In this article, we discuss some fundamental results concerning modules over a principal ideal domain (PID). These results generalize several classical properties of abelian groups. Recall that every abelian group can be viewed as a module over the ring \( \mathbb{Z} \), and since \( \mathbb{Z} \) is a PID, many theorems about abelian groups are special cases of more general results about modules over PIDs.

Throughout this discussion, we assume that \( R \) is a principal ideal domain with unity, and that all \( R \)-modules considered are unitary.

Proposition

Let \( F \) be a free module over \( R \) and let \( M \) be a submodule of \( F \). Then \( M \) is also a free \( R \)-module, and \[ \dim_R M \le \dim_R F. \]

This result shows that every submodule of a free module over a PID is itself free, and its dimension does not exceed that of the original module.

Proof 

For simplicity, we assume that \( F \) has finite dimension. Let \[ B = \{x_1, x_2, \dots, x_n\} \] be a basis of \( F \).

We aim to show that the submodule \( M \) has a basis with at most \( n \) elements.

Define \[ M_r = M \cap \operatorname{Span}(x_1, \dots, x_r), \quad 1 \le r \le n. \] Clearly, \( M_n = M \). We proceed by induction on \( r \).

Base Case (\( r = 1 \))

Every element of \( M_1 \) has the form \( a x_1 \), where \( a \in R \).

Let \[ I = \{ a \in R \mid a x_1 \in M_1 \}. \]

Then \( I \) is an ideal of \( R \).  (Easy to prove!)

Since \( R \) is a PID, there exists \( a_1 \in R \) such that \( I = (a_1) \).

Thus, \[ M_1 = (a_1 x_1). \]

If \( a_1 = 0 \), then \( M_1 = \{0\} \).

Otherwise, \( M_1 \) is generated by the single element \( a_1 x_1 \).

Hence, in both cases, \( M_1 \) is free.

Induction Hypothesis

Assume that for some \( r < n \), the module \( M_r \) is free and has a basis \( B_r \) with at most \( r \) elements.

Inductive Step

Consider the submodule \( M_{r+1} \).

Define \[ J = \left\{ a \in R \;\middle|\; a x_{r+1} + \sum_{i=1}^{r} a_i x_i \in M_{r+1} \text{ for some } a_i \in R \right\}. \]

Then \( J \) is an ideal of \( R \),        (verify!)

and hence \( J = (a_{r+1}) \) for some \( a_{r+1} \in R \).

If \( a_{r+1} = 0 \), then no new element is obtained, and we have \( M_{r+1} = M_r \).

Otherwise, choose \[ x_0 = a_{r+1} x_{r+1} + \sum_{i=1}^{r} a_i x_i \in M_{r+1}. \]

It can be easily verified that every element of \( M_{r+1} \) can be written as a linear combination of elements of \( M_r \) and \( x_0 \), and that \[ M_r \cap R x_0 = \{0\}. \] Therefore, \[ M_{r+1} = M_r \oplus R x_0. \]

Hence, a basis for \( M_{r+1} \) is given by \[ B_{r+1} = B_r \cup \{x_0\}. \]

Conclusion

By induction, we conclude that \( M = M_n \) is free and has a basis with at most \( n \) elements. Therefore, \[ \dim_R M \le \dim_R F. \]

This completes the proof.

Summary