nLab principal ideal domain

Definition

An integral domain $R$ is a principal ideal domain if it satisfies any of these equivalent conditions:

1. Every ideal in $R$ is a principal ideal.

2. $R$ is a Bézout unique factorization domain

3. $R$ is a Noetherian Bézout domain

4. $R$ is a Bézout domain which satisfies the ascending chain condition on principal ideals

5. $R$ is a Bézout atomic domain

6. $R$ has a Dedekind-Hasse norm

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

To show that definitions (2-4) are equivalent, consider that unique factorization domains and Noetherian domains both satisfy the ascending chain condition on principal ideals. In addition, every GCD domain which satisfies the ascending chain condition on principal ideals is a unique factorization domain. Since Bézout domains are GCD domains, definitions (2-4) are equivalent to each other.

Now to show that definitions 1 is equivalent to (2-4). Every integral domain whose ideals are principal is a Bézout domain by considering the ideals which are finitely generated. To show that these are Noetherian domains and unique factorization domains, we have the following:

Lemma

In a integral domain $R$ whose ideals are principal, an element $x$ is irreducible iff $(x)$ is a maximal ideal.

Proof

Suppose $x$ is irreducible and $a \notin (x)$. Then the ideal generated by $a, x$ is principal, say $(b)$. Then $x = b y$ and since $x$ is irreducible, one of $b, y$ is a unit; if $y$ is a unit then $(x) = (b) = (a, x)$ and thus $a \in (x)$, contradiction. So then $b$ must be a unit, i.e., $(b)$ is the improper ideal. Thus $(x)$ has no proper extension: $(x)$ is maximal.

For any ring $R$, if $(x)$ is maximal and $x = a b$ for a non-unit $a$, then the inclusion $(x) \subseteq (a)$ is an equality by maximality, so $a = x v$ for some $v$. Then $x = x v b$. In an integral domain we conclude $1 = v b$; thus $b$ is a unit.

Theorem

An integral domain $R$ whose ideals are principal is Noetherian.

Proof

The union of an ascending chain of ideals $I_1 \subseteq I_2 \subseteq \ldots$ is an ideal $I$; if $I = (x)$, then $x$ belongs to one of the $I_n$, whereupon $I = I_n$.

Thus, an integral domain $R$ whose ideals are principal is a Noetherian Bézout domain.

Similarly,

Theorem

An integral domain $R$ whose ideals are principal is a unique factorization domain.

Proof

Working in classical logic, Proposition says that the reverse inclusion relation on the set of nonzero ideals is well-founded. Let $A$ be the subset of ideals $(a)$ that are products of finitely many (maybe zero!) maximal principal ideals. For any proper ideal $(x) \neq (0)$, if every $(t)$ properly containing $(x)$ can be factored into maximals, then so can $(x)$. (Spelling this out: either $(x)$ is maximal/irreducible, or factors as $(s)(t)$ where both $s$ and $t$ are non-units; $(s)$ and $(t)$ factor into maximals by hypothesis, and therefore so does $(x)$.) Thus $A$ is an inductive set, so by well-foundedness it contains every ideal $(x) \neq (0)$, i.e., $x$ can be factored into irreducibles.

For uniqueness of the factorization, we first remark that if $p$ is irreducible and $p|a b$, then $p|a$ or $p|b$. (For $R/(p)$ is a field and thus a fortiori an integral domain, so that if $a b \equiv 0 \; mod p$, then $a \equiv 0 \; mod p$ or $b \equiv 0 \; mod p$.) Thus if $p_1 p_2 \ldots p_m = q_1 q_2 \ldots q_n$ are two factorizations into irreducibles of the same element, then $p_1$ divides one of the irreducibles $q_i$, in which case $(p_1) = (q_i)$ and each is a unit times the other, meaning we can cancel $p_1$ on both sides and argue by induction.

Thus, an integral domain $R$ whose ideals are principal is a Bézout unique factorization domain.

It remains to show using classical logic that every ideal of a Noetherian Bézout domain or a Bézout UFD is a principal ideal… (to be done)

Examples

• any field

• the ring $\mathbb{Z}$ 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)

• any Euclidean domain

• in the ring of entire holomorphic functions on $\mathbb{C}$ every finitely generated ideal is principal (Helmer 40), but the ring is only a Bézout domain.

