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I am trying to figure out the structure of the group of square classes $F^{\times}/(F^{\times})^{2}$ where $F=\mathbb{Q}[i]$ (Gaussian numbers).

I was trying when a gaussian integers is a perfect square by following the arguments on Perfect squares and cubes in quadratic number fields for the case $d=-1$, however I could not solve anything from that. Instead I was thinking if I can say something about the quotient like Guillot's "A Gentle Course in Local Class Field Theory", which states that $\mathbb{Q}^{\times}/(\mathbb{Q}^{\times})^{2}=\{{\pm}1\}A/2A$ where where $A$ is a free abelian group, basis the set of prime numbers.

Any idea about this would be appreciate it.

Thank you for your time!

Liddo
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    What is the answer for $F=\mathbb{Q}$? Then recall that $\mathbb{Q}[i]=\mathbb{Q}(i)$ has unique factorization in Gaussian primes. – lhf Nov 01 '21 at 19:11

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Let $P$ be a set of representatives for the prime elements of $\Bbb Z[i]$ up to associates. Then by unique factorization, we have $F^\times \cong \Bbb Z[i]^\times \times \Bbb Z^{(P)}$, where $\Bbb Z^{(P)}$ denotes the free abelian group on $P$. Now $\Bbb Z[i]^\times = \{\pm 1, \pm i\}$, thus $F^\times/(F^\times)^2 \cong \{\pm 1, \pm i\}/\{\pm 1\} \times \Bbb Z^{(P)}/2\Bbb Z^{(P)} \cong \Bbb Z/2\Bbb Z \times (\Bbb Z/2\Bbb Z)^{(P)}$

Lukas Heger
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