nLab
principal ideal domain

Definition

A ring $R$ is a principal ideal domain if

1. it is an integral domain;

2. every ideal in $R$ is a principal ideal.

Often pid is used as an abbreviation of “principal ideal domain”.

Examples

• any field

• the ring of integers

• a polynomial ring with coefficients in a field (in fact, for any commutative ring $R$, the ring $R[x]$ is a pid if and only if $R$ is a field)

• a discrete valuation ring (for example, a ring of formal power series over a field)

• the ring of entire holomorphic functions on $ℂ$

Structure theory of modules

Free and projective modules

Theorem

If $R$ is a pid, then any submodule $M$ of a free module $F$ over $R$ is also free. (For the converse statement, see here.)

Proof

By freeness of $F$, there exists an isomorphism $F\cong {\sum }_{j\in J}{R}_j$, a coproduct of copies $R_j$ of $R$ (as a module over the ring $R$) indexed over a set $J$, which we assume well-ordered using the axiom of choice. Define submodules of $F$:

$$F_{\le j}≔\sum _{i\le j}{R}_i,F_{<j}≔\sum _{i<j}{R}_i.$$F_{\leq j} \coloneqq \sum_{i \leq j} R_i, \qquad F_{\lt j} \coloneqq \sum_{i \lt j} R_i\,.

Any element of $M\cap F_{\le j}$ can be written uniquely as $(x,r)$ where $x\in F_{<j}$ and $r\in R_j$. Define a homomorphism

$$p_j:M\cap F_{\le j}\to R$$p_j \colon M \cap F_{\leq j} \to R

by $p_j((x,r))=r$. The kernel of $p_j$ is $M\cap F_{<j}$, and we have an exact sequence

$$0\to M\cap F_{<j}\to M\cap F_{\le j}\to im{p}_j\to 0$$0 \to M \cap F_{\lt j} \to M \cap F_{\leq j} \to \im p_j \to 0

where $im{p}_j$ is a submodule (i.e., an ideal) of $R$, hence generated by a single element $r_j$. Let $K\subseteq J$ consist of those $j$ such that $r_j\ne 0$, and for $k\in K$, choose $m_k$ such that $p_k(m_k)=r_k$. We claim that $\{m_k:k\in K\}$ forms a basis for $M$.

First we prove linear independence of $\{m_k\}$. Suppose ${\sum }_{i=1}^n{a}_i{m}_{k_i}=0$, with $k_1<k_2<\dots <k_n$. Applying $p_{k_n}$, we get $a_n{p}_{k_n}(m_{k_n})=a_n{r}_{k_n}=0$. Since $r_{k_n}\ne 0$, we have $a_n=0$ (since we are working over a domain). The assertion now follows by induction.

Now we prove that the $m_k$ generate $M$. Assume otherwise, and let $j\in J$ be the least $j$ such that some $m\in M\cap F_{\le j}$ cannot be written as a linear combination of the $m_k$, for $k\in K$. If $j\notin K$, then $m\in M\cap F_{<j}$, so that $m\in M\cap F_{\le i}$ for some $i<j$, but this contradicts minimality of $j$. Therefore $j\in K$. Now, we have $p_j(m)=r\cdot r_j$ for some $r$; put $m\prime =m-r{m}_j$. Clearly

$$p_j(m\prime )=p_j(m)-r\cdot p_j(m_j)=0$$p_j(m') = p_j(m) - r \cdot p_j(m_j) = 0

and so $m\prime \in M\cap F_{<j}$. Thus $m\prime \in M\cap F_{\le i}$ for some $i<j$. At the same time, $m\prime$ cannot be written as a linear combination of the $m_k$; again, this contradicts minimality of $j$. Thus the $m_k$ generate $M$, as claimed.

Since the integers $\mathbb{Z}$ for a pid, and abelian groups are the same as $\mathbb{Z}$-modules, we have

Corollary

(Dedekind) A subgroup of a free abelian group is also free abelian.

Remark

The analog of this statement for possibly non-abelian groups is the Nielsen-Schreier theorem.

Also, since projective modules are retracts of free modules, we have

Corollary

Projective modules over a pid are free. In particular, submodules of projective modules are projective.

Torsion-free modules

Proposition

A finitely generated torsionfree module $M$ over a pid $R$ is free.

Proof

Let $S$ be a finite set of generators of $M$, and let $T\subseteq S$ be a maximal subset of linearly independent elements. (Unless $M=0$, then $T$ has at least one element, because $M$ is torsionfree.) We claim that $M$ can be embedded as a submodule of the free module $F$ generated by $T$ (which in turn is the span of $T$ as a submodule $F\subseteq M$). By Theorem 1, it follows that $M$ is free.

Let $x_1,\dots ,x_n$ be the elements of $T$. It follows from maximality of $T$ that for any $m\in S-T$, there is a linear relation

$$r_{m}m+r_1{x}_1+\dots +r_n{x}_n=0$$r_m m + r_1 x_1 + \ldots + r_n x_n = 0

with $r_m\ne 0$. For each $m$ in the complement $S-T$, pick such an $r_m$, and form $r={\prod }_{m\in S-T}{r}_m$. Then the image of the scalar multiplication ${\lambda }_r:M\to M$ factors through $F\subseteq M$, and $M\to {\lambda }_r(M)$ is monic because $M$ is torsionfree. This completes the proof.

Proposition

Let $R$ be a pid. Then an $R$-module $M$ is torsionfree if and only if it is flat.

Proof

Suppose $M$ is flat. Let $K$ be the field of fractions of $R$; since $R$ is a domain, we have a monic $R$-module map $R\to K$. By flatness, we have an induced monomorphism $M\cong R{\otimes }_{R}M\to K{\otimes }_{R}M$. For any nonzero $r\in R$, the naturality square

$$\begin{array}{ccc}M& \to & K{\otimes }_{R}M\\ {\lambda }_r\downarrow & & \downarrow 1\otimes {\lambda }_r\\ M& \to & K{\otimes }_{R}M\end{array}$$\array{ M & \to & K \otimes_R M \\ \mathllap{\lambda_r} \downarrow & & \downarrow \mathrlap{1 \otimes \lambda_r} \\ M & \to & K \otimes_R M }

commutes, and since the map $1\otimes {\lambda }_r$ is multiplication by a non-zero scalar on a vector space, it follows that this map and therefore also ${\lambda }_r$ is monic, i.e., $M$ is torsionfree.

In the other direction, suppose $M$ is torsionfree. Any module is the filtered colimit over the system of finitely generated submodules and inclusions between them; in this case all the finitely generated submodules of $M$ are torsion-free and hence free, by Proposition 1. Thus $M$ is a filtered colimit of free modules; it is therefore flat by a standard result proved here.

Structure theory of finitely generated modules

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