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 $M$ which are linearly independent the LI rank, and for a free $R$-module $M$ call the size of a basis of $M$ the free rank.

As you observe, since a basis must be linearly independent, then free rank $\leq$ LI rank. Proposition 3 of $\S12.1$ of Dummit and Foote (p. 459) implies the reverse inequality.

Proposition 3. Let $R$ be an integral domain and let $M$ be a free $R$-module of rank $n < \infty$. Then any $n+1$ elements of $M$ are $R$-linearly dependent, i.e., for any $y_1, \ldots, y_{n+1} \in M$ there are elements $r_1, \ldots, r_{n+1} \in R$, not all zero, such that $$r_1 y_1 + \cdots + r_{n+1} y_{n+1} = 0 \, .$$

If you know about localization, you can give a quick proof of this by passing to the field of fractions $F$ of $R$. Let $S = R \setminus \{0\}$ so $S^{-1}R = F$. Then $S^{-1} M$ is a vector space over the field $F$, so a maximal linearly independent set is a basis by results about vector spaces. There is an alternative proof included in Dummit and Foote.

Just as a note, on p. 460 Dummit and Foote make the same definition of rank as your professor and remark on the equivalence of the definitions.