Let $M$ be a finitely generated torsion $R$-module. Show that we can write $M \cong R/(a_1) \oplus ... \oplus R/(a_n)$ such that $a_1 | a_2 | ... | a_n$. Show that $a_1,...,a_n$ are uniquely determined up to a unit
With the help of Lukas Heger, I was able to show that $M$ has such a decomposition. Now I am working on the uniqueness part. Here is an argument from Lukas Heger which I am having trouble following:
Okay so maybe an induction argument can be made to work. We want to prove that if $M=R/(a_1) \oplus \dots R/(a_n) \cong N=R/(b_1) \oplus \dots R/(b_m)$ with $a_1 \mid a_2 \mid \dots$ and $b_1 \mid b_2 \mid \dots$, then $n=m$ and $a_i=b_i$ up to units. We induct on the number of prime factors (with multiplicity) of $\mathrm{Ann}_R(M)$. As a base case if $\mathrm{Ann}_R(M)=(p)$ is prime, then $M$ is a vector space over $M/(p)$ and so we get that $M \cong (R/(p))^n$ where $n=\mathrm{dim}_{R/(p)}(M)$. Clearly this $n$ is unique. Now the inductive step: let $p$ be any prime divisor of $\mathrm{Ann}_R(M)=(a)$. Now consider $pM$. If $p \nmid a_i$, then $p R/(a_i)=R/(a_i)$. If $a_i=a'_ip$, then $p R/(a_i) \cong R/(a'_i)$. Define $a'_i=a_i$ if $p \nmid a_i$ and $a'_i=a_i/p$ if $p \mid a_i$. Do the same thing for the $b_i$s. So from $M \cong N$, we get $pM \cong pN$: $R/(a'_1) \oplus \dots \oplus R/(a'_n) \cong R/(b'_1) \oplus \dots \oplus R/(b'_m)$ The inductive step gives $n=m$ and $a'_i=b'_i$ We can also consider $M/pM$ and $N/pN$ which are vector spaces over $R/(p)$, the dimension will be the number of $a_i$ that are divisible by $p$. Putting these informations together we can reconstruct $a_i$ from $a'_i$ and $b_i$ from $b'_i$ since we know at which point we need to multiply with $p$
For the base case, I have a few questions. Why does $\dim_{R/(p)} (M) = n$ (the number of $a_i$'s in $M$'s decomposition), and why does uniqueness of the dimension imply that $a_i$ and $b_i$ differ by a unit? If it's the case that $\dim_{R/(p)}(M) = n$, then I can see why $m=n$, but I don't see why $a_i$ and $b_i$ differ by a unit.
For the inductive part, presumably $p$ is a divisor of $a$, which we're assuming has $\ell+1$ primes in its factorization. Is this right? Also, the notation $pR/(a_i)$ means $\{pr + (a_i) \mid r \in R \}$, and $p R/(a_i) \cong R/(a'_i)$ is an isomorphism of $R$-modules with the isomorphism given by $pr + (a_i) \mapsto r + (a_i')$?
Also, the inductive step can be invoked because $Ann_{R}(pM) = (\frac{a}{p})$ and $\frac{a}{p}$ has $\ell$ prime factors?
Also, when he says,
We can also consider $M/pM$ and $N/pN$ which are vector spaces over $R/(p)$, the dimension will be the number of $a_i$ that are divisible by $p$.
Is this just an alternative way of showing $n=m$, or is it crucial to the proof? I ask because I am unable to establish it.
Finally, from what I understand,
Putting these informations together we can reconstruct $a_i$ from $a'_i$ and $b_i$ from $b_i'$ since we know at which point we need to multiply with $p$.
encompasses several cases. We could have $a_i = a_i'$ and $b_i = b_i', or $a_i = a_i'p and $b_i = b_i'$, etc. The only case which I am unable to verify is when, without loss of generality, $a_i = a_i'p$ and $b_i = b_i'$. From the induction step, we know that $a_i'$ and $b_i'$ differ by a unit; that is, $a_i' = ub_i'$ for some unit $u \in R$. Multiplying both sides by $p$, $a_i' p = u pb_i'$ or $a_i = up b_i$ (after substitution)...But I don't see how go from this to saying that $a_i$ and $b_i$ differ by a unit.
As of now, I think these are the only parts I don't understand. I could use some help.