# nLab principal ideal domain

### Context

#### Algebra

higher algebra

universal algebra

# Contents

## 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 $R\left[x\right]$ 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},\phantom{\rule{2em}{0ex}}{F}_{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 $\left(x,r\right)$ where $x\in {F}_{ 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 ${p}_{j}\left(\left(x,r\right)\right)=r$. The kernel of ${p}_{j}$ is $M\cap {F}_{, and we have an exact sequence

$0\to M\cap {F}_{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 ${p}_{k}\left({m}_{k}\right)={r}_{k}$. We claim that $\left\{{m}_{k}:k\in K\right\}$ forms a basis for $M$.

First we prove linear independence of $\left\{{m}_{k}\right\}$. 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 ${a}_{n}{p}_{{k}_{n}}\left({m}_{{k}_{n}}\right)={a}_{n}{r}_{{k}_{n}}=0$. 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}_{, so that $m\in M\cap {F}_{\le i}$ for some $i, but this contradicts minimality of $j$. Therefore $j\in K$. Now, we have ${p}_{j}\left(m\right)=r\cdot {r}_{j}$ for some $r$; put $m\prime =m-r{m}_{j}$. Clearly

${p}_{j}\left(m\prime \right)={p}_{j}\left(m\right)-r\cdot {p}_{j}\left({m}_{j}\right)=0$p_j(m') = p_j(m) - r \cdot p_j(m_j) = 0

and so $m\prime \in M\cap {F}_{. Thus $m\prime \in M\cap {F}_{\le i}$ for some $i. 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 $ℤ$ for a pid, and abelian groups are the same as $ℤ$-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 $M\to {\lambda }_{r}\left(M\right)$ 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

$\begin{array}{ccc}M& \to & K{\otimes }_{R}M\\ {\lambda }_{r}↓& & ↓1\otimes {\lambda }_{r}\\ M& \to & K{\otimes }_{R}M\end{array}$\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

Revised on October 24, 2012 01:32:45 by Toby Bartels (64.89.53.177)