nLab principal ideal domain




A ring RR is a principal ideal domain if

  1. it is an integral domain (hence in particular a commutative ring)

  2. every ideal in RR is a principal ideal.

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

A pid can also be characterized as an integral domain wich possesses a Dedekind-Hasse norm, making very clear the fact that it is a generalization of an Euclidean domain.


That both the integers and the polynomial rings 𝔽 q[x]\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.



In a pid RR, an element xx is irreducible iff (x)(x) is a maximal ideal.


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

For any ring RR, if (x)(x) is maximal and x=abx = a b for a non-unit aa, then the inclusion (x)(a)(x) \subseteq (a) is an equality by maximality, so a=xva = x v for some vv. Then x=xvbx = x v b. In an integral domain we conclude 1=vb1 = v b; thus bb is a unit.


A pid is a Noetherian ring.


The union of an ascending chain of ideals I 1I 2I_1 \subseteq I_2 \subseteq \ldots is an ideal II; if I=(x)I = (x), then xx belongs to one of the I nI_n, whereupon I=I nI = I_n.


A PID is a unique factorization domain.


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

For uniqueness of the factorization, we first remark that if pp is irreducible and p|abp|a b, then p|ap|a or p|bp|b. (For R/(p)R/(p) is a field and thus a fortiori an integral domain, so that if ab0modpa b \equiv 0 \; mod p, then a0modpa \equiv 0 \; mod p or b0modpb \equiv 0 \; mod p.) Thus if p 1p 2p m=q 1q 2q np_1 p_2 \ldots p_m = q_1 q_2 \ldots q_n are two factorizations into irreducibles of the same element, then p 1p_1 divides one of the irreducibles q iq_i, in which case (p 1)=(q i)(p_1) = (q_i) and each is a unit times the other, meaning we can cancel p 1p_1 on both sides and argue by induction.

Structure theory of modules

Free and projective modules


If RR is a pid, then any submodule MM of a free module FF over RR is also free. (For the converse statement, see here.)


By freeness of FF, there exists an isomorphism F jJR jF \cong \sum_{j \in J} R_j, a coproduct of copies R jR_j of RR (as a module over the ring RR) indexed over a set JJ, which we assume well-ordered using the axiom of choice. Define submodules of FF:

F j ijR i,F <j i<jR 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 MF jM \cap F_{\leq j} can be written uniquely as (x,r)(x, r) where xF <jx \in F_{\lt j} and rR jr \in R_j. Define a homomorphism

p j:MF jRp_j \colon M \cap F_{\leq j} \to R

by p j((x,r))=rp_j((x, r)) = r. The kernel of p jp_j is MF <jM \cap F_{\lt j}, and we have an exact sequence

0MF <jMF jimp j00 \to M \cap F_{\lt j} \to M \cap F_{\leq j} \to \im p_j \to 0

where imp j\im p_j is a submodule (i.e., an ideal) of RR, hence generated by a single element r jr_j. Let KJK \subseteq J consist of those jj such that r j0r_j \neq 0, and for kKk \in K, choose m km_k such that p k(m k)=r kp_k(m_k) = r_k. We claim that {m k:kK}\{m_k: k \in K\} forms a basis for MM.

First we prove linear independence of {m k}\{m_k\}. Suppose i=1 na im k i=0\sum_{i=1}^{n} a_i m_{k_i} = 0, with k 1<k 2<<k nk_1 \lt k_2 \lt \ldots \lt k_n. Applying p k np_{k_n}, we get a np k n(m k n)=a nr k n=0a_n p_{k_n}(m_{k_n}) = a_n r_{k_n} = 0. Since r k n0r_{k_n} \neq 0, we have a n=0a_n = 0 (since we are working over a domain). The assertion now follows by induction.

