Tag Archives: principal ideal domain

Over a PID which is not a field, no finitely generated module is injective

Let R be a principal ideal domain which is not a field. Prove that no finitely generated R-module is injective.


Let M be a finitely generated R-module. By Theorem 5 (FTFGMPID), we have M \cong_R R^t \oplus \mathsf{Tor}(M) for some t \in \mathbb{N}. By this previous exercise, M is injective if and only if R^t and \mathsf{Tor}(M) are injective. We claim that if M is injective, then M = 0.

We begin with a lemma.

Lemma: If M is a finitely generated torsion module over a principal ideal domain, then there exists a nonzero element r \in R such that rM = 0. Proof: By FTFGMPID, we have M \cong \bigoplus_i R/(p_i^{k_i}) for some primes p_i \in R and nonnegative natural numbers k_i. Now r = \prod_i p_i^{k_i} is nonzero and clearly annihilates M. \square

By Proposition 36 on page 396 of D&F, \mathsf{Tor}(M) (if nontrivial) cannot be injective, since we have rM = 0 for some r. So if M is injective, then \mathsf{Tor}(M) = 0.

Consider now R^t; if t \geq 1, it suffices to consider a single copy of R. If p \in R is not a unit, then pR = (p) \subsetneq R. So again by Proposition 36 on page 396, R is not injective, so that R^t is not injective if t \geq 1.

So if M is injective, then M = 0.

Conversely, the zero module is trivially injective since every short exact sequence 0 \rightarrow 0 \rightarrow M \rightarrow N \rightarrow 0 splits (trivially).

So over a principal ideal domain R which is not a field, no nontrivial module is injective.

Characterize the irreducible torsion modules over a PID

Let R be a principal ideal domain and let M be a torsion R-module. Prove that M is irreducible if and only if M = Rm for some nonzero element m \in M where the annihilator of m in R is a nonzero prime ideal P.


Suppose M is an irreducible, torsion R-module. By this previous exercise, we have M \cong_R R/I where I \subseteq R is a maximal ideal. Since R is a principal ideal domain, in fact I = (p) for some prime element p \in R. Let \psi : R/(p) \rightarrow M be an isomorphism, and let m = \psi(1 + (p)). Now M = Rm, and moreover \mathsf{Ann}_R(m) = \mathsf{Ann}_R(1 + (p)) = (p).

Conversely, consider the module M = Rm, where \mathsf{Ann}_R(m) = (p) is a prime ideal. Define \psi : R \rightarrow (m)_R by r \mapsto r \cdot m. Certainly we have \mathsf{ker}\ \psi = \mathsf{Ann}_R(m) = (p). By the First Isomorphism Theorem for modules, we have \overline{\psi} : R/(p) \rightarrow M an isomorphism. Again by this previous exercise, R/(p) is irreducible.

Describe M/Tor(M) for a finitely generated module M over a PID

Let M be a finitely generated module over a principal ideal domain R. Describe the structure of M/\mathsf{Tor}(M).


We begin with a lemma.

Lemma: Let R be a domain, and let M and N be (left, unital) R-modules. Then \mathsf{Tor}(M \oplus N) = \mathsf{Tor}(M) \oplus \mathsf{Tor}(N). Proof: (\subseteq) Let (m,n) \in \mathsf{Tor}(M \oplus N). Then there exists a nonzero element r \in R such that r(m,n) = (0,0). Then (rm,rn) = (0,0), and so rm = 0 and rn = 0. So m \in \mathsf{Tor}(M) and n \in \mathsf{Tor}(N). (\supseteq) Let (m,n) \in \mathsf{Tor}(M) \oplus \mathsf{Tor}(N). Then there exist nonzero elements r,s \in R such that rm = 0 and sn = 0. Now rs is nonzero in R (since it is a domain), and we have rs(m,n) = (s(rm), r(sn)) = 0. So (m,n) \in \mathsf{Tor}(M \oplus N). \square

Now by the Fundamental Theorem of Finitely Generated Modules over a PID (FTFGMPID), if M is such a module, we have M \cong_R R^t \oplus \bigoplus R/(p_k^{e_k}) for some primes p_k \in R and natural numbers e_k. Now \mathsf{Tor}(M) = \mathsf{Tor}(R^t) \oplus \mathsf{Tor}(\bigoplus R/(p_k^{e_k})) = \bigoplus R/(p_k^{e_k}), since R^t is torsion-free (since R is a domain) while… the stuff inside the big sum is all torsion.

