A Module That Has A Basis Shorter Than Its Rank?
Answer : This is actually proven in Dummit and Foote. To distinguish the two definitions of rank, call the maximum number of elements of M M M which are linearly independent the LI rank , and for a free R R R -module M M M call the size of a basis of M M M the free rank . As you observe, since a basis must be linearly independent, then free rank ≤ \leq ≤ LI rank. Proposition 3 of § 12.1 \S12.1 §12.1 of Dummit and Foote (p. 459) implies the reverse inequality. Proposition 3. Let R R R be an integral domain and let M M M be a free R R R -module of rank n < ∞ n < \infty n < ∞ . Then any n + 1 n+1 n + 1 elements of M M M are R R R -linearly dependent, i.e., for any y 1 , … , y n + 1 ∈ M y_1, \ldots, y_{n+1} \in M y 1 , … , y n + 1 ∈ M there are elements r 1 , … , r n + 1 ∈ R r_1, \ldots, r_{n+1} \in R r 1 , … , r n + 1 ∈ R , not all zero, such that r 1 y 1 + ⋯ + r n + 1 y n + 1 = 0 . r_1 y_1 + \cdots + r_{n+1} y_{n+1} = 0 \, . r 1 y 1 ...