces a homomorphism of the groups of projective classes $\phi ^ {*} : K _ {0} R \rightarrow K _ {0} k \approx \mathbf Z$, and the rank of a projective module $P$ is by definition the image of a representative of $P$ in $\mathbf Z$. Such a homomorphism $\phi$ exists for any commutative ring $R$.

#### References

 [1] P.M. Cohn, "Free rings and their relations" , Acad. Press (1971) [2] J.W. Milnor, "Introduction to algebraic -theory" , Princeton Univ. Press (1971)

The rank of a projective module $P$, as defined here, depends on the choice of $\phi$.