# Lang's proof of the fact that a finitely generated $p$-module is the direct sum of cyclic $p$-modules

The question refers to a proof of the theorem that a finitely generated $p$-module is the direct sum of cyclic $p$-modules. In particular, refer to Lang's "Algebra" p. 151, right above Theorem 7.7. I can not understand what he is inducting on, even though i see that $dim\bar{E}_p < dimE_p$. I can't see how we obtain that $E$ is generated by an independent set.

Alternatively, could we not construct a proof by induction on the cardinality of a minimal generating set of $E$, instead of resorting to the vector space $E_p$ induced by $E$?

Thank you :-)

• Dear Manos: What is a $p$-module? – Pierre-Yves Gaillard Aug 27 '11 at 4:30
• Hi Piere: a $p$-module is a torsion module for which each element has period a power of $p$. ($p$ is a prime element of the underlying principal ideal domain). – Manos Aug 27 '11 at 14:00
• Thanks! I think Lang is almost the only one to use this terminology, but never mind. - Is it the existence or the uniqueness part of the proof that you don’t understand? - I looked at various proofs of this theorem. I reproduced the one I find the simplest here. (See Theorem 3.) - I got the notification, but it’s safer I think if you use an @Pierre. – Pierre-Yves Gaillard Aug 27 '11 at 14:46

## 1 Answer

The excerpt in question can be viewed here.

George Bergman writes here:

P. 151, statement of Theorem III.7.7: Note where Lang above the display refers to the $R/(q_i)$ as nonzero, this is equivalent to saying that the $q_i$ are nonunits.

P. 151, next-to-last line of text: After "i = 1, ... , l" add, ", with some of these rows 'padded' with zeros on the left, if necessary, so that they all have the same length r as the longest row".

Here is the link to George Bergman's A Companion to Lang's Algebra.

And here is a statement of the main results.

Let $A$ be a principal ideal domain and $T$ a finitely generated torsion module. Then there is a unique sequence of nonzero ideals $I_1\subset I_2\subset\cdots$ such that $T\simeq A/I_1\oplus A/I_2\oplus\cdots$ (Of course we have $I_j=A$ for $j$ large enough.)

The proper ideals appearing in this sequence are called the invariant factors of $T$.

Let $P_1,\dots,P_n$ be the distinct prime ideals of $A$ which contain $I_1$, and for $1\le i\le n$ let $T_i$ be the submodule of $T$ formed by the elements annihilated by a high enough power of $P_i$. Then $T=T_1\oplus\cdots\oplus T_n$, and the sequence of invariant factors of $T_i$ has the form $$P_i^{r(i,1)}\subset P_i^{r(i,2)}\subset\cdots$$ with $r(i,1)\ge r(i,2)\ge\cdots\ge0$. (Of course we have $r(i,j)=0$ for $j$ large enough.)

The $P_i^{r(i,j)}$ called the elementary divisors of $T$.

We clearly have $I_j=P_1^{r(1,j)}\cdots P_n^{r(n,j)}$.

Let $M$ be a finitely generated $A$-module and $T$ its torsion submodule. Then there a unique nonnegative integer $r$ satisfying $M\simeq T\oplus A^r$.

The simplest proof of these statements I know is in this answer (which I wrote without any claim of originality).

• Thank you very much. Regarding Bergman's companion to Lang's algebra, how do i open the files? Can they be found in a pdf form? Regarding the terminology, Steven Roman in "Advanced Linear Algebra" refers to these modules as $p$-primary modules as well. Lang skips the "primary". I find it convenient because it refers to the same concept as the terminology "$p$-group". Regarding the theorem, my question actually refers to theorem 7.5 and part of its proof on the upper half on page 151. More precisely, i can not see what he is inducting upon... – Manos Aug 30 '11 at 13:29
• ...In particular, it is claimed that if $E \neq 0$, then there exist elements $\bar{x}_2, \cdots, \bar{x}_s$ that are independent, and then he invokes Lemma 7.6. I can see that by using the lemma, $x_1, x_2, \cdots, x_s$ are independent. What i can't see is why independent elements $\bar{x}_2, \cdots, \bar{x}_s$ exist and also what we are inducting on to prove the decomposition. – Manos Aug 30 '11 at 13:34
• @Manos: You're welcome. Here is what I think: He is inducting on $\dim E_p$. Since $\dim\overline E_p < \dim E_p$, the elements $\overline{x}_2, \cdots, \overline{x}_s$ exist by induction assumption. – Pierre-Yves Gaillard Aug 30 '11 at 14:30
• @Manos: Sorry I forgot about Bergman's files. I only found them in ps format. My browser (Safari) can read this format. You may try to write to Bergman. I don't what kind of computer you're using, but I think you should be able to find free software capable of reading ps. (If you study Lang's book, Bergman's Companion can be very helpful.) – Pierre-Yves Gaillard Aug 30 '11 at 15:32
• On unux/linux, the "ps2pdf" utility will convert Postscript to PDF. The downloadable freeware "TeXShop" for Mac OS X includes ps2pdf, also. – paul garrett Nov 22 '11 at 17:55