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What's the 2-adic value of $\sqrt3 + \dfrac1{1+\sqrt3}$?

To start off, I have that to admit a square root, a 2-adic unit must be of the form $4^a (8b+1)$ for some $a \in \mathbb{N}$, $b \in \mathbb{Z}$.

I don't think $3$ is of that form, so I can only make this work if I extend $\Bbb Z_2$

I'm tentatively interpreting this answer as saying that I can simply add $\omega=\sqrt3$ as an additional unit without too much going wrong.

From there I'm running blind. I feel like I need to turn it into a polynomial then Hensel lift to find my answer. Total guesswork here but I have that my number will be a solution to the quotient of the polynomials $\omega^2+\omega+1$ and $\omega+1$ so if I can find the 2-adic value of those I'll be done.

I have that $\omega^2+\omega+1$ will be odd and $\omega+1$ will be even so I'll get a number in $\Bbb Q_2\setminus\Bbb Z_2$, and from there I'm even out of guesswork.

it's a hire car baby
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    I agree with you that one must extend $\Bbb Z_2$ and that $\sqrt3$ is a unit. Given that $(1+\sqrt3)(1-\sqrt3) = -2$, what do you think the $2$-adic valuation of $1+\sqrt3$ is? If you know that answer, can you work out the $2$-adic valuation of $\frac{4+\sqrt3}{1+\sqrt3}$? – Greg Martin Oct 20 '23 at 20:16
  • @GregMartin I'm sure this is wrong but I'm getting that $\lvert(1+\sqrt3)\rvert=\lvert(1-\sqrt3)\rvert$ and therefore $\lvert(1+\sqrt3)\rvert=\sqrt2$. That would make my answer $\nu_2(\frac1{\sqrt2})=-\frac12$ based on the factorisation into polynomials I have in the question. – it's a hire car baby Oct 20 '23 at 21:44
  • @GregMartin actually I can see that my deduction $\omega^2+\omega+1$ will be odd was incorrect so my above solution is wrong. Now I'm getting that $\lvert\omega^2+\omega+1\rvert=\sqrt2$ and $\lvert\sqrt2\rvert=\sqrt2$ and the quotient of those is $1$ so the answer is $\nu_2(1)=0$ – it's a hire car baby Oct 20 '23 at 21:55
  • Just for clarity note that $\sqrt{3}$ is, as usual outside of real numbers, not a perfectly well-defined symbol. However the answer to your question should not depend on which of the two possible meanings one attaches to that symbol, as follows from Greg Martin's comment and the valuation being nonarchimedean so that the value of a sum is the maximum of the summands' values if those are not equal. – Torsten Schoeneberg Oct 20 '23 at 23:39
  • I downvoted because everything written about $\omega$'s in the question looks both wrong and irrelevant. – Torsten Schoeneberg Oct 20 '23 at 23:41
  • @TorstenSchoeneberg that's not part of the question. It's part is my own attempt at the question rather than the question itself, which I am required to include due to site policy. It doesn't seem sensible to require users to attempt a question they can't answer then downvote the question for them doing so. – it's a hire car baby Oct 21 '23 at 03:21
  • @TorstenSchoeneberg are you saying that in this setting $\omega+\frac1{1+\omega}\neq (\omega^2+\omega+1)\div(1+\omega)$ Because that doesn't look all that different to Greg's $(4+\sqrt3)/(1+\sqrt3)$ to me, it looks identical. So if everything about $\omega$s is wrong so is Greg's comment, correct? And I need to disregard it? – it's a hire car baby Oct 21 '23 at 03:45
  • Also @TorstenSchoeneberg where tou write "the value of a sum..." do you mean the valuation of the sum, or the result of the summation itself? It's ambiguous to me. – it's a hire car baby Oct 21 '23 at 03:52
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    I don't personally agree with the down-vote (as the OP states, the incorrectness is because of their confusion regarding the problem), but I must commend @TorstenSchoeneberg for at least stating the reason behind the down-vote. Too many drive-by down-votes these days. – Brian Tung Oct 21 '23 at 16:40
  • By "the value of the sum ..." I meant that $\lvert a + b \rvert = \max(\lvert a \rvert, \lvert b \rvert)$ if $\lvert a\rvert \neq \lvert b \rvert$, for any nonarchimedean value $\lvert \cdot \rvert$. --- OK, not everything about $\omega$ is wrong, and I guess I should actually appreciate writing $\omega$ instead of $\sqrt{3}$. However, the last line/paragraph of the question makes no sense to me. Even if one somehow gave meaning to calling elements in question "even" or "odd", why would anything here come out in $\mathbb Q_2 \setminus \mathbb Z_2$? – Torsten Schoeneberg Oct 21 '23 at 16:57
  • @TorstenSchoeneberg my reasoning was that a square root of a unit would be a unit, and a sum of three units would be a unit. And a sum of a unit plus 1 would be even. So I'd have an odd number divided by an even one. My guess is that my mistake was to think that a square root of a unit must be a unit. – it's a hire car baby Oct 21 '23 at 17:39
  • No, that part of your reasoning is correct. I just veto calling $1+\omega$, or any other element of $\mathbb Z_2[\omega]$, "even" instead of using precise terminology either that it's "divisible by $2$" (which it is not) or "not a unit" (which it is). I also still don't see what any of that would have to do with $\mathbb Q_2$. – Torsten Schoeneberg Oct 21 '23 at 19:07

1 Answers1

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Let me try to cast some light on the issue.

First, don’t use omega here for an irrationality, ’cause half the people reading your question will immediately think of a primitive cube root of unity, $\omega^2+\omega+1=0$. If you don’t want the pain of repeately typing “$\sqrt3$”, then call this irrationality, say, $r$, for “root”.

Second, you should choose either the multiplicative valuation $|*|_2$ for which $|2|_2=\frac12$, or the additive valuation $v_2(*)$, for which $v_2(2)=1$. I prefer the additive one, partly because of typographical simplicity. You seem to have had this in mind in your question.

You know that $3$ is a $2$-adic unit, so that $v(3)=0$. Since $r^2=3$, you have $v(r)=\frac12v(3)=0$ as well. But a crucial quantity here is $1+r$, whose $\Bbb Z_2$-minimal polynomial is $X^2-2X-2$: the (field-theoretic) norm of $1+r$ is $-2$, and the rule for extending valuations is: if $\alpha$ is of degree $m$ over $\Bbb Q_p$, then $$ v(\alpha)=\frac1mv_p\bigl(\mathbf N^{\Bbb Q_p(\alpha)}_{\Bbb Q_p}(\alpha)\bigr)\,. $$ That’s the norm there in the formula, a function from the superscripted field to the subscripted field. In our case $\frac12v_2(-2)=\frac12$. Much more directly, in this case, you certainly expect that $v_2(1+r)=v_2(1-r)$, just because they’re conjugates, so that $v_2(1+r)=\frac12v_2\bigl((1+r)(1-r)\bigr)=\frac12v(-2))=\frac12$.

As to the quantity that stimulated this question, namely $$ \sqrt3+\frac1{1+\sqrt3}=r+\frac1{1+r}\,, $$ your two terms have valuation $0$ and $-\frac12$, so that the sum has valuation $-\frac12$

Lubin
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  • Thank-you for this. I will digest over the coming days. What astounds me the most is that my first comment here got the right answer! Was my logic sound there? https://math.stackexchange.com/questions/4790785/whats-the-2-adic-value-of-sqrt3-frac11-sqrt3/4792035#comment10186056_4790785 – it's a hire car baby Oct 23 '23 at 09:39