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Let $\mathbb Q_p$ be a $p$-adic field and $f,g \in \mathbb Q_p[x]$. Suppose $v$ is a $p$-adic valuation, and let $f \circ g$ be the formal composition.

Is $v(f \circ g) \geq v(f)+v(g)$ ?

For the product, we know that $v(fg)=v(f)+v(g)$ is holds.

Suppose, $f(x)=px+p^2x^2$ and $g(x)=x+px^3$, then \begin{align} v(f(x))=v(px+p^2x^2)\geq \min \{v(px), v(p^2x^2) \}&=\min \{v(p)+v(x), v(p^2)+v(x^2) \} \\ &=\min\{1+v(x), 2+2v(x) \} \\ &=1+v(x). \end{align} \begin{align} v(g(x))=v(x+px^3)\geq \min \{v(x), v(px^3) \}&=\min \{v(x), v(p)+v(x^3) \} \\ &=\min\{v(x), 1+3v(x) \} \\ &=v(x). \end{align} While $$v(f \circ g)=v(p^4x^6+2p^3x^4+p^2x^3+p^2x^2+px) \geq \min \{4+6v(x),3+4v(x),2+3v(x),2+2v(x),1+v(x) \}=1+v(x).$$ This doesn't give any conclusion. The question reduces to define $v(x)$ on $\mathbb Q_p[x]$.

How to define valuation $v$ on $\mathbb Q_p[x]$ extending the valuation $v_p$ on $\mathbb Q_p$ ?

Suppose $h(x)=a_ix^i$ is a monomial of $f(x)$. Then define $v(h)=v_p(a_i)+i$. This satisfies the definition of valuation. Am I right ?

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    $\begingroup$ I don’t think you want to define $v(f)$ for a polynomial. Would you define $|f|$ for a polynomial (or any other function) in the real domain? $\endgroup$
    – Lubin
    Jun 22, 2023 at 22:14
  • $\begingroup$ @Lubin, I had in mind $|f(x)|=p^{-v(f(x))}$. Indeed, I want to show that a sequence of functions, say $\{h_n(x)\}$ converges $p$-adically. in other words, I want to show $v(h_{m+n}(x)-h_n(x)) \to \infty$ so that $\{h_n\}$ becomes Cauchy sequence. Anyway, regarding my original question, is it valid to define $v(g(x)):=v_p(a_i)+i$ for a monomial $g(x)=a_ix^i$ of $f(x)$, where $v_p$ is valuation on $\mathbb Q_p$ ? $\endgroup$
    – MAS
    Jun 23, 2023 at 0:14
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    $\begingroup$ I gave my best stab at answering your question, below. $\endgroup$
    – Lubin
    Jun 23, 2023 at 1:24

2 Answers 2

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Let’s be a little more general than in your setup. Let $k$ be any finite (algebraic) extension of $\Bbb Q_p$, and let $\Omega$ be an algebraic closure, or, if you like, the $p$-adic completion $\Bbb C_p$ of this. Like you, I’ll let $v$ be the (additive) valuation on $\Bbb Q_p$, normalized so that $v(p)=1$; you probably know that $v$ extends uniquely to $k$ and to $\Omega$. Now let’s start:

You don’t want to associate to a polynomial $f(x)\in k[x]$ just a real number, but rather a real-valued function of a real variable. This has been done a long time ago, in prehistory almost. It has been called the “valuation function” of $f$, and I’ll denote it $V_f\colon\Bbb R\to\Bbb R$. For most elements $z\in\Omega$, we have $v(f(z))=V_f(v(z))$. That is, to know how big $v(f(z))$ is, you just plug the valuation of $z$ into $V_f$.

And how is $V_f$ defined? Let $f(x)=\sum_ia_ix^i$, the $a_i$ being in $k$. Then draw, in the Cartesian $(\xi,\eta)$-plane, all the closed half-planes $H_i: \eta\le i\xi+v(a_i)$. And then you get a convex set $\mathcal V_f$ as the intersection of all the $H_i$. And now we have it: $V_f$ is the function whose graph is the boundary of $\mathcal V_f$. You see that $V_f$ is a concave function (“concave down” in the language I learned in college). If I had world enough and time, I would draw an example, say for $f(x)=x^2+px + p^3$. In this case, there is a half-line going all the way to the left, of slope $2$; a segment of slope $1$ between $(1,2)$ and $(2,3)$; and a horizontal half-line going all the way to the right, height $3$. This tells you, for instance, that $v(f(\sqrt p\,))=1$, $v(f(p\sqrt p\,))=\frac52$, and $v(f(p^2\sqrt p\,))=3$.

The rules that the valuation function satisfies are pretty much what you expect: \begin{align} V_{f+g}&\ge\min\left\{V_f,V_g\right\}\\ V_{fg}&=V_f+V_g\,, \end{align} and most interestingly, if $g$ has no constant term, then $$ V_{f\circ g}=V_f\circ V_g\,. $$ I think the last rule is what you were looking for.

There’s some really interesting mathematics getting called in here, having to do with convex sets in Euclidean space. Suffice it to say that $\mathcal V_f$ is in some sense a dual convex body to the Newton polygon of $f$, when the latter is correctly defined. But that’s a tale for another evening.

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  • $\begingroup$ Thank you for the beautiful answer. I solved my my original purpose behind the question. I had some idea about valuations but your beautiful exposition is just brilliant! I learned a lot $\endgroup$
    – MAS
    Jun 23, 2023 at 12:06
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If you take $f(x)=g(x)=x$, then your inequality fails for any $x$ such that $\nu(x)\geq 1$, and $\nu(f\circ g)<\nu(f)+\nu(g)$

On the other hand, if $\nu(x)\leq -1$ then $\nu(f\circ g)>\nu(f)+\nu(g)$.

From this I don't think there is any true relation between $\nu(f\circ g)$ and $\nu(f)+\nu(g)$.

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  • $\begingroup$ Yes, I also think so. How to extend $v_p$ on $\mathbb Q_p$ to a new valuation $v(x)$ on $\mathbb Q_p[x]$ ? $\endgroup$
    – MAS
    Jun 22, 2023 at 16:11

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