## 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.