Using this previous exercise, we have that M/\mathsf{Tor}(M) \cong R^t.

Over a PID, isomorphic finitely generated torsion modules have the same elementary divisors

Let R be a principal ideal domain and let p \in R be a prime.

  1. Let M be a finitely generated torsion R-module, let k \in \mathbb{N}, and let n_{p,k} denote the number of elementary divisors of M of the form p^\alpha with \alpha > k. Prove that (p^k)M/(p^{k+1})M \cong_R (R/(p))^{n_{p,k}}.
  2. Suppose M_1 and M_2 are isomorphic finitely generated torsion R-modules. Use part (a) to prove that M_1 and M_2 have the same set of elementary divisors.

By the existence portion of FTFGMPID, M \cong_R \bigoplus_i R/(p_i^{\alpha_i}) where p_i ranges over the set of primes in R (up to associates). Now (p^k)M/(p^{k+1})M \cong_R \left[ \bigoplus (p^k)(R/(p_i^{\alpha_i}) \right] / \left[ \bigoplus (p^{k+1})(R/(p_i^{\alpha_i}) \right] \cong_R \bigoplus (p^k)(R/(p_i^{\alpha_i})/(p^{k+1})(R/(p_i^{\alpha_i}) by this previous exercise. By this previous exercise, this is module-isomorphic to \bigoplus T_i where T_i = R/(p_i) if p_i = p and \alpha_i > k and 0 otherwise. Which, by construction, is isomorphic to (R/(p))^{n_{p,k}}.

Now suppose M_1 and M_2 are isomorphic finitely generated torsion R-modules.

We claim that (p^k)M_1/(p^{k+1})M_1 \cong_R (p^k)M_2/(p^{k+1})M_2 for all primes p \in R and all natural numbers k. To see this, suppose \theta : M_1 \rightarrow M_2 is a module isomorphism. Certainly the relation \{ (p^k \cdot m, p^k \cdot \theta(m)) \ |\ m \in M_1 \} is well defined, since \theta is an isomorphism. So we have a surjective mapping \psi : (p^k)M_1 \rightarrow (p^k)M_2 given by \psi(p^k \cdot m) = p^k \cdot \theta(m). Moreover, this is an R-module homomorphism, since \psi(p^k \cdot m + r \cdot (p^k \cdot n)) = \psi(p^k \cdot (m+r \cdot n)) = p^k \cdot \psi(m) + r \cdot p^k \cdot \psi(n). Now the composite \overline{\psi} = \pi \circ \psi : (p^k)M_1 \rightarrow (p^k)M_2/(p^{k+1})M_2, where \pi is the natural projection, is also a module homomorphism. We claim that \mathsf{ker}\ \overline{\psi} = (p^{k+1})M_1. It is certainly the case that (p^{k+1})M_1 \subseteq \mathsf{ker}\ \overline{\psi}. To see the other inclusion, suppose p^k \cdot a \in \mathsf{ker}\ \overline{\psi}. Then p^k \theta(a) \in (p^{k+1})M_2, so that p^k \theta(a) = p^{k+1}b for some b \in M_2. (Note that if we took an element from (p^k) with a non-unit factor other than p^{k+1}, then that factor can be attached to b.) Now p^k(\theta(a) - pb) = 0. Since \theta is an isomorphism, we have c \in M_1 with \theta(c) = b. Then p^k(\theta(a) - p\theta(c)) = 0, so that \theta(p^k(a - pc)) = 0, and thus p^k(a - pc) = 0. So p^ka = p^{k+1}c, as desired. (There is probably a better way to do this.) By the First Isomorphism Theorem, we have (p^k)M_1/(p^{k+1})M_1 \cong_R (p^k)M_2/(p^{k+1})M_2.

