How to Determine π-adic Order of Divisibility in Polynomial Rings

Hello! This is a fantastic question about algebraic number theory and p-adic valuations. Let's work through this together!

What We're Solving:

We want to understand how to define the π-adic order (or valuation) in the quotient ring $\mathbb{Z}_{p^q}[t]/(t^2 - \delta)$, where $\delta = pl$ with $p \nmid l$. This is essentially asking: how do we measure "divisibility by π" in this particular ring structure?

The Approach:

The key insight is that $\mathbb{Z}_{p^q}[t]/(t^2 - \delta)$ is a quotient ring where we've adjoined a "square root" of $\delta$ to $\mathbb{Z}_{p^q}$. Since $\delta = pl$, we need to identify what plays the role of a "prime element" π in this extended ring, then define the valuation based on how many times elements are divisible by this π.

Step-by-Step Solution:

Step 1: Understand the ring structure In $\mathbb{Z}_{p^q}[t]/(t^2 - \delta)$, we can think of $t$ as $\sqrt{\delta} = \sqrt{pl}$. Every element can be written uniquely as $a + bt$ where $a, b \in \mathbb{Z}_{p^q}$.

Step 2: Identify the prime element π Since $t^2 = pl$ and $p \nmid l$, we have $t^2 \equiv 0 \pmod{p}$. This suggests that $t$ itself should be our prime element π. Indeed, $t = \sqrt{pl}$ contains "half" the p-adic content of $p$.

Step 3: Define the π-adic valuation For any non-zero element $\alpha = a + bt \in \mathbb{Z}_{p^q}[t]/(t^2 - \delta)$, the π-adic order $v_π(\alpha)$ is the largest integer $k$ such that $π^k$ divides $\alpha$ in this ring.

Step 4: Computational approach To find $v_π(\alpha)$ for $\alpha = a + bt$:

• If $b \not\equiv 0 \pmod{p}$, then $v_π(\alpha) = 0$
• If $b \equiv 0 \pmod{p}$ but $a \not\equiv 0 \pmod{p}$, then $v_π(\alpha) = 0$
• If both $a, b \equiv 0 \pmod{p}$, write $\alpha = p(a' + b't)$ and repeat
• The process continues, counting how many factors of π can be extracted

Step 5: Alternative perspective using norms Another approach uses the norm map. The norm of $\alpha = a + bt$ to $\mathbb{Z}_{p^q}$ is $N(\alpha) = (a + bt)(a - bt) = a^2 - b^2\delta$. Then $v_π(\alpha) = \frac{1}{2}v_p(N(\alpha))$ when this is an integer.

The Answer:

The π-adic order in $\mathbb{Z}_{p^q}[t]/(t^2 - \delta)$ is defined with $π = t = \sqrt{\delta}$. For any element $\alpha = a + bt$, the π-adic order $v_π(\alpha)$ is the largest power $k$ such that $t^k$ divides $\alpha$. This can be computed by repeatedly factoring out $t$ or by using the norm formula $v_π(\alpha) = \frac{1}{2}v_p(N(\alpha))$ when applicable.

Memory Tip:

Think of it this way: when you adjoin $\sqrt{pl}$ to $\mathbb{Z}_{p^q}$, you're "splitting" the prime $p$ in a sense. The element $t = \sqrt{pl}$ becomes your new "prime-like" element, and it has half the p-adic weight of the original prime $p$. That's why the norm formula involves dividing by 2!

This is a beautiful example of how valuations extend when we pass to larger rings. Keep exploring these connections - they're fundamental in algebraic number theory!