# Showing $GL_n^1(\mathbb Z_p)$ with usual $p$-valuation is $p$-saturated.

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:

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

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).

• 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 $I+pM_n(\mathbb Z_p) \subset GL_n(\mathbb Z_p)$? – user366818 Apr 9 at 20:28
• 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? – Alvaro Martinez Apr 9 at 20:40
• 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 $I + p^2M_n(\mathbb Z_p)$ – user366818 Apr 9 at 20:41
• Ah okay, I think I understand how to show that works with that hint, thank you. – user366818 Apr 9 at 20:42
• 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. – user366818 Apr 10 at 19:38