LEC ALG4 7

We carry over the global environment from LEC ALG4 6. Here’s the story so far:

1. $R$ is a PID, $M$ is a finitely generated $R$-module. We are after a structure theorem for such $M$.
2. We saw that such a module can be written as a direct sum of a free module and a torsion module.
3. We saw that a torsion module can be expressed as a direct sum of submodules annihilated by primes.
4. Hence, to describe the structure of finitely generated modules over PIDs, we need to describe the structure of finitely generated modules annihilated by a prime $p$ - called $p$-primary modules.

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Finitely generated p-primary modules

Definition 398.1($p$-primary modules).

$M$ is a $p$-primary module if every element of $M$ is annihilated by some prime $p\in R$. When $M$ is Noetherian, there exists a single power $e$ that works uniformly for all $m\in M$, i.e, $p^{e}M=0$ (by the proof of Prp 394.3). If $e$ is such that $p^{e-1}M\ne 0$, we call $e$ the exponent of $M$. We define the exponent of individual elements $x\in M$ analogously and denote them by $\exp (x)$.

Remark 398.2(Remark).

For $x\in M$ and $c\in R$, if $c\notin ⟨p⟩$, $x$ and $cx$ have the same exponent. Indeed, suppose $e$ and $k$ are exponents of $x$ and $cx$ respectively. Clearly, the only possibility is $k⩽e$. Suppose $k<e$. Since $c$ and $p^{e-k}$ are coprime, we can write $1=\lambda c+\mu p^{e-k}$. Thus, $p^{k}x=\lambda (p^{k}cx)+\mu (p^{e}x)=0$, a contradiction.

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Proposition 398.3(Proposition).

If $M$ is a finitely generated $p$-primary $R$-module with exponent $e$, then ${\text{Ann}}_R(M)=⟨p^e⟩$.

Proof.

Since $M(p)=M\ne 0$, ${\text{Ann}}_R(M)\subseteq ⟨p⟩$ by Prp 394.4. For any $q\ne p$, $M(q)\cap M(p)=M(q)\cap M=0$ by Prp 394.6. Thus, $M(q)=0$, and ${\text{Ann}}_R(M)\not\subset ⟨q⟩$. Thus, ${\text{Ann}}_R(M)$ is generated by some power of $p$. Since $p^e\in {\text{Ann}}_R(M)$ and $p^{e-1}\notin {\text{Ann}}_R(M)$, we must have ${\text{Ann}}_R(M)=⟨p^e⟩$.□

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Corollary 398.4(Corollary).

If $M$ is a singly generated $p$-primary $R$-module with exponent $e$, then $M\cong R/⟨p^e⟩$.

Proof.

If $M$ is singly generated by $x_1$, the $R$-module homomorphism $\phi :R\to M$ defined by $r\mapsto r{x}_1$ becomes surjective. By Rmk 2 and Prp 3, $\ker \phi ={\text{Ann}}_R(x_1)={\text{Ann}}_R(M)=⟨p^e⟩$. Thus,

$$\begin{array}{r}M\cong \frac{R}{\ker \phi }=\frac{R}{⟨p^e⟩}.\end{array}$$ □

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Remark 398.5(Remark).

Let $M$ be an $R$-module generated by $\{x_1,\dots ,x_n\}$. Let $I$ be an $R$-ideal. Then ${\overline{x}}_1,\dots ,{\overline{x}}_n\in M/IM$ generate $M/IM$ as an $R$-module. By Rmk 386.8, they also generate $M/IM$ as an $R/I$-module.

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If $I$ is a prime ideal and $M$ is $I$-primary, the converse holds:

Proposition 398.6(Proposition).

Let $M$ be a finitely generated $p$-primary $R$-module. Let $x_1,\dots ,x_m\in M$ be such that ${\overline{x}}_1,\dots ,{\overline{x}}_n\in M/pM$ generate $M/pM$ as an $R/⟨p⟩$-vector space. Then, $x_1,\dots ,x_n$ generate $M$ as an $R$-module¹.

Proof.