That both the integers and the polynomial rings $\mathbb{F}_q[x]$ over finite fields are principal integral domains with finite group of units is one aspect of the close similarity between the two that is the topic of the function field analogy. That also the holomorphic functions on the complex plane form a Bézout domain may then be viewed as part of the further similarity that relates the previous two to topics such as geometric Langlands duality. See at function field analogy – table for more on this.

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 $F$:

$$F_{\leq j} \coloneqq \sum_{i \leq j} R_i, \qquad F_{\lt j} \coloneqq \sum_{i \lt j} R_i\,.$$

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

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

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

$$0 \to M \cap F_{\lt j} \to M \cap F_{\leq j} \to \im p_j \to 0$$

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 $r_j \neq 0$, 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 $k_1 \lt k_2 \lt \ldots \lt k_n$. Applying $p_{k_n}$, we get $a_n p_{k_n}(m_{k_n}) = a_n r_{k_n} = 0$. Since $r_{k_n} \neq 0$, 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 $m \in M \cap F_{\leq j}$ cannot be written as a linear combination of the $m_k$, for $k \in K$. If $j \notin K$, then $m \in M \cap F_{\lt j}$, so that $m \in M \cap F_{\leq i}$ for some $i \lt j$, but this contradicts minimality of $j$. Therefore $j \in K$. Now, we have $p_j(m) = r \cdot r_j$ for some $r$; put $m' = m - r m_j$. Clearly

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

and so $m' \in M \cap F_{\lt j}$. Thus $m' \in M \cap F_{\leq i}$ for some $i \lt j$. At the same time, $m'$ 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 $\mathbb{Z}$ form 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.

Normal forms

Proposition

(matrices over principal ideal domains equivalent to Smith normal form)

For $R$ a commutative ring which is a principal ideal domain, every matrix $A \in Mat_{n \times m}(R)$ with entries in $R$ is matrix equivalent to a diagonal matrix filled up with zeros:

There exist invertible matrices $P \in Mat_{n \times n}(R)$ and $Q \in Mat_{m \times m}(R)$ such that the product matrix $P A Q$ is a diagonal matrix filled up with zeros:

$$P A Q \;=\;$$

such that, moreover, each $a_i$ divides $a_{i+1}$.

For matrices with coefficients in the pid of integers see also

• Hermite normal form.

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 , it follows that $M$ is free.

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

$$r_m m + r_1 x_1 + \ldots + r_n x_n = 0$$

with $r_m \neq 0$. 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 $\lambda_r \colon M \to M$ 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

$$\array{ M & \to & K \otimes_R M \\ \mathllap{\lambda_r} \downarrow & & \downarrow \mathrlap{1 \otimes \lambda_r} \\ M & \to & K \otimes_R M }$$

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 . 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

structure theorem for finitely generated modules over a principal ideal domain

In constructive mathematics

First of all, in constructive mathematics, there are various notions of integral domains, such as Heyting integral domains and discrete integral domains. These already result in multiple notions of principal ideal domains, such as Heyting principal ideal domains and discrete principal ideal domains.

In addition, the various definitions of principal ideal domains do not coincide with each other in constructive mathematics. For example, consider the classically true equivalence of discrete Bézout unique factorization domains and discrete integral domains where every ideal is principal:

Theorem

Given a discrete Bézout unique factorization domain $R$, every ideal of $R$ is a principal ideal if and only if the law of excluded middle holds.

Proof

In one direction, the usual proofs rely on being able to decide whether any particular element of $R$ belongs to the ideal or not. For the converse, let $\varphi$ be an arbitrary proposition. Consider the ideal $\{ x \in R | (x = 0) \vee \varphi \}$. By assumption, it is generated by the elements of $R$. Since $R$ is a discrete integral domain, it holds that $x = 0$ or $x \neq 0$. In the first case $\neg\varphi$ holds, in the second $\varphi$.

It is unknown whether in the absence of excluded middle, there are Heyting integral domains where every ideal is a principal ideal. But with Heyting integral domains, maybe we should be talking about principal anti-ideal domains instead of principal ideal domains.

See also

• principal ideal

• principal ideal ring

References

• O. Helmer, Divisibility properties of integral functions, Duke Math. J. 6 (1940), 345-356.

• Wikipedia, Principal ideal domain

• Eric Wofsey, Principal Ideal Domains, (written for Mathcamp 2009) pdf

• Henri Lombardi, Claude Quitté (2010): Commutative algebra: Constructive methods (Finite projective modules) Translated by Tania K. Roblo, Springer (2015) (doi:10.1007/978-94-017-9944-7, pdf)