In Iwasawa theory, one of the fundamental results is the following structure theorem for finitely generated modules over the ring $\Lambda = \mathbf{Z}_p[[T]]$.

If $M$ is a finitely generated torsion $\Lambda$-module, then there is a pseudo-isomorphism $$M \to \left( \bigoplus_{i=1}^s \Lambda/(p^{n_i}) \right) \oplus \left( \bigoplus_{j=1}^t \Lambda/(f_i) \right) $$ where the $f_i \in \Lambda$ are distinguished polynomials.

My question is the following: are the numbers $s$ and $t$ in the above theorem uniquely determined by $M$? That is, is the number of direct summands in the above decomposition uniquely determined by $M$?

As to why I'm asking this, I'm in a situation where I have a finitely generated torsion $\Lambda$ module $M$ and I happen to know that $M$ is generated by $n$ elements $x_1, \dots, x_n$. (EDIT: I also know that the elements $\{x_1,...,x_n\}$ form a minimal generating set, so no strict subset of $\{x_1,...,x_n\}$ generates all of $M$.) I'd then like to use this to conclude that there are $n$ summands in the direct-sum composition above (i.e: that $s+t=n$). To do this, however, one needs to know that $s$ and $t$ are uniquely determined by $M$, which is why I'm asking this question.

Any tips would be appreciated. Thanks!

  • $\begingroup$ Surely you cannot conclude this just from the information you've given, since I can always increase $n$ by tossing in redundant generating elements? $\endgroup$
    – LSpice
    Dec 10, 2021 at 1:48
  • $\begingroup$ Being generated by $n$ elements does not tell you enough. After all, $\mathbf Z$ is generated as a $\mathbf Z$-module by $8$ elements: $3, 5, 7, 9, 11, 13, 15, 17$. That does not make $\mathbf Z$ a direct sum of $8$ cyclic subgroups. Can you prove some minimality condition on your $n$ that you forgot to mention in your question? $\endgroup$
    – KConrad
    Dec 10, 2021 at 1:51
  • $\begingroup$ @KConrad Thanks for the comment. I've edited the question to include the fact that the generating set is minimal. Hopefully that should clarify things. $\endgroup$ Dec 10, 2021 at 3:45
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    $\begingroup$ Minimality in the sense of subset containment also is not enough: in the spirit of @KConrad's example, $\{2, 3\}$ is a minimal generating set for $\mathbb Z$. $\endgroup$
    – LSpice
    Dec 10, 2021 at 4:18
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    $\begingroup$ As Will Sawin points out below: $\Lambda/\mathfrak{m}_{\Lambda}=\mathbb{Z}/p\mathbb{Z}$ and $\Lambda/(f)$ and $\Lambda/p$ all are generated by one element, but $s+t$ is zero in the first case. $\endgroup$ Dec 10, 2021 at 9:16

1 Answer 1


By Nakayam's lemma, since $M$ is finitely generated, $x_1,\dots, x_n$ generate $M$ if and only if they generate $M/ (p, T)M$ (where $(p,T)$ is the maximal ideal of the local ring $\mathbb Z_p[[T]]$.)

So $x_1,\dots, x_n$ are a minimal generating set if and only if they are a basis of $M/ (p,T)M$, and thus in this case $n$ is the rank of $M/(p,T)M$.

If your pseudo-isomorphism were an isomorphism, you'd be done at this point, as that rank would be $s+t$. However, for a general torsion Iwasawa module, we can have, for example $s=t=0$ with $M$ and thus its minimal generating set nonempty, as happens when $M$ is finite.

So you need some additional condition on $M$.

  • $\begingroup$ Wonderful, this was exactly what I was looking for. The issue of "pseudo-isomorphism" vs "isomorphism" is what was getting in the way. Thanks! $\endgroup$ Dec 10, 2021 at 11:49

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