Let $N={\sum }_{i=1}^{m}R{x}_i\subseteq M$. We have to show that $N=M$. Let $y\in M$. There exist $a_1,\dots ,a_m\in R$ such that $\overline{y}={\sum }_{i=1}^m{a}_i{\overline{x}}_i$. Write $z={\sum }_{i=1}^m{a}_i{x}_i$. Then, $y-z\in pM$. Thus, for all $y\in M$, there exists $z\in N$ such that $y-z\in pM$. We can now write

$$\begin{array}{rll}y-z& =p{y}_1& \in pM\\ y-z-p{z}_1& =p^2{y}_2& \in p^{2}M\\ y-z-p{z}_1-p^2{z}_2& =p^3{y}_3& \in p^{3}M\\ & ⋮& \\ y-z-p{z}_1-\cdots -p^{e-1}{z}_{e-1}& =p^e{y}_e& \in p^{e}M=0,\end{array}$$

where $z,z_1,\dots ,z_{e-1}\in N$. It follows that $y\in N$.□

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Corollary 398.7(Corollary).

Every minimal generating set of a finitely generated $p$-primary $R$-module $M$ has the same cardinality, ${\dim }_{R/p}M/pM$.

Proof.

Let$X_1$ and $X_2$ be two minimal generating sets for $M$, with $|X_1|⩽|X_2|$. By Rmk 5, both ${\overline{X}}_1$ and ${\overline{X}}_2$ generate $M/pM$. Since $M/pM$ is a vector space, there exists a subset ${\overline{X}}_3\subseteq {\overline{X}}_2$ with $|{\overline{X}}_3|=|{\overline{X}}_1|$ which also generates $M/pM$. By Prp 6, $X_3\subseteq X_2$ generates $M$, contradicting the minimality of $X_2$.□

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Definition 398.8(Definition).

A subset $\{x_1,\dots ,x_m\}\subseteq M$ is called independent if for all $a_1,\dots ,a_n$,

$$\sum _{i=1}^m{a}_i{x}_i=0⟹a_i{x}_i=0\forall i.$$

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Remark 398.9(Remark).

$y_1,\dots ,y_m$ are independent iff the module $⟨y_1,\dots ,y_m⟩$ has the direct sum decomposition

$$⟨y_1,\dots ,y_m⟩=⟨y_1⟩\oplus \cdots \oplus ⟨y_m⟩$$

in terms of the cyclic modules $⟨y_i⟩$.

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Lemma 398.10(Lemma).

Let $M$ be a finitely generated $p$-primary $R$-module with exponent $e$. Let $x_1\in M$ have exponent $e$. Let $z_1,\dots ,z_m\in M/R{x}_1$ be independent elements of $M/R{x}_1$. Then, for all $i$ there exist lifts $y_i\in M$ of $z_i$ such that $y_i$ and $z_i$ have the same exponent. Moreover, $x_1,y_1,\dots ,y_m$ are independent in $M$.

Proof.

Let $f$ be the exponent of $z\in \{z_1,\dots ,z_m\}$. Let $y$ be a preimage of $z$. Then, $p^{f}y\in R{x}_1$. Write $p^{f}y=c{p}^g{x}_1$, where $g⩾0$ and $p\nmid c$.

1. If $g⩾e$, then $p^{f}y=0$. So, $\exp (y)⩽f$. However, $\exp (y)$ cannot be less than $f$, since $z^{\exp y}=0$. Thus, $\exp y=f$.
2. If $g<e$, we have $p^{f+e-g}y=c{p}^e{x}_1=0$ and, by Rmk 2, $p^{f+e-g-1}y=c{p}^{e-1}{x}_1\ne 0$. Thus, $\exp y=f+e-g$. Since $\exp y⩽e$, we have $f⩽g$. Now consider $y^{\prime }=y-c{p}^{g-f}{x}_1$, which is also a preimage of $z$. We have $p^f(y-c{p}^{g-f}{x}_1)=0$. Again, $\exp y^{\prime }$ cannot be less than $f$, so we have $\exp y=f$.

□

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Lemma 398.11(Lemma).

Let $p,b\in R$ with $p$ prime. Then, $R/⟨p⟩\cong bR/⟨pb⟩$.

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Theorem 398.12(Lang (2002) 3.7.5, p2).

Let $M$ be a finitely generated $p$-primary $R$-module. Let $s={\dim }_{R/p}M/pM$. Then, there exist unique $e_1⩾e_2⩾\cdots ⩾e_s$ such that

$$M\cong \frac{R}{⟨p^{e_1}⟩}\oplus \cdots \oplus \frac{R}{⟨p^{e_s}⟩}.$$

Proof.

$s$ is the cardinality of any minimal generating of $M$ by Cor 7. We induct on $s$. For $s=1$, we are done by Cor 4.

