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Let $p$ be a prime number, and let $X$ be an abelian pro$-p-$group (i.e for some indexing set $I,$ we have $X=\varprojlim X_i$ where $X_i$ is a finite, abelian $p-$group for each index $i \in I .$) admits a continuous action by the group $\Gamma$ as a $\Gamma-$module ($\Gamma$ is a multiplicative topological group isomorphic with the additive group of $p-$adic integers $\mathbb Z_p.$)

I want to show that we can view $X$ as a module over the ring $\Lambda=\mathbb Z_p[[T]].$

I found this proof in R. Greenberg's book:

Let $\gamma_0$ denote a topological generator for $\Gamma,$ and regarding $T$ as the endomorphism $\gamma_0-1.$ For each $i \in I,$ we have $p^{a_i}X_i= 0$ for some $a_i > 0.$ Also, $\gamma_0-1$ defines an endomorphism of $X_i$ which has a nontrivial kernel if $X_i$ is nontrivial$^{(1)}$. Consequently, $(\gamma_{0}-1)X_i$ is a proper subgroup of $X_i$ if $X_i\neq 0.$ It follows that $(\gamma_0-1)^{b_i} X_i=0 $ for some $b_i>0.$ Thus, we can regard $X_i$ as a module over the finite ring $\mathbb Z_p[T]/(p^{a_i},T^{b_i})$ where we let $T$ act on $X_i$ as the endomorphism $\gamma_0-1.$ However, we obviously have $\mathbb Z_p[T]/(p^{a_i},T^{b_i})\cong \mathbb Z_p[[T]]/(p^{a_i},T^{b_i}),$ and so we can regard $X_i$ as a $\Lambda-$module which is annihilated by the ideal $(p^{a_i},T^{b_i}).$ Taking $t_i = a_i + b_i,$ it is obvious that $\frak{m}^{t_i}\subset$$ (T^{b_i},p^{a_i})$ (where $\frak{m}=$$(T,p)$) and so each $X_i$ can be regarded as a module over $\Lambda/\frak{m}^{t_i}.$ Regarding the $X_i$’s as $\Lambda-$modules, it is clear $^{(2)}$ that the maps defining the projective limit are $\Lambda-$module homomorphisms. Thus $X$ becomes a topological $\Lambda-$module.

1) I don't understand what he means by "the endomorphism $\gamma_0-1$ of $X$ or of $X_i$."

2) in $^{(1)}$ I don't understand why $\gamma_0-1$ defines an endomorphism of $X_i$ which has a nontrivial kernel if $X_i$ is nontrivial, and why in this case $(\gamma_{0}-1)X_i$ is a proper subgroup of $X_i$ and this imply that there is $b_i>0$ verify $(\gamma_0-1)^{b_i} X_i=0.$

3) in $^{(2)}$ i don't see why the maps defining the projective limit are $\Lambda-$module homomorphisms.

Thanks!

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up vote 3 down vote accepted
  1. Since $X$ is a $\Gamma$-module, it is automatically a module for the ring $\mathbb{Z}[\Gamma]$ (even $\mathbb{Z}_p[\Gamma]$). Explicitly, write $X$ multiplicatively, and the action of $\Gamma$ on $X$ exponentially. Then $$ x^{\gamma_0 - 1} = x^{\gamma_0}/x $$ Similarly for $X_i$.

  2. The kernel of $\gamma_0 - 1$ is the set of $x$ such that $x^{\gamma_0} = x$. Since all finite quotients of $\Gamma$ have cardinality a power of $p$, the action of $\Gamma$ breaks up $X_i$ into a bunch of orbits, each of which has cardinality a power of $p$. At least one of those orbits is a singleton (namely the orbit of the trivial element of $X_i$). I claim there is more than one. Indeed, $X_i$ is also a $p$-group, and so we can't have $$ |X_i| = 1 + p*\text{stuff}. $$ Therefore the kernel of $\gamma_0 - 1$ is not trivial. Now that we know the kernel is non-trivial, we conclude that the image is not all of $X_i$ (since $X_i$ is a finite abelian group, the cokernel of a map from $X_i$ to itself has the same size as the kernel). Now we study the map $\gamma_0 - 1$ on $(\gamma_0 - 1)X_i$. But the same argument applies word for word (we used nothing about $X_i$ except that it was a $p$-group!) The conclusion is that if we iterate the map $\gamma_0 - 1$ enough times, we kill $X_i$.

  3. By definition these maps are compatible with the $\Gamma$-action, and so are compatible with multiplication by $T$.

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thank you so much, but I don't understand what you mean in 1. –  Med Dec 27 '13 at 16:45
    
$ x^{\gamma - 1}:= x^\gamma/x $ –  hunter Dec 27 '13 at 16:56
    
aaah, if I understand you mean $\gamma_0-1$ is the endomorphism defined by $x\mapsto x^{\gamma}/x$ ?? –  Med Dec 27 '13 at 17:01
    
right! exactly. –  hunter Dec 27 '13 at 17:21
    
Now I understand, perfect proof , thank you very much :-) –  Med Dec 28 '13 at 3:40
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