We saw in Part 1 that (p^k)M_1/(p^{k+1})M_1 \cong_R (R/(p))^n_{k+1}, where n_{k+1} is the number of elementary divisors of M_1 which are of the form p^\alpha for some \alpha > k. In particular, we have that m_k = n_k - n_{k+1} is the number of elementary divisors of M_1 of the form p^k. The isomorphism above demonstrates that these numbers are the same for M_1 and M_2. Note that R/(p) is a field if p is prime (since R is a PID) and that if F^a \cong F^b as F-vector spaces, then a = b by this previous exercise.

Compute a quotient module

Let R be a principal ideal domain, let a \in R be nonzero, and let M = R/(a). Given a prime p \in R, say a = p^nq, where p does not divide q. Prove that (p^k)M/(p^{k+1})M is module-isomorphic to R/(p) if k < n and to 0 if k \geq n.


We begin with some lemmas.

Lemma 1: Let C \subseteq A,B be ideals of a ring R, and consider A/C as an R-module. Then B(A/C) = (BA)/C. Proof: (\subseteq) If x \in B(A/C), then x = \sum b_i(a_i + C) = \sum (b_ia_i + C) = (\sum b_ia_i) + C \in (BA)/C. (\supseteq) If x \in (BA)/C, then x = (\sum b_ia_i) + C = \sum (b_ia_i + C) = \sum b_i(a_i+C). Thus x \in B(A/C). \square

Lemma 2: Let R be a principal ideal domain and let a,b,c \in R be nonzero with b|c. Note that (ac) \subseteq (ab). Prove that (ab)/(ac) \cong_R (b)/(c). Proof: Let \psi: (b) \rightarrow (ab)/(ac) be given by bx \mapsto \overline{abx}. (This is well-defined since R is a domain, and is clearly an R-module homomorphism.) Certainly \psi is surjective. Now if bx \in \mathsf{ker}\ \psi, then abx \in (ac), so that bx \in (c). Conversely, if bx \in (c), then bx = cy for some y, and so \overline{abx} = \overline{acy} = 0. By the First Isomorphism Theorem, (b)/(c) \cong_R (ab)/(ac). \square

Lemma 3: Let R be a principal ideal domain and a,b,c \in R such that b|c. If (a,c/b) = (1), then (a)((b)/(c)) = (b)/(c). Proof: Say c = bd. Note that ax+dy = 1 for some x,y \in R, so that abx + cy = b. Now b+(c) = abx + (c) = a(bx+(c)) \in (a)((b)/(c)), so that (a)((b)/(c)) = (b)/(c). \square

If k < n, then p^{k+1}|a, so that (a) \subseteq (p^k), (p^{k+1}). By Lemma 1, we have (p^k)(R/(a))/(p^{k+1})(R/(a)) = ((p^k)/(a))/((p^{k+1})/(a)) \cong_R (p^k)/(p^{k+1}) by the Third Isomorphism Theorem. Using Lemma 2, this is isomorphic to (1)/(p) \cong_R R/(p).

If k \geq n with (say) k = n+t, then ((p^k)(R/(a)))/((p^{k+1})/(R/(a)) = ((p^t)((p^n)/(p^nq)))/((p^{t+1})((p^n)/(p^nq))) = ((p^n)/(p^nq))/((p^n)/(p^nq)) = 0, using Lemma 3 and the Third Isomorphism Theorem.

Every torsion module over a principal ideal domain is the direct sum of its p-primary components

Let R be a principal ideal domain and let N be a torsion R-module. Prove that the p-primary component of N is a submodule for every prime p \in R. Prove that N is the direct sum of its p-primary components.


We proved this in a previous exercise.

A fact about the annihilators of torsion modules over a PID

Let R be a principal ideal domain, let B be a torsion (left, unital) R-module, and let p \in R be prime. Prove that if pb = 0 for some nonzero b \in B, then \mathsf{Ann}(B) \subseteq (p).


