Problem of modules over PID, Exercise $9.13(2)$ in Rotman's Advanced Modern Algebra $2$nd Edition

Question

Rotman's Advanced Modern Algebra ($3$rd Edition), Exercise B-$3.24(2)$, also Exercise $9.13(2)$ in $2$nd Edition.

$R$ is PID, $M$ is $P$-primary $R$-module. $P=(p)$ is non-zero prime ideal. $k_P:=R/P$ is field.

Define $M[p]:=\{m\in M, pm=0 \}$ and $V_P(n,M):=\text{dim}_{k_P}\frac{p^nM\ \cap M[p]}{p^{n+1}M\ \cap M[p]}.$

Suppose $M$ is direct sum of cyclic modules $C_i$, prove the number of summands $C_i$ having order ideal/annihilator $\color{blue}{(p^{n+1})}$ is $V_P(n,M)$.

Attention: The book ($3$rd Edition) originally writes "$\ldots$$C_i$ having order ideal/annihilator $(p^{n})$ is $V_P(n,M)$", but that's wrong and should be $\color{red}{(p^{n+1})}$ instead.

I wrote my thoughts in my answer below.

Answer 1

A few things. I found the second edition online, so I'm working from that.

I found your question a little unclear, so the first part of this is me working through it trying to figure out precisely what you're asking and why. Edit It has since been clarified, and the contradiction resolved.

Regardless of your work, there is an obvious contradiction in the statements. We have

1. $U_P(n,M)=\#\{\text{cyclic summands w/ order $p^{n+1}$}\}$ (Theorem 9.14)
2. $U_P(n,M)=V_P(n,M)$ for $M$ f.g. (Ex 9.13 (i))
3. $V_P(n,M)=\#\{\text{cyclic summands w/ order $p^{n}$}\}$ (Ex 9.14 (ii))

Now for a finitely generated module $M$, this presents a clear contradiction. This is I presume why you asked the question, you find yourself in the situation of having apparently proved a contradiction, so you need to resolve this situation.

Now as with most textbooks, minor typos are not uncommon. It's not likely for statement 2 to contain a typo, so let's see if we can find an example showing statements 1 or 3 to be false.

Definitions

We recall the definitions:

1. $U_P(n,M)=d(p^nM)-d(p^{n+1}M)$, where $d(M)=\dim(M/pM)$ (definition immediately after corollary 9.13)
2. $$V_P(n,M)=\dim \frac{p^nM\cap M[p]}{p^{n+1}M\cap M[p]}$$

Consider $M=R/P$. Then $d(p^nM)=\delta_{0n}$, so $U_P(n,M)=d(p^nM)-d(p^{n+1}M)=\delta_{0n}-\delta_{0,n+1}=\delta_{0n}$. Thus statement 1 holds.

On the other hand, $M[p]=M$, so $V_P(n,M)=\dim(p^nM/p^{n+1}M)=d(p^nM)=d(p^nM)-d(p^{n+1}M)=\delta_{0n}$, since $d(p^{n+1}M)=0$ for all $n\ge 0$. Thus statement 3 is violated.

How I would prove the corrected version of (3)

Note that everything is linear across direct sums, so it suffices to prove the statement for a cyclic module.

Consider $M=R/P^{e+1}$. We want to show that $V_P(n,M)=\delta{ne}$.

Now $M[p]=P^e/P^{e+1}$, and $p^nM = P^n/P^{e+1}$ when $n \le e$ and $0$ otherwise. Then $p^nM\cap M[p]= P^e/P^{e+1}$ for $n\le e$, and $0$ otherwise. Hence if $n < e$, $$p^nM\cap M[p] = P^e/P^{e+1} = p^{n+1}M\cap M[p],$$ if $n > e$, then $$p^nM\cap M[p] = 0 = p^{n+1}M\cap M[p],$$ and if $n=e$, then $$p^nM\cap M[p] = P^e/P^{e+1},\text{ but } p^{n+1}M\cap M[p]=0,$$ so $V_P(e,M)=1\dim P^e/P^{e+1}=1$.

Thus $V_P(n,M)=\delta_{ne}$ as desired.

Answer 2

Suppose $M=M_P \cong \bigoplus_{e_m\geq1} R/P^{e_m},$ then $M[p]\cong \bigoplus_{e_n\geq0}P^{e_n}/P^{e_n+1}.$

$p^nM = \bigoplus_{e_m\geq1} P^n/P^{e_m}\cong \bigoplus_{e_m\geq n+1} P^n/P^{e_m}$.

$p^nM \cap M[p]\cong \bigoplus_{e_m\geq n+1}P^{e_m-1}/P^{e_m}$.

$p^{n+1}M = \bigoplus_{e_m\geq1} P^{n+1}/P^{e_m}\cong \bigoplus_{e_m\geq n+2} P^n/P^{e_m}$.

$p^{n+1}M \cap M[p]\cong \bigoplus_{e_m\geq n+2}P^{e_m-1}/P^{e_m}$.

Thus $(p^nM \cap M[p])/(p^{n+1}M \cap M[p])\cong\bigoplus_{e_m=n+1}P^{e_m-1}/P^{e_m}\cong(R/P)^{\oplus \#\{e_m|e_m=n+1\}}$.

dim$(p^nM \cap M[p])/(p^{n+1}M \cap M[p]) = V_P(n,M)=\#\{e_m|e_m=n+1\}$.

Thus there're $ V_P(n,M)$ $R/P^{n+1}$ as direct summands in $M_P$.