I just want to know what properties of valuations extend to $\mathbb R$...

Denote an extension of the 2-adic valuation from $\mathbb Q$ to $\mathbb R$ by $\nu$. Suppose $\nu(x)=\nu(y)=0$.

Is it true that $\nu(x+y)\ne 0$?

What about $\nu(x^2+y^2)\le 1$?

I'm interested in knowing both whether these are true for every extension, as well as knowing whether there is some extension for which they are true (for every $x$ and $y$).

  • 2
    $\begingroup$ I changed the title to something more appropriate, so no more clickbait. $\endgroup$ – KConrad Apr 26 at 15:23
  • 8
    $\begingroup$ I saw nothing wrong with the title actually. Nothing wrong with a bit of humour. $\endgroup$ – RP_ Apr 26 at 16:03

No. The important thing to know is that, if $K \subseteq L$ is a field extension and $v: K \to \mathbb{R}$ is a valuation, then $v$ can be extended to $L$. So I can answer all of your questions by working in some easy to handle subfield of $\mathbb{R}$. I'll work in $K = \mathbb{Q}(\sqrt{5})$ for the first question and in $K = \mathbb{Q}(\sqrt{3})$ for the second.

The ring of integers in $\mathbb{Q}[\sqrt{5}]$ is $\mathbb{Z}[\tau]$ where $\tau = \tfrac{1+\sqrt{5}}{2}$, with minimal polynomial $\tau^2=\tau+1$. Note that $\mathcal{O}_K/(2 \mathcal{O}_K)$ is the field $\mathbb{F}_4$ with four elements. Your first statement is true in $\mathbb{Q}$ only because $\mathbb{Z}/(2 \mathbb{Z})$ has two elements.

Specifically, both $1$ and $\tau$ are in $\mathcal{O}_K$ but not $2 \mathcal{O}_K$, so $v(1) = v(\tau) = 0$, but $1+\tau$ is also not in $2 \mathcal{O}_K$ so $v(1+\tau)=0$ as well.

Similarly, the ring of integers in $\mathbb{Q}(\sqrt{3})$ is $\mathbb{Z}[\sqrt{3}]$ and the prime $2$ is ramified, with $2 = (1+\sqrt{3})^2 (2-\sqrt{3})$ (note that $2-\sqrt{3}$ is a unit). We have $v(1) = v(\sqrt{3}) = 0$, but $v(1+\sqrt{3}^2) = 2$. In this case, the result is true in $\mathbb{Q}$ because $2$ is unramified.

  • 2
    $\begingroup$ "the field $\Bbb F_4$ with four elements"? $\endgroup$ – Greg Martin Apr 26 at 16:47
  • $\begingroup$ Thanks for the correction! @GregMartin $\endgroup$ – David E Speyer Apr 26 at 20:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.