[Note: I’m not sure why D&F assume that B is torsion. This proof doesn’t use that fact as far as I can tell. Read with caution.]

Recall that \mathsf{Ann}(B) \subseteq R is an ideal (by this previous exercise). Since R is a principal ideal domain, we have \mathsf{Ann}(B) = (t) for some t \in R.

Consider the cyclic submodule (b)_R generated by b. Again, \mathsf{Ann}((b)_R) \subseteq R is an ideal, so \mathsf{Ann}((b)_R) = (s) for some s. By our assumption, p \in (s), so that p = sr for some r \in R. Since p is prime, either p|r or p|s. If p|r, then we have r = pq for some q \in R. Now pqs = p, so that qs = 1; in particular, s is a unit, and so \mathsf{Ann}((b)_R) = R. But then 0 = 1 \cdot b = b, a contradiction. So p|s, and thus (s) \subseteq (p).

Since \mathsf{Ann}(B) \subseteq \mathsf{Ann}((b)_R), we have (t) \subseteq (s), and so (t) \subseteq (p), as desired.

Every ideal in ZZ[i] is principal

Prove that every ideal in \mathbb{Z}[i] is principal.


We showed here that \mathbb{Z}[i] is a Euclidean domain. Thus it is a principal ideal domain.

Over a PID, flat and torsion free are equivalent

Let R be a principal ideal domain. Prove that a right unital R-module A is flat if and only if it is torsion free.


Suppose A is flat as a right R-module. Given r \in R nonzero, define \psi_R : R \rightarrow R by \psi_r(a) = ra. Because R is a commutative ring, \psi_r is a left module homomorphism. Because R is a domain, \psi_r is injective. Because A is flat, 1 \otimes \psi_r : A \otimes_R R \rightarrow A \otimes_R R is injective. Now suppose ar = 0, with r \neq 0. In A \otimes_R R, we have a \otimes r = 0 \otimes r, so that (1 \otimes \psi_r)(a \otimes 1) = (1 \otimes \psi_r)(0 \otimes 1). Since 1 \otimes \psi_r is injective, we have a \otimes 1 = 0 \otimes 1. Recall that A \otimes_R R \cong A via the multiplication map a \otimes r \mapsto ar. Thus we have a = 0, and so A is torsion free as a right R-module.

Conversely, suppose A is torsion free as a right R-module, and let I \subseteq R be a nonzero ideal. Since R is a PID, say I = (r) with r \in R nonzero. Note that the map \theta_r : A \rightarrow A given by \theta_r(a) = ar is a module homomorphism. Moreover, \theta_r has a trivial kernel since A is torsion free- so \theta_r is injective. Now consider the map \rho : R \rightarrow I given by \rho(x) = rx; this is a module homomorphism. Certainly \rho is surjective, and moreover since R is a domain, \rho is injective. Thus \rho is a module isomorphism. Now the homomorphism 1 \otimes \rho : A \otimes_R R \rightarrow A \otimes_R I is also an isomorphism since (1 \otimes \rho) \circ (1 \otimes \rho^{-1}) = 1 \otimes 1 = 1 and (1 \otimes \rho^{-1}) \circ (1 \otimes \rho) = 1 \otimes 1 = 1. Recall that A and A \otimes_R R are isomorphic as abelian groups via the mappings \iota : A \rightarrow A \otimes_R R and \tau : A \otimes_R R \rightarrow A given by \iota(a) = a \otimes 1 and \tau(a \otimes r) = ar. Consider the map 1 \otimes \iota : A \otimes_R I \rightarrow A \otimes_R R.

Now (\tau \circ (1 \otimes \iota) \circ (1 \otimes \rho) \circ \iota)(a) = \tau((1 \otimes \iota)((1 \otimes \rho)(a \otimes 1))) = \tau((1 \otimes \iota)(a \otimes r)) = \tau(a \otimes r) = ar = \theta_r(a). Thus we have \tau \circ (1 \otimes \iota) \circ (1 \otimes \rho) \circ \iota = \theta_r, and so 1 \otimes \iota = \tau^{-1} \circ \theta_r \circ \iota^{-1} \circ (1 \otimes \rho). Thus 1 \otimes \iota is injective.

