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First a few definitions:

$GL_n^1(\mathbb Z_p) = \ker(GL_n(\mathbb Z_p) \rightarrow GL_n(\mathbb Z/p\mathbb Z))$

$\omega: G \rightarrow \mathbb R^{>0} \cup \{\infty\}$ is a $p$-valuation (or a $p$-adic filtration using just the first two properties) on a group $G$ if:

  • $\omega(xy^{-1}) \geq \min(\omega(x), \omega(y)), \; \forall x,y \in G$
  • $\omega(x^{-1}y^{-1}xy) \geq \omega(x) + \omega(y), \; \forall x,y \in G$
  • $\omega(g) > \frac{1}{p-1}, \forall g \in G$
  • $\omega(g^p) = \omega(g) + 1, \; \forall g \in G)$

We define the $p$-valuation $\omega$ on $G = GL_n^1(\mathbb Z_p)$ coming from the $p$-adic filtration on $M_n(\mathbb Z_p)$ given by taking the minimum over the usual $p$-adic filtration on all entries.

Finally:

We say a $p$-valued group, $(G, \omega)$ is $p$-saturated if $\forall g \in G, \omega(g) > 1 + \frac{1}{p-1}, \; \; \exists h \in G, h^p = g $

I asked to show that our $(G, \omega)$ is $p$-saturated, where $p$ is an odd prime.

I have reasoned that $G = I + pGL_n(\mathbb Z_p)$, and so if $A \in G$ is such that $\omega(A) > \frac{p}{p-1}$, then $A \in I + p^2 GL_n(\mathbb Z_p)$

So now I just want to show that there is some $C \in GL_n(\mathbb Z_p)$ such that:

$(I + pC)^p = I + p^2B$, where $B \in GL_n(\mathbb Z_p)$ is such that $I+ p^2B = A$

In other words:

$$A - I = \sum_{n=1}^{p} {{p}\choose{n}}p^nC^n$$

However, I'm not really sure how to show this. Any help to point me in the right direction for how I might go about showing this would be very much appreciated, thank you.

[EDIT]: Noticed a mistake in my working:

$GL_n^1(\mathbb Z_p) = (I + pM_n(\mathbb Z_p))\cap GL_n(\mathbb Z_p)$

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Note that if $p=2$, then $\mathrm{GL}_n^1(\mathbb{Z}_p)$ is not a $p$-valued group: the diagonal matrix with entries $(p+1,1,...,1)$ has valuation $1=\frac{1}{2-1}$, so you probably want $p$ to be odd (or $G=\mathrm{GL}_n^2(\mathbb{Z}_p)$ for $p=2$).

Also observe that $G$ is actually $I+pM_n(\mathbb{Z_p})$, not $I+p\mathrm{GL}_n(\mathbb{Z}_p)$ (for starters, you want the identity in the group!).

[EDIT] The exercise reduces to showing that there exists some $C\in M_n(\mathbb{Z}_p)$ such that $(I+pC)^p=I+p^2A$. Here it would be useful to have a matrix analog of Hensel's lifting algorithm to lift the solution $C=A$ in mod $p^3$. In this particular case we can argue as follows.

First note that since $M_n(\mathbb{Z}_p)$ is a complete filtered ring, it will suffice to successively find solutions $C_i$ in mod $p^i$ converging to some element $C\in M_n(\mathbb{Z}_p)$. We start by taking $C_3=A$. We define $C_4$ as follows. Write $(I+pC_3)^p=I+p^2A+p^3D$ for some $D\in M_n(\mathbb{Z}_p)$, then let $C_4=C_3-pD$. Note that it commutes with $A$ (since $D$ does), so we can verify:

$(I+pC_4)^p=(I+pC_3-p^2D)^p=(I+pC_3)^p-p^3D(I+pC_3)^{p-1}\equiv I+p^2A \pmod{p^4}$

We can continue in this fashion by writing $(I+pC_i)^p=I+p^2A+p^{i}D'$ for some $D'\in M_n(\mathbb{Z}_p)$, then defining $C_{i+1}=C_i-p^{i-2}D'$. From this it is clear that the $C_i$ converge to some element in $M_n(\mathbb{Z}_p)$. Finally note that by induction $D'$ commutes with $C_i$ and $A$ (because $C_i$ commutes with $A$) and therefore $C_{i+1}$ commutes with both $C_i$ and $A$ (this fact is crucial to check the congruence).

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  • $\begingroup$ Thank you for your answer, I've had made an edit to mention $p$ is odd. However, I think there is now a slight issue: I can see that $GL_n^1(\mathbb Z_p) = (I + pM_n(\mathbb Z_p)) \cap GL_n(\mathbb Z_p)$, but do we immediately know that $I+pM_n(\mathbb Z_p) \subset GL_n(\mathbb Z_p)$? $\endgroup$ – user366818 Apr 9 at 20:28
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    $\begingroup$ Ah I agree that's a bit subtle. It boils down to showing that $\sum_{i\ge 0}(-pA)^i$ makes sense as an element in $M_n(\mathbb{Z}_p)$. Do you know what a complete $p$-valued group is? $\endgroup$ – Alvaro Martinez Apr 9 at 20:40
  • $\begingroup$ I'm also having another problem seeing how we are done with the observation, since that doesn't actually demonstrate that $(I + pC)^p = A$, and rather just that the $p$th power of anything in $I + pM_n(\mathbb Z_p)$ lies inside $I + p^2M_n(\mathbb Z_p)$ $\endgroup$ – user366818 Apr 9 at 20:41
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    $\begingroup$ Ah okay, I think I understand how to show that works with that hint, thank you. $\endgroup$ – user366818 Apr 9 at 20:42
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    $\begingroup$ Yes thank you for the correction. I apologise for the pedantic questions, I had initially thought about shifting the power like you have but could not quite wrap my head around why this was not a problem. Coming back to it now, I can see more clearly that it is okay. $\endgroup$ – user366818 Apr 10 at 19:38

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