Remark

If $A$ is an integral domain, a projective $A$-*module admits a well-defined rank* (in the sense of Definition 2), as has been observed above; moreover, this rank coincides with the rank defined in Algebra, Chapter II, § 7, no. 2; it is sufficient to apply Theorem 2 (c) with $p = (0)$.

Let $E$ and $F$ be two finitely generated projective $A$-modules. We know (Algebra, Chapter II, §§ 2 and 3) that $E \times F$, $E \otimes_A F$, $\mathrm{Hom}_A(E, F)$ and the dual $E^*$ of $E$ are projective and finitely generated; so is the exterior power $\bigwedge^k E$ for every integer $k > 0$ (Algebra, Chapter 111). Also, it follows immediately from Definition 1 and § 2, no. 7, Propositions 18 and 19 and no. 8, that, for every prime ideal $p$ of $A$:

(1)

(2) $\mathrm{rg}_p(E \otimes_A F) = \mathrm{rg}_p(E) \cdot \mathrm{rg}_p(F)$

(3) $\mathrm{rg}_p(\mathrm{Hom}_A(E, F)) = \mathrm{rg}_p(E) \cdot \mathrm{rg}_p(F)$

(4) $\mathrm{rg}_p(E^*) = \mathrm{rg}_p(E)$

(5) $\mathrm{rg}_p(\wedge^k E) = \binom{\mathrm{rg}_p(E)}{k}$.

If the ranks of E and F are defined, so are those of $E \times F, E \otimes_A F,$ $\mathrm{Hom}_A(E, F), E^*$ and $\wedge^k E$ and the above equations also hold with the index $p$ omitted. Moreover: