Elementary proof that $\frac{\mathbb{K}^*}{{\mathbb{K}^*}^2}$ is infinite when $\mathbb{K}$ is a number field Let $\mathbb K$ be a number field, i.e. a finite extension of $\mathbb Q$. In the (multiplicative) abelian group $A={\mathbb K}^{*}$, we can consider the quotient $\frac{A}{2A}$ (or $\frac{{\mathbb K}^{*}}{{{\mathbb K}^{*}}^2}$ if you prefer ) which will always be a direct sum of copies of $\frac{\mathbb Z}{2\mathbb Z}$.
It easily follows from the Cebotarev density theorem that there are infinitely many primes in ${\mathcal O}_{\mathbb K}$, and hence $\frac{A}{2A}$ is infinite.
My question : can we show that $\frac{A}{2A}$ is infinite in a more elementary way ?
My guess would be  that if $\pi$ is the canonical projection ${\mathbb K}^{*} \to \frac{{\mathbb K}^{*}}{{{\mathbb K}^{*}}^2}$ and $\cal P$ is the set of ordinary primes in $\mathbb N$, then $\pi({\cal P})$ is infinite. Perhaps an even stronger property holds, that the $\pi(p)$ are ${\mathbb F}_2$-linearly independent for large enough $p$.
 A: Your guess regarding the canonical projection $\pi \colon K^* \to K^* / {K^*}^2$ is correct, there is an infinite subset of primes in $\mathbb{N}$ whose image under $\pi$ is also infinite.

Let $\mathbb{P}$ be the set of primes in $\mathbb{N}$. For any $p,q \in \mathbb{P}$, we have that
$$
p{K^*}^2 = q{K^*}^2 \iff \sqrt{p/q} \in K^*. \tag{1}
$$
So, let $S := \{ p/q \in \mathbb{Q} : p,q \in \mathbb{P}, \sqrt{p/q} \in K^*\}$, and $\sqrt{S} := \{ \sqrt{r} : r \in S \}$. Consider the intermediate field $L := \mathbb{Q}(\sqrt{S})$.
Claim: If $\lvert S \rvert = \infty$, then $[L : \mathbb{Q}] = \infty$.
Proof. Suppose that $\lvert S \rvert = \infty$. We show that if $Y = \{ p_1/q_1, \dotsc, p_N/q_N \} \subseteq S$ and $p/q \in S$ is an element such that $p \not\in X := \{p_1,\dotsc,p_N,q_1,\dotsc,q_N\}$, then
$[\mathbb{Q}(\sqrt{Y})(\sqrt{p/q}):\mathbb{Q}(\sqrt{Y})] = 2$.
So, suppose for the sake of contradiction that $\sqrt{p/q} \in \mathbb{Q}(\sqrt{Y})$. Then, $\sqrt{p} \in \mathbb{Q}(\sqrt{X},\sqrt{q})$. But, $p \not\in X \cup \{ q\}$, so $[\mathbb{Q}(\sqrt{X}, \sqrt{q})(\sqrt{p}):\mathbb{Q}(\sqrt{X}, \sqrt{q})] = 2$, by the result here. In particular, $\sqrt{p} \not\in \mathbb{Q}(\sqrt{X}, \sqrt{q})$, a contradiction.
Now, we show that there exists a sequence $(p_i/q_i)_{i \in \mathbb{N}}$ in $S$ such that $p_{N+1} \not\in \{ p_1,\dotsc,p_N,q_1,\dotsc,q_N \}$ for each $N \in \mathbb{N}$. This can be done by induction, and using the hypothesis that $\lvert S \rvert = \infty$.
Pick $p_1/q_1 \in S$ arbitrarily. Assume that $p_1/q_1,\dotsc,p_N/q_N$ have been chosen for some $N \geq 1$. Since $$\{ p_1,\dotsc,p_N,q_1,\dotsc,q_N \}\tag{2}$$ is a finite set of primes, and $S$ is infinite, there must exist a prime that "occurs" in some element of $S$ that is distinct from the ones in the displayed set. Since $p/q \in S \iff q/p \in S$, we can choose an element $p/q \in S$ as required.
Hence, $[L:\mathbb{Q}] = \infty$, because for every $n \in \mathbb{N}$, there is an intermediate field $L_n := \mathbb{Q}(\sqrt{p_1/q_1},\dotsc,\sqrt{p_n/q_n})$ such that $[L_n : \mathbb{Q}] = 2^n$. $\blacksquare$
Thus, if $\lvert S \rvert = \infty$, then $[L : \mathbb{Q}] = \infty$, so $[K : \mathbb{Q}] = \infty$, which contradicts that $K$ is a number field.
Hence, $S$ is a finite set. In other words, only finitely many of the cosets $p{K^*}^2$ collapse onto each other. There remain infinitely many distinct cosets of the form $p{K^*}^2$ for $p \in \mathbb{P}$. Hence, $K^*/{K^*}^2$ is infinite.
A: Consider the natural extension homomorphism $\eta : Q^*/{Q^*}^2 \to K^*/{K^*}^2$ defined by $[a]_Q \to [a]_K$, where $[a]$ denotes the class of $a$ modulo squares. We are interested in the image Im $\eta$. Recall that $[a]_Q=[b]_Q$ iff $Q(\sqrt a)=Q(\sqrt b)$, and Ker $\eta$ consists of the classes $[a]_Q$ s.t. $Q(\sqrt a)\subset K$. Since $[K:Q]$ is finite, it ensues from Kummer theory that Ker $\eta$ is finite, so $K^*/{K^*}^2$ is infinite iff $Q^*/{Q^*}^2$ is.
The question about the infiniteness of $Q^*/{Q^*}^2$ has been asked (finitely) many times in these columns. The answer lies of course in the infiniteness of the rational primes. I think that the most concise and precise proof is the following (I give all the details): show that $k_n=Q(\sqrt p_1,..., \sqrt p_n)$, where the $p_i$'s are $n$ distinct primes, has degree $2^n$over $Q$. In additive notation, $Q^*/{Q^*}^2$ could be viewed as a vector space over the finite field $F_2$, and the problem can be rephrased as dim ($k_n^*$/ ${k_n^*}^2)=2^n$. But a (multiplicative) relation of $F_2$-linear dependence in $k_n^*$/ ${k_n^*}^2$ between the  $\sqrt p_{i}$ would read
${\sqrt p_1}^{\epsilon_1}...{\sqrt p_n}^{\epsilon_n}=1$, with $\epsilon_i=0$ or $1$. Squaring this relation and coming back to Z, one would get ${ p_1}^{\epsilon_1}...{ p_n}^{\epsilon_n}=$ a square in Z: impossible except if all the ${\epsilon_i}$ vanish.
