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Let $F$ be a non-archimedean local field or a general global field.

Has $F^\times / (F^\times)^2$ cardinality $2$?

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up vote 7 down vote accepted

For $F = \mathbb Q_2$ we have $\mathbb Q_2^\times / (\mathbb Q_2^\times)^2 \cong (\mathbb Z / 2\mathbb Z)^3$, so the cardinality is 8, not 2.

A 2-adic number $x =\mathbb Q_2^\times$ can be written as $x = 2^k u$ with $k \in \mathbb Z$ and $u \in \mathbb Z_2^\times$, so $\mathbb Q_2^\times \cong \mathbb Z \times \mathbb Z_2^\times$ . One can easily show that $x$ is a square in $\mathbb Q_2^\times$ iff $k$ is even and $u \equiv 1 \pmod {2^3}$, so $(\mathbb Q_2^\times)^2 \cong 2\mathbb Z \times U^{(3)}$ where $U^{(3)} = \{ u \in \mathbb Z_2^\times | u \equiv 1 \pmod{2^3} \}$ is the third group of principal units. Thus $$\mathbb Q_2^\times / (\mathbb Q_2^\times)^2 \cong \mathbb Z/2\mathbb Z \times \mathbb Z_2^\times/U^{(3)} \cong (\mathbb Z/2\mathbb Z)^3.$$

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