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Hello,

By the local square theorem I know that $1+4\alpha$ is square if $|\alpha|<1$ ($\alpha$ is not a unit). Now, Can I always get a unit $\alpha$ such that $1+4\alpha$ is not a square ?? For example, taking $1 \in \mathbb{Q}_2$ you get 5, a unit of minimal quadratic defect in $\mathbb{Q}_2$.

Thanks !

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I realised that there exist always an $\alpha \in \mathcal{O}_k$ ($k$ dyadic) such that $1+4\alpha$ is not square because there exist always a unit $\delta$ with quadratic defect $4\mathcal{O_k}$, so if $1+4\alpha$ is square for all unit $\alpha$, then such a unit $\delta$ could not exist. Now I change my question: If $\alpha \in \mathcal{O}_k$ is a unit. Can I always find $\beta \in k(\delta)$ ($\delta with quadratic defect 4\mathcal{O}_k$) such that $tr(\beta)=\beta+\bar{\beta}=\alpha$ ?? –  user18497 Nov 10 '11 at 16:53
    
sorry, is $\beta \in k(\sqrt{\delta})$ and ($\delta$ with quadratic defect $4\mathcal{O}_k$) –  user18497 Nov 10 '11 at 16:59
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3 Answers 3

You expect that when you try to take square root of $1+4u$, you're led to an unramified extension, just as ${\mathbb{Q}}_2(5^{1/2})$ is unramified over $ {\mathbb Q}_2$. Indeed, from $x^2 - (1+4u)$ you are led (by an appropriate change of variables) to $X^2 + X - u$, clearly either irreducible with roots in an unramified extension of your $2$-adic ground field, or reducible, depending on whether the corresponding polynomial in characteristic $2$ doesn't or does have roots in the residue field. That drops out of Hensel's Lemma.

The upshot is that you can ``always get a unit $u$ such that $1+4u$ is not a square'' if and only if the residue field has quadratic extensions. Always, in particular, if the residue field is finite.

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Maybe it is not a bad idea to understand all this in terms of the filtration on the quotient $K^\times/K^{\times 2}$, where $K$ is a finite extension of ${\bf Q}_2$ and the filtration on the quotient is induced by the filtration $...\subset U_2\subset U_1\subset K^\times$ of the multiplicative group by units of various levels. It can be shown that the image of $U_{2e}$ has order $2$ whereas the image of $U_{2e+1}$ is trivial, where $e$ is the ramification index of $K$ over ${\bf Q}_2$.

In particular, you can always find a unit $x$ such that $1+4x$ generates the image of $U_{2e}$, which is the same as saying that $1+4x$ is not a square in $K^\times$.

You should try to work out the filtration on $K^\times/K^{\times p}$ for any finite extension of ${\bf Q}_p$ ($p$ any prime), or look it up here.

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A complete description is as follows (Serre, Course of Arithmetic, Chapter 2). An element $x=2^nu\in \mathbb Q_2$ is a square if and only if $n$ is even and $u\equiv 1 \mod 8$.

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I know the answer in that case. Thanks. –  user18497 Nov 10 '11 at 17:01
    
A description of squares for the general dyadic case is given in the book by W. Narkiewicz, Elementary and analytic theory of algebraic numbers, Warsaw, 1974 (page 215). It is more complicated, but perhaps you will be able to deduce the property you need. –  Anatoly Kochubei Nov 10 '11 at 18:25
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