Question in p-adic integration (Igusa type)

Question

I am trying to learn how to solve Igusa type local zeta function. Ex. $$\int_{\mathbb{Z}_{p}}||x^3,x^2y,y^2||d\mu(x,y)$$ A nice method I was recently introduced to was to substitute $x=a+px'$ and $y=b+py'$, where $a$ and $b \in \{0,1,\cdots,p-1\}$ . Hence, we should obtain $$\frac{1}{p^2}(\sum_{a=0}^{p-1}\sum_{b=0}^{p-1}\int_{\mathbb{Z}_p}||(a+px)^3,(a+px)^2(b+py),(b+py)^2||d\mu(x,y)) \\ = \frac{1}{p^2}(p^2-1)+p^{2s-2}\int_{\mathbb{Z}_p}||px^3,px^2y,y^2||d\mu(x,y).$$

Here I am trying to find a trick I can use to solve it.

Answer 1

Nice, I solve it using your method and I checked my answer using Zeta SageMath.

Here is my answer, first

$$\int_{\mathbb{Z}_p^2}||x^3,x^2y,y^2||_p^sd\mu(x,y) = \int_{\mathbb{Z}_p^2}||x^3,y^2||_p^sd\mu(x,y).$$

Since, if $|x^2y|_p^s\ge|x^3|_p^s \Rightarrow |y|_p^s\ge|x|_p^s\Rightarrow |y^2|_p^s\ge|xy|_p^s\ge|x^2y|_p^s$, since $|x|_p^s\le 1$. $$ $$ Similarly, if $|x^2y|_p^s\ge|y^2|_p^s\Rightarrow |x^2|_p^s\ge|y|_p^s\Rightarrow|x^3|_p^s\ge|xy|_p^s\ge|x^2y|_p^s$. Hence, $$\int||x^3,y^2||_p^s d\mu(x,y)= 1-p^{-2}+p^{-2s-2}\int||px^3,y^2||_p^s d\mu(x,y)\cdots(1)$$ Now consider, $$\int||px^3,y^2||_p^s d\mu(x,y)=1-p^{-1}+p^{-s-1}\int||x^3,py^2||_p^s d\mu(x,y)\cdots(2)$$ Similarly, $$\int||x^3,py^2||_p^s d\mu(x,y)=1-p^{-1}+p^{-s-1}\int||p^2x^3,y^2||_p^s d\mu(x,y)\cdots(3)$$ Finally, $$\int||p^2x^3,y^2||_p^s d\mu(x,y)=1-p^{-1}+p^{-2s-1}\int||x^3,y^2||_p^s d\mu(x,y)\cdots(4).$$ Now all I have to do is to put (4) in (3) in (2) in (1) and obtain $$L= 1-p^{-2}+p^{-2s-2}[1-p^{-1}+p^{-s-1}[1-p^{-1}+p^{-s-1}[1-p^{-1}+p^{-2s-1}[L]]]],$$ where $L = \int||x^3,y^2||_p^s d\mu(x,y)$. Hence, $$ L = \frac{1-p^{-2}+(1-p^{-1})(p^{-2}T^2+p^{-3}T^3+p^{-4}T^{4})}{1-p^{-5}T^{6}},$$ where $T=p^{-s}$.

Thank you very much for this method it made it much easier. I still need to learn another method which uses vertex geometry, it is more general but pain in the head :)

Answer 2

I think this might end up being a bit longer than I expected and may not be the best solution; I originally made a mistake when I solved it the first time so it's not as tricky as I would have hoped, but I think there should still be some idea to take away at least. I'll go ahead and put an exponent of $s$ on the integral just out of habit, you can always replace it with $s=1$. I will omit $\mathbb{Z}_p^2$ when completely integrating over it.

$$\int||x^3,x^2y,y^2||^sd\mu$$

The first step you did was good, but then it leads to another, similar integral. If you play around and try to push that integral forward you'll start to see a pattern, and it's often nice to write down the more general integral so in this case,

$$J_k(s) = \int||p^kx^3,p^kx^2y,y^2||^sd\mu$$

So our integral to solve is $J_0(s)$.

Spliting at $y$ alone, when $y \in \mathbb{Z}_p^\times$ the integral is 1 regardless of $x$. $$ =1-p^{-1} + p^{-1}\int||p^kx^3,p^kx^2py,p^2y^2||^sd\mu$$ Now let's split $x$ in just this integral,

$$\int||p^kx^3,p^{k+1}x^2y,p^2y^2||^sd\mu = (1-p^{-1})\int||p^k,p^{k+1}y,p^2y^2||^sd\mu + p^{-1}\int||p^{k+3}x^3,p^{k+3}x^2y,p^2y^2||^sd\mu$$

There are two integrals here to deal with, I'll name the first integral $G_k(s)$ and start working it in a moment, but the second integral here simplifies nicely.

$$\int||p^kx^3,p^{k+1}x^2y,p^2y^2||^sd\mu = (1-p^{-1})G_k(s) + p^{-1-2s}J_{k+1}(s)$$

$$G_k(s) = \int||p^k,p^{k+1}y,p^2y^2||^sd\mu$$

Its middle term can be removed since $|p^{k+1}|< |p^k|$ so multiplying it by a p-adic integer $|y|\le 1$ can never make it larger.

$$\int||p^k,p^2y^2||^sd\mu$$ $$ = p^{-2s}\int||p^{k-2},y^2||^sd\mu$$ Since I pulled out that, I'm assuming $k\ge 2$ for my next step here so we'll have to work out the $G_0(s)$ and $G_1(s)$ cases of this integral separately. $$= p^{-2s}\int_{\mathbb{Z}_p\setminus p^{k-2} \mathbb{Z}_p}||p^{k-2},y^2||^sd\mu + p^{-2s}\int_{p^{k-2} \mathbb{Z}_p}||p^{k-2},y^2||^sd\mu $$

This is starting to get a bit cumbersome, but $G_0(s)$, $G_1(s)$, and $G_k(s)$ for $k\ge 2$ are simple so I'll just stop there and focus on the main overview of the integral.

$$J_k(s) = 1-p^{-1} + (1-p^{-1})p^{-1}G_k(s) + p^{-2-2s}J_{k+1}(s)$$

To better convey the form that tends to happen when solving integrals in this way by spotting the pattern I'll define

$$A_k(s) := 1-p^{-1} + (1-p^{-1})p^{-1}G_k(s)$$

So now we can see it as,

$$J_k(s) = A_k(s) + p^{-2-2s} J_{k+1}(s)$$

Since we wanted to solve $J_0(s)$, we can simply imagine plugging this into itself by algebra over and over, each $A_k(s)$ gets another $p^{-2-2s}$ on it the deeper down it goes, giving us:

$$J_0(s) = \sum_{k=0}^\infty A_k(s) (p^{-2-2s})^k$$

At this point, we're effectively done. Everything you see here boils down to a geometric series when you expand out the $A_k(s)$. It might have been handy to have defined it with $(1-p^{-1})$ factored out, but it's not too serious. I can type it up more completely if there are parts you want me to elaborate on but I don't want to dilute the main flow (and I also don't trust myself to not make algebra mistakes haha).