By the flatness criterion for modules (proved here), A is flat as a right R-module.

Every torsion module over a PID is the internal direct sum of its p-primary components

Let R be a principal ideal domain and let M be a torsion unital left R-module. If p is a prime in R, the p-primary component of M is the set M_p = \{m \in M \ |\ p^k \cdot m = 0\ \mathrm{for\ some}\ k \geq 1 \}.

  1. Prove that M_p is a submodule of M.
  2. Prove that this definition of p-primary component agrees with the definition given in this previous exercise if M has a nonzero annihilator.
  3. If P \subseteq R is the set of all primes in R (up to associates), prove that R = \bigoplus_{p \in P} M_p. (Where \bigoplus denotes an internal direct sum.)

  1. It is clear that M_p = \bigcup_{k \in \mathbb{N}^+} \mathsf{Ann}_M(p^k). Note also that if k \leq \ell and p^k \cdot m = 0, then p^\ell \cdot m = 0. Hence \mathsf{Ann}_M(p^k) \subseteq \mathsf{Ann}_M(p^\ell). That is, \{\mathsf{Ann}_M(p^k)\}_{k \in \mathbb{N}^+} is a chain of submodules of M. Thus M_p is a submodule of M.
  2. Suppose \mathsf{Ann}_R(M) = (a) is nonzero; say a = \prod p_i^{k_i}. In this previous exercise we defined the p_i-primary component of M to be \mathsf{Ann}_M(p_i^{k_i}). Certainly \mathsf{Ann}_M(p_i^{k_i}) \subseteq M_{p_i}. Now suppose m \in M_{p_i}; say p_i^k \cdot m = 0. Since a \cdot m = 0 and R is a PID, (a, p_i^k) = (p_i^t) is in \mathsf{Ann}_R(m) where t \leq k_i$. So m \in \mathsf{Ann}_M(p_i^{k_i}), and so M_p = \mathsf{Ann}_M(p_i^{k_i}). Thus if M has a nonzero annihilator, the two notions of p-primary component agree.
  3. Now let K \subseteq P be a finite set of primes in R, and fix some p \in K. Suppose m \in M_p \cap (\sum_{P \setminus p} M_q). Say m = m_p = \sum m_q, where m_p \in M_p and m_q \in M_q for each q. Since M_p = \bigcup_{k \in \mathbb{Z}^+} \mathsf{Ann}_M(p^k), we have e_p and e_q (for each q such that p^{e_p} \cdot m_p = 0 and q^{e_q} \cdot m_q = 0. Note that (\prod q^{e_q}) \cdot m = \sum (\prod q^{e_q}) \cdot m_q = 0 and p^{e_p} \cdot m = p^{e_p} \cdot m_p = 0. Now (p^{e_p}, \prod q^{e_q}) \subseteq \mathsf{Ann}_R(m). Since R is a PID and p^{e_p} and \prod q^{e_q} are relatively prime (as we know their factorizations), \mathsf{Ann}_R(m) = R. In particular, m = 1 \cdot m = 0. Thus M_p \cap (\sum M_q) = 0. Moreover, note that if m \in M, then a \cdot m = 0 for some a \in R since M is torsion. Say a = \prod_T p_i^{k_i}. Let q_t = \prod_{i \neq t} p_i^{k_i}. Note that (q_t \ |\ t \in T) = R, so that 1 = \sum x_tq_t for some x_t \in R. Finally, p_i^{k_i} \cdot (x_tq_t \cdot m) = 0, so that x_tq_t \cdot m \in M_{p_i}. Finally, m = 1 \dot m = \sum x_tq_t \cdot m \in \sum_T M_{p_i}. Thus M = \bigoplus_P M_p, by this characterization of arbitrary internal direct sums.