LEC ALG4 6

Finitely generated torsion modules over PIDs

1. $R$ is a PID, $M$ is a finitely generated torsion $R$-module with generating set $X=\{x_1,\dots ,x_n\}$.
2. We select once and for all a system of representatives for the prime elements of $R$, modulo units.

Immediate consequences:

1. $R$ is Noetherian, and it follows that $M$ is Noetherian by Cor 389.4.
2. $R$ is a UFD, by Prp 114.11.

Proposition 394.1(Proposition).

${\text{Ann}}_R(M)$ is a nonzero (principal) ideal.

Proof.

It is clear that

$$\bigcap _{x\in X}{\text{Ann}}_R(Rx)={\text{Ann}}_R(M).$$

${\text{Ann}}_R(R{x}_i)\ne 0$ for all $i$, since $M$ is a torsion module. Let $c_i\in {\text{Ann}}_R(R{x}_i)$ for each $i$. Then, $c_1\dots c_n$ lies in the intersection.□

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Definition 394.2(Definition).

Let $a\in R$, $a\ne 0$.

$$M(a):=\{x\in M:\exists e⩾0\text{ such that }{a}^{e}x=0\}.$$

Proposition 394.3(Proposition).

There exists $e^{\prime }⩾0$ such that

$$M(a)=\{x\in M:a^{e^{\prime }}x=0\}$$

Proof.

The ascending chain

$$\{x\in M:ax=0\}\subseteq \{x\in M:a^{2}x=0\}\subseteq \dots$$

stabilizes since $M$ is Noetherian.□

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Proposition 394.4(Proposition).

TFAE:

1. $M(p)\ne 0$;

2. There exists $x\ne 0$ such that $px=0$;

3. ${\text{Ann}}_R(M)\subseteq ⟨p⟩$.

Proof.

$(1⟺2)$ is clear.

$(2⟹3)$ Let $x\ne 0$ be such that $px=0$. Clearly, ${\text{Ann}}_R(M)\subseteq {\text{Ann}}_R(Rx)$ and ${\text{Ann}}_R(Rx)\supseteq ⟨p⟩$. Since prime ideals in a PID are maximal and ${\text{Ann}}_R(Rx)\ne R$, we have ${\text{Ann}}_R(Rx)=⟨p⟩$.

$(3⟹2)$ By Prp 1, write ${\text{Ann}}_R(M)=⟨a⟩$. Then, $a=pb$ for some $b\in R$. It suffices to prove the existence of a $y\in M$ such that $by\ne 0$. Since the exponent of $p$ in $b$ is one less than in $a$, we have $⟨b⟩\not\subset ⟨a⟩$. Thus, $b\notin {\text{Ann}}_R(M)$, so there must exist $y\in M$ such that $by\ne 0$.□

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Remark 394.5(Remark).

Let ${\text{Ann}}_R(M)=⟨a⟩$. The primes $p\in R$ for which $M(p)\ne 0$ are precise those which divide $a$, by Prp 4. Since $a$ has a unique finite prime factorization, there exist only finitely many such $p$.

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Proposition 394.6(Proposition).

Let $b,c\in R$ be relatively prime nonzero nonunits. Then, $M(bc)=M(b)\oplus M(c)$.

Proof.

Let $x\in M(b)\cap M(c)$. There exist $m,n⩾1$ such that $b^{m}x=0=c^{n}x$. Write $1=\lambda b^m+\mu c^n$. Then, $x=\lambda b^{m}x+\mu c^{n}x=0$.

Note that $M(b)\subseteq M(bc)$, and $M(c)\subseteq M(bc)$. Thus, the sum $M(b)+M(c)$ is a direct sum inside $M(bc)$. It remains to show that $M(bc)=M(b)+M(c)$. Let $x\in M(bc)\setminus \{0\}$. There exists $N⩾1$ such that $(bc{)}^{N}x=0$. Write $1=\lambda b^N+\mu c^N$. Then, $x=\lambda b^{N}x+\mu c^{N}x$, where $b^{N}x\in M(c)$ and $c^{N}x\in M(b)$.□

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Theorem 394.7(Lang (2002) 3.7.5, p1).

Let $R$ be a PID, $M$ a finitely generated torsion module. Let ${\text{Ann}}_R(M)=⟨a⟩$, and $\{p_1,\dots ,p_m\}$ be the prime divisors of $a$ ¹. Then,

$$M=\underset{i=1}{\overset{m}{⨁}}M(p_i).$$

Proof.

Induct on $m$. Clear for $m=1$. Write $a=bc$ where $b,c$ are nonunits and $(b,c)=1$.

$$a=\underset{b}{\underbrace{p_1^{e_1}\dots p_n^{e_n}}}\underset{c}{\underbrace{p_{n+1}^{e_{n+1}}\dots p_m^{e_m}}}.$$

By Prp 6, $M=M(a)=M(b)\oplus M(c)$. Using the induction hypothesis, we can write

$$\begin{array}{r}M(b)=\underset{p_i|b}{⨁}(M(b))(p_i),M(c)=\underset{p_i|c}{⨁}(M(c))(p_i).\end{array}$$

Let $p_i|b$. Clearly, $x\in M(p_i)$ iff $x\in (M(b))(p_i)$ (ditto for $p_i|c$). Thus, we have required result.□

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Footnotes

1. Or equivalently, by Prp 4, those primes $p\in R$ for which $M_p$ is nonzero. ↩

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References

Lang, S. (2002). Algebra (Vol. 211). Springer New York. https://doi.org/10.1007/978-1-4613-0041-0