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principal ideal domain

Contents

Definition

A ring RR is a principal ideal domain if

  1. it is an integral domain;

  2. every ideal in RR is a principal ideal.

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

Examples

Structure theory of modules

Free and projective modules

Theorem

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.)

Proof

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} 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 MM over a pid RR is free.

Proof

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 1, 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.

Proposition

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

Proof

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 1. 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

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