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A valuation ring for a nontrivial real-valued rank-one valuation is a principal ideal domain exactly when its value group is discrete. A least positive value makes the value group cyclic and lets every nonzero ideal be generated by an element of smallest valuation. Conversely, if the maximal ideal is principal, its generator has the least positive value. Nontriviality matters: a field with the trivial valuation is a principal ideal domain, but not a discrete valuation ring in the usual nontrivial convention.

A principal ideal ring is a ring whose left ideals and right ideals each have one generator. For a commutative ring, this says that every ideal is a principal ideal. The examples here use commutative rings. Unlike a principal ideal domain, it may have zero divisors: $\mathbb{Z}/4\mathbb{Z}$ has this property, with ideals $(0)$, $(2)$, and $(1)$. A finite direct product of rings that are principal ideal rings is again a principal ideal ring. Indeed, the coordinate idempotent elements show that every ideal is a product of coordinate ideals; a tuple of their generators generates the product.

For a module $M$ over an integral domain $R$ with field of fractions $K$, its rank is the dimension of $K{\otimes }_{R}M$. The tensor product of modules here extends scalars to a field, turning the question into a dimension calculation. For a free module $R^r$, the result is $r$.

For a finitely generated module over a principal ideal domain, the structure theorem for finitely generated modules over a principal ideal domain gives $M\cong R^r\oplus {⨁}_{i}R/(d_i)$, with $d_i$ nonzero nonunits. Each torsion element of a module vanishes after extension of scalars, because its annihilating nonzero scalar becomes invertible in $K$. Thus $K{\otimes }_{R}M\cong K^r$, proving that the rank is the number of free summands. The rank does not record the torsion submodule: for example, $R\oplus R/(d)$ and $R$ have the same rank.

For every matrix $A$ over a principal ideal domain, there are invertible matrices $U,V$ such thatThe nonzero diagonal entries are unique up to multiplication by units. Over the integers, $U$ and $V$ are unimodular and the $d_i$ may be chosen positive; their products $d_1\cdots d_k$ are the greatest common divisors of the $k$-rowed minors.

Let $M$ be an $m$-by-$n$ matrix over a principal ideal domain $R$. Choose Smith normal form $PMQ=D$ with exactly $r$ nonzero diagonal entries. Since $P$ and $Q$ are invertible matrices, $Mx=0$ is equivalent to $D(Q^{-1}x)=0$. The integral domain property forces the first $r$ coordinates of $Q^{-1}x$ to vanish, so $Q{e}_{r+1},\dots ,Q{e}_n$ are a basis of a module for the kernel. These vectors form part of a basis of a module of $R^n$, which is stronger than merely finding a nonzero vector in the kernel over the fraction field.

A module $M$ over an integral domain $R$ is torsion-free when $rm=0$ with $0\ne r\in R$ implies $m=0$, equivalently when its torsion submodule is zero. Every free module over an integral domain is torsion-free, as one can check in each coefficient. Every vector space is torsion-free over its scalar field.

The converse requires hypotheses. A finitely generated module over a principal ideal domain is free if it is torsion-free, but the module $\mathbb{Q}$ over $\mathbb{Z}$ is torsion-free and not free. Any two nonzero rational numbers have a nontrivial integer relation, so a possible basis of a module could have at most one element; no single $q$ generates all rational numbers, since $q/2\notin \mathbb{Z}q$ for $q\ne 0$. The finite-generation hypothesis cannot be dropped. See Keith Conrad's Modules over a PID, Corollary 2.6.

Suppose the ring of integers of a number field is a principal ideal domain, $a,b$ are integral, and $b$ is a unit. On $y^2=x(x^2+ax+b)$ every nonzero $x$-coordinate has even valuation at every finite prime: positive valuation gives $2v(y)=v(x)$, and negative valuation gives $2v(y)=3v(x)$. Thus the two-torsion square-class homomorphism takes values among unit square classes. For the Gaussian integers, their unit group modulo squares has two elements, represented by $1,i$. This can reduce a two-isogeny descent to a very small calculation even though the field contains no ordering.