Now we prove that the m km_k generate MM. Assume otherwise, and let jJj \in J be the least jj such that some mMF jm \in M \cap F_{\leq j} cannot be written as a linear combination of the m km_k, for kKk \in K. If jKj \notin K, then mMF <jm \in M \cap F_{\lt j}, so that mMF im \in M \cap F_{\leq i} for some i<ji \lt j, but this contradicts minimality of jj. Therefore jKj \in K. Now, we have p j(m)=rr jp_j(m) = r \cdot r_j for some rr; put m=mrm jm' = m - r m_j. Clearly

p j(m)=p j(m)rp j(m j)=0p_j(m') = p_j(m) - r \cdot p_j(m_j) = 0

and so mMF <jm' \in M \cap F_{\lt j}. Thus mMF im' \in M \cap F_{\leq i} for some i<ji \lt j. At the same time, mm' cannot be written as a linear combination of the m km_k; again, this contradicts minimality of jj. Thus the m km_k generate MM, as claimed.

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


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


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


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

Normal forms


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

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

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

PAQ= P A Q \;=\;

such that, moreover, each a ia_i divides a i+1a_{i+1}.

For matrices with coefficients in the pid of integers see also

Torsion-free modules


A finitely generated torsionfree module MM over a pid RR is free.


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

Let x 1,,x nx_1, \ldots, x_n be the elements of TT. It follows from maximality of TT that for any mSTm \in S - T, there is a linear relation

r mm+r 1x 1++r nx n=0r_m m + r_1 x_1 + \ldots + r_n x_n = 0

with r m0r_m \neq 0. For each mm in the complement STS - T, pick such an r mr_m, and form r= mSTr mr = \prod_{m \in S - T} r_m. Then the image of the scalar multiplication λ r:MM\lambda_r \colon M \to M factors through FMF \subseteq M, and Mλ r(M)M \to \lambda_r(M) is monic because MM is torsionfree. This completes the proof.


Let RR be a pid. Then an RR-module MM is torsionfree if and only if it is flat.


Suppose MM is flat. Let KK be the field of fractions of RR; since RR is a domain, we have a monic RR-module map RKR \to K. By flatness, we have an induced monomorphism MR RMK RMM \cong R \otimes_R M \to K \otimes_R M. For any nonzero rRr \in R, the naturality square

M K RM λ r 1λ r M K RM\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 1λ r1 \otimes \lambda_r is multiplication by a non-zero scalar on a vector space, it follows that this map and therefore also λ r\lambda_r is monic, i.e., MM is torsionfree.

In the other direction, suppose MM 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 MM are torsion-free and hence free, by Proposition . Thus MM 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

In constructive mathematics, many important rings may fail to be principal ideal domains in the naïve sense; the notion of Bézout domain, in which only the finitely generated ideals are required to be principal, is better behaved.

For instance, the ring of integers is a principal ideal domain if and only if the law of excluded middle holds: In one direction, the usual proofs rely on being able to decide whether any particular integer belongs to the ideal or not. For the converse, let φ\varphi be an arbitrary proposition. Consider the ideal {x|(x=0)φ}\{ x \in \mathbb{Z} | (x = 0) \vee \varphi \}. By assumption, it is generated by some number nn. Since the integers are discrete, it holds that n=0n = 0 or n0n \neq 0. In the first case ¬φ\neg\varphi holds, in the second φ\varphi.

However, this ideal cannot be proved to be finitely generated either. If an ideal is generated by n 1,,n kn_1, \ldots, n_k, then we may form their gcd one step at a time, which we can do algorithmically. Therefore, \mathbb{Z} remains a Bézout domain.

On the other hand, we could try to modify the concept of principal ideal domain to recover a concept that is identical to the usual one in classical mathematics but also includes \mathbb{Z}. For instance, we could demand that every decidable ideal is principal, or (potentially more strongly) that any ideal generated by a decidable subset is principal. While these seem to work at first, they are too weak to prove that every PID is a Bézout domain, so we should try to think of something better.

Some authors have tried to define a principal ideal domain as a Noetherian Bézout domain, but it is unknown if this still coincides in constructive mathematics with the definition of principal ideal domain above.

See also


Last revised on August 8, 2022 at 00:28:08. See the history of this page for a list of all contributions to it.