Extending $2$-adic valuation to a finite extension of $\mathbb Q$.

Question

Say we want to extend the $2$-adic valuation of $\mathbb Q$ to $\mathbb Q[x]/x^3-3$. First, we determine the possible valuations of $x$. Since $x^3=3$ and since $3$ has $2$-adic valuation $0$, $x$ must have valuation $0$ as well.

But I'm not sure how to get the valuation of $x^2-3$. Since $x^2$ and $3$ both have valuation $0$, the valuation of their difference can potentially be any nonpositive real. Is there a way to determine what the valuation of $x^2-3$ must be?

Answer 1

Denote your valuation by $v$. Using the fact that $v(x) = 0$, $$v(x^2-3) = v\big(x(x^2-3)\big)=v(3-3x) = v(1-x).$$

Now, from the fact that $X^3-3 = 0$, we can deduce that $$(1-x)^3 -3(1-x)^2+3(1-x)+2=0$$and that $$1=v(2)=v(1-x) + v\big((1-x)^2-3(1-x)+3\big).$$It follows that either $$v(1-x) =1\text{ and } v\big((1-x)^2-3(1-x)+3\big)=0$$or$$v(1-x) =0\text{ and } v\big((1-x)^2-3(1-x)+3\big)=1.$$Which possibility will depend on which choice of valuation you take extending $v$. Letting $\mathcal O=\mathbb Z[\sqrt[3]3]$, we have $$2\mathcal O = \langle1-\sqrt[3]3\rangle\langle (1-\sqrt[3]3)^2-3(1-\sqrt[3]3)+3\rangle =:\mathfrak{p_1p_2}$$and each $\mathfrak p_i$ gives a different valuation.

Answer 2

A different approach to the problem of extending the $2$-adic valuation to $K=\Bbb Q[x]/(x^3-3)$, simultaneously more abstract and more computational:

The possible $2$-adic valuations on $K$ correspond to the $\Bbb Q_2$-irreducible factors of $X^3-3$. Since $\Bbb Q_2$ has a perfectly good cube root of $3$ in it, let’s call it $\lambda$ for now, we have $X^3-3=(X-\lambda)(X^2+\lambda X+\lambda^2)$. Your first $2$-adic valuation on $K$ is gotten by embedding it into $\Bbb Q_2$ directly, and to find the $2$-adic valuation of $a+bx+cx^2$ for given rational numbers $a$, $b$, and $c$, you plug in your numerical value of $\lambda$ (see below) for $x$ and see what the resulting $2$-adic number turns out to be.

The second valuation embeds $K$ into the field $\Bbb Q_2[y]/(y^2+\lambda y+\lambda^2)$ (don’t forget that $\lambda\in\Bbb Z_2\subset\Bbb Q_2$) by sending $x$ to $y$; the $2$-adic valuation of $a+by$ is $\min(v(a),v(b))$ for $2$-adic numbers $a$ and $b$ and the standard $2$-adic valuation $v$, so that we get $$ v_2(a+bx+cx^2)=v\bigl(a+by+c(-\lambda^2-\lambda y)\bigr)=\min\bigl(v(a-c\lambda^2),v(b-\lambda)\bigr)\,. $$ I promised the numerical value of $\lambda$. There are many ways of finding it, but Newton-Raphson is quickest. To twenty-four places of $2$-adic accuracy, we have $\lambda=\dots21f17b;$, where I’m using hexadecimal notation, right to left. That is, $\lambda$ is $11+7\cdot16+1\cdot16^2+\dots$. If you want the valuation(s) of $x^2-3$, well, for $v_1$ you’re talking about $\lambda^2-3=\dots7ed116;$, so $v_1(x^2-3)=1$, and $y^2-3=(-\lambda^2-3)-\lambda y$, which is a unit, so that the $v_2$-value of $x^2-3$ is $0$.