Now suppose $s⩾1$. Let $e_1$ be the exponent of $M$. Let $x_1\in M$ have exponent $e_1$. Denote $M/pM$ by $M_p$. Clearly $x_1\notin pM$, since if $x_1=pm$, we have a contradiction by $p^{e-1}{x}_1=p^{e}m=0$. Now, consider the image of ${\tilde{x}}_1$ under the projection $M\twoheadrightarrow M_p$. Since $M_p$ is a $R/p$-vector space of dimension $s$, we can choose ${\tilde{y}}_2,\dots ,{\tilde{y}}_s\in M_p$ such that $\{{\tilde{x}}_1,{\tilde{y}}_2,\dots ,{\tilde{y}}_s\}$ is a $R/p$-basis for $M_p$. It follows from Prp 6 that $x_1,y_2,\dots ,y_s$ generate $M$ as an $R$-module.

Next, consider consider the projection $M\twoheadrightarrow M/x_{1}R$. Note that $M/x_{1}R$ is generated by the $s-1$ generators ${\overline{y}}_2,\dots ,{\overline{y}}_s$. Using the induction hypothesis, we can write

$$\frac{M}{x_{1}R}=\frac{R}{⟨p^{e_2}⟩}\oplus \cdots \oplus \frac{R}{⟨p^{e_s}⟩},$$

with $e_2⩾\cdots ⩾e_s$. Let ${\overline{z}}_2,\dots ,{\overline{z}}_s\in M/x_{1}R$ be generators of the $s$ components of the above direct sum with exponents $e_2,..,e_s$ respectively. They are necessarily independent. Use Lem 10 to lift these to $z_2,\dots ,z_s\in M$ such that $x_1,z_2,\dots ,z_s$ are independent and have exponents $e_1⩾e_2⩾\cdots ⩾e_s$ (recall that $e_1$ is the highest exponent an element of $M$ can have).

These generate $M$: for $m\in M$, we have

$$\begin{array}{rl}m+x_{1}R& =\sum _{i=2}^s{\alpha }_i(z_i+x_{1}R)\\ & =\sum _{i=2}^s{\alpha }_i{z}_i+x_{1}R\\ ⟹m& ={\alpha }_1{x}_1+\sum _{i=2}^s{\alpha }_i{z}_i.\end{array}$$

Thus, by Rmk 9, we have

$$\begin{array}{r}M=\frac{R}{⟨p^{e_1}⟩}\oplus \cdots \oplus \frac{R}{⟨p^{e_s}⟩}.\end{array}$$

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We now prove uniqueness. Consider the submodules $pM,p^{2}M,\dots ,p^{e_1-1}M$ of $M$. Let $n_k$ denote the number of $e_i$ for which $e_i>k$. Let $e_{i_k}$ be the smallest $e_i$ greater than $k$. Observe that the direct sum

$$\begin{array}{r}{p}^{k}M=p^k\frac{R}{p^{e_1}R}\oplus \cdots \oplus p^k\frac{R}{p^{e_s}R}=p^k\frac{R}{p^{e_1}R}\oplus \cdots \oplus p^k\frac{R}{p^{e_{i_k}}R}\end{array}$$

has exactly $n_k$ nontrivial summands, since $p^{k}R/p^{e_i}R$ is $0$ for all $k⩾e_i$. Now consider the quotient $p^{k}M/p^{k+1}M$. If $n_k=n_{k+1}$, we have

$$\frac{p^{k}M}{p^{k+1}M}=\underset{i=1}{\overset{n_k}{⨁}}\left(\frac{\frac{p^{k}R}{p^{e_i}R}}{\frac{p^{k+1}R}{p^{e_i}R}}\right)\cong \underset{n_k}{⨁}\frac{R}{pR}$$

by the third isomorphism theorem and Lem 11. It is easy to show that even in the case $n_k>n_{k+1}$, the above isomorphism continues to hold. Thus,

$${\dim }_{R/p}\frac{p^{k}M}{p^{k+1}M}=n_k.$$

The dimensions of these quotient vector spaces are an invariant of $M$. Since the sequence $e_1,\dots ,e_s$ is completely determined by $n_0,\dots ,n_{e_1-1}$, we conclude that our decomposition is unique.□

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Example 398.13(Example).

Let $R=\mathbb{Z}$, $p=2$. Let $M$ be a finitely generated $\mathbb{Z}$-module with $2^{2}M=0$. Then, there exist $m_1,m_2$ such that

$$M\cong {\left(\frac{\mathbb{Z}}{2\mathbb{Z}}\right)}^{m_1}\oplus {\left(\frac{\mathbb{Z}}{4\mathbb{Z}}\right)}^{m_2}.$$

Footnotes

1. Incidentally, the same result can be proven for finitely generated modules over local rings using Nakayama’s lemma. ↩

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References

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