TITLE: Lang's proof of the fact that a finitely generated $p$-module is the direct sum of cyclic $p$-modules
QUESTION [1 upvotes]: 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 :-)

REPLY [1 votes]: 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).
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