A commutative ring \(R\) with identity is local if it has exactly one maximal ideal. Let \(R\) be a local ring which is a PID, but is not a field.
Classify the possible finitely generated \(R\)-modules, in terms of their invariant factors (or elementary divisors). Show that there is only a countably infinite number of finitely generated \(R\)-modules, up to isomorphism.
Prove that if \(M\) and \(N\) are nonzero, finitely generated, torsion \(R\)-modules, then there exists a nonzero \(R\)-module homomorphism \(\phi: M \to N\).
Solution
Let \(\mathfrak m\) be the unique maximal ideal of \(R\). Since \(R\) is a PID and is not a field, \[ \mathfrak m=(\pi) \] for some nonzero nonunit \(\pi\in R\).
1. Every irreducible element of \(R\) is associate to \(\pi\).
Let \(q\in R\) be irreducible. Then \((q)\) is a nonzero prime ideal of the PID \(R\), hence a maximal ideal. Since \(R\) has only one maximal ideal, \[ (q)=\mathfrak m=(\pi). \] Thus \(q\) and \(\pi\) are associates.
2. Every finitely generated \(R\)-module is isomorphic to\[ R^r\oplus \bigoplus_{i=1}^t R/(\pi^{e_i}) \]for integers \(r,t\ge0\) and positive integers \(e_i\); after reordering one may assume\[ 1\le e_1\le e_2\le\cdots\le e_t. \]
The structure theorem for finitely generated modules over a PID gives \[ M\cong R^r\oplus\bigoplus_j R/(d_j), \] where the nonzero nonunits \(d_j\) may be chosen in invariant-factor form, or equivalently decomposed into powers of irreducibles in elementary-divisor form. By 1, every irreducible of \(R\) is associate to \(\pi\). Hence every elementary divisor is associate to a power \(\pi^e\), giving the displayed decomposition. Conversely every such finite direct sum is a finitely generated \(R\)-module.
3. There are only countably infinitely many isomorphism classes of finitely generated \(R\)-modules.
By 2, an isomorphism class is determined by a nonnegative integer \(r\) and a finite nondecreasing sequence \[ (e_1,\ldots,e_t) \] of positive integers. The set of all finite sequences of integers is countable, so there are at most countably many isomorphism classes. There are infinitely many, since \[ R/(\pi),\ R/(\pi^2),\ R/(\pi^3),\ldots \] are pairwise nonisomorphic: the annihilator of \(R/(\pi^n)\) is \((\pi^n)\), and these ideals are distinct. Thus there are exactly countably infinitely many isomorphism classes.
4. For all integers \(a,b\ge1\), there is a nonzero homomorphism\[ R/(\pi^a)\longrightarrow R/(\pi^b). \]
If \(a\ge b\), define \[ \varphi(\overline r)=\overline r. \] This is well-defined because \[ (\pi^a)\subseteq(\pi^b), \] and it is nonzero.
If \(a<b\), define \[ \varphi(\overline r)=\overline{r\pi^{b-a}}. \] If \(r-r'\in(\pi^a)\), then \[ (r-r')\pi^{b-a}\in(\pi^b), \] so the map is well-defined. It is nonzero because \[ \pi^{b-a}\notin(\pi^b). \]
5. If \(M\) and \(N\) are nonzero finitely generated torsion \(R\)-modules, then there is a nonzero homomorphism \(M\to N\).
By 2, choose cyclic direct summands \[ R/(\pi^a)\subseteq M, \qquad R/(\pi^b)\subseteq N \] with \(a,b\ge1\). Let \[ p:M\twoheadrightarrow R/(\pi^a) \] be the projection onto the chosen summand and \[ i:R/(\pi^b)\hookrightarrow N \] the inclusion of the chosen summand. By 4 there is a nonzero map \[ \varphi:R/(\pi^a)\to R/(\pi^b). \] Then \[ i\circ\varphi\circ p:M\to N \] is nonzero: choose \(x\) in the selected summand of \(M\) with \(\varphi(x)\ne0\). Thus the required homomorphism exists.