2
$\begingroup$

I am using Lang's Algebra to prove the Structure Theorem for finitely generated modules over PIDs, and I am having difficulties understanding the proof of the existence of the decomposition for $E(p)$. $E$ is a torsion module over a PID $R$, $p \in R$ prime element, and

$E(p)=\{m \in E\, ;\, p^nm=0 \, \textrm{for some positive interger n}\}$.

Lang starts with Lemma 7.6 which I understand just fine. The next step is what I do not grasp.

From what I could get, the idea is to show that there is an independent generator of $E(p)$, lets say $\{y_1,...,y_1\}$ and by having that, since it is independent you could see it as a direct sum of cyclic modules, i.e.

$(y_1,..,y_n)= (y_1)\bigoplus...\bigoplus (y_n) \cong\frac{R}{(p^{r_1})} \bigoplus....\bigoplus \frac{R}{(p^{r_n})}$

where $p^{r_i}$ is $y_i$'s period.

This is how he does it (its a copy paste):

Lang's Proof (Sorry for the sloppiness of the picture, but it was the best I could do with my knowledge)

Apart from the fact that I don't get the overall proof, here are some doubts that pop to mind:

  1. Is $\overline{E_p}$ well defined? ($E_p=\{m \in E \, ; pm=0\}$)

Because he defines $\overline{E}$ using an element $x_1$ with a maximal period, but $x_1$ is not necessarily in $E_p$

  1. Lang says he does it by induction, induction over what?

Sorry if the question is not very well formulated. Thanks in advanced.

$\endgroup$
2
  • 1
    $\begingroup$ Do you need to understand Lang's version of the proof? There are better and more understandable proofs elsewhere IMO. $\endgroup$
    – Qudit
    Dec 7 '17 at 21:46
  • $\begingroup$ I would prefer Lang's version, the thing is that in the course I am taking, most of it is a build up to this theorem, and probably other proofs use tools that escape the course. I have seen a really short proof using exact sequences and Jordan matrix, but it seems like ''cheating'' if you will. $\endgroup$
    – Temporary
    Dec 8 '17 at 0:19
2
$\begingroup$
  1. It's actually $\overline E_p$ and not $\overline{E_p}$ : you take $F= \overline{E}$ and then $F_p$, not $F=E_p$ and then $\overline F$.

  2. The induction is over the number of generators : you assume the result for all modules that have less than $r$ generators, and prove it for those that have $r$.

$\endgroup$
1
  • $\begingroup$ Hi, thank you for your answer! Do you think you can elaborate on Lang's proof for more clarity? (The OP does say that they don't get the overall proof, even apart from the two specific questions, and I am in the same boat as the OP.) $\endgroup$
    – user279515
    Aug 3 '19 at 10:46
2
+250
$\begingroup$

Since Brahadeesh indicated a desire for further elaboration in the comment on Max's answer, I'll give it a shot.

Following Lang, I'll assume "without loss of generality that $E = E(p)$." Lang proves the theorem via induction on $\dim E_p$ where $E_p = \{x\in E: px=0\}$ is considered as a vector space over $R/pR$.

Given $E$, Lang denotes $x_1 \in E$ some element with period $p^{r_1}$ such that $r_1$ is maximal. Then Lang defines $\bar E = E/(x_1)$. Lang's goal is to apply Lemma 7.6 in the following way: if we already know $\bar E$ has a decomposition

$$ \bar E = (\bar y_1) \oplus (\bar y_2) \oplus \dots \oplus (\bar y_k) $$ then Lemma 7.6 guarantees that $E$ has the decomposition

$$ E = (x_1) \oplus (y_1) \oplus (y_2) \oplus \dots \oplus (y_k) $$

But the only way we can already know $\bar E$ has the direct sum decomposition is via the inductive hypothesis, which applies only if $\dim \bar E_p < \dim E_p$ (where $\bar E_p = \{\bar x \in \bar E : p\bar x =0\}$ is considered as a vector space over $R/pR$). So Lang must show $\dim \bar E_p < \dim E_p$. To do this, Lang uses Lemma 7.6 in a different way. Let $m = \dim \bar E_p$. Then we can find $m$ linearly independent elements $\bar y_1, \bar y_2, \dots, \bar y_m \in \bar E_p$. By Lemma 7.6, we can decompose $E = (x_1) \oplus (y_1, y_2, \dots ,y_m)$, and since there is some element in $(x_1)$ of period $p$, that element will give rise to a nonzero element of $E_p$ linearly independent of the elements which come from $(y_1, y_2, \dots, y_m)$. Thus $\dim E_p \geq m+1 > \dim \bar E_p$.

This means, by the inductive hypothesis, $\bar E$ has a decomposition $$\bar E = (\bar x_2) \oplus (\bar x_3) \oplus \dots \oplus (\bar x_s) = R/(p^{r_2}) \oplus R/(p^{r_3}) \oplus\dots\oplus R/(p^{r_s})$$ with $r_2 \geq r_3 \geq \dots \geq r_s$. Now, by Lemma 7.6,

$$ E = (x_1) \oplus (x_2) \oplus (x_3) \oplus \dots \oplus (x_s) = R/(p^{r_1}) \oplus R/(p^{r_2}) \oplus R/(p^{r_3}) \oplus\dots\oplus R/(p^{r_s})$$ Since $r_1$ was chosen to be maximal, we also have $r_1 \geq r_2 \geq r_3 \geq \dots \geq r_s$, which is the last thing we needed.

$\endgroup$
3
  • $\begingroup$ Thank you very much for taking the time and effort to write this up! It has helped me very much :) $\endgroup$
    – user279515
    Aug 6 '19 at 17:32
  • $\begingroup$ Can you elaborate a bit more on the induction hypothesis? I didn't get why $dim(E_p)>dim(E¯_p)$ helps us to get independent generators of $E¯$. $\endgroup$
    – Sushant
    May 8 '21 at 16:55
  • $\begingroup$ @Sushant The inductive hypothesis is that the theorem already holds if $\dim E_p < k$, and we want to prove it holds if $\dim E_p = k$. We use the inductive hypothesis by looking at a space $\bar E$ with $\dim \bar E_p < k$. $\endgroup$
    – BallBoy
    May 10 '21 at 3:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.