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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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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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