Squares in rings of algebraic integers Let $K$ be a cubic field, and let $\alpha \in \mathcal{O}_K$ be an irrational algebraic integer. Does there exist a rational integer $m \ne 0$ such that $m \alpha$ is a square in $K$? What about number fields of other degrees?
 A: Let $L$ be the Galois closure of $K/\mathbb{Q}$, then Chebotarev's density theorem or this argument enable us to find a prime $p$ that splits completely in $L$. So $p$ also split completely in $K$, and we denote
$$p\mathcal{O}_K=\prod_{i=1}^{[K:\mathbb{Q}]}\mathfrak{p}_i.$$
Since the class group of $K$ is finite, it is easy to find an ideal $\mathfrak{a}\subseteq\mathcal{O}_K$ coprime to $p\mathcal{O}_K$ such that $\mathfrak{p_1}\mathfrak{a}$ is principal. Let $\alpha\mathcal{O}_K=\mathfrak{p_1}\mathfrak{a}$, then $\nu_{\mathfrak{p}_1}(\alpha)=1$ while $\nu_{\mathfrak{p}_2}(\alpha)=0$.
However, for any non-zero integer $m\in\mathbb{Z}$, we have $\nu_{\mathfrak{p}_1}(m)=\nu_{p}(m)=\nu_{\mathfrak{p}_2}(m)$. Therefore, one of $\nu_{\mathfrak{p}_1}(m\alpha)$ and $\nu_{\mathfrak{p}_2}(m\alpha)$ must be odd, thus $m\alpha$ can not be a square in $K$.
This construction works for any finite extension $K/\mathbb{Q}$ with $[K:\mathbb{Q}]\geqslant2$.
A: Contrary to @Mercury, my answer will not be number theoretic, but Galois theoretic. It will be specific to the cubic case, but will give necessary and sufficient criteria. Let $K$ be the given cubic field over $\mathbf Q$. If $\alpha \in K$ and $\notin \mathbf Q$, necessarily $K= \mathbf Q(\alpha)$ because $3$ is a prime number. Let $\phi_{K/\mathbf Q}$ be the natural map $\mathbf Q^*/{\mathbf Q^*}^2 \to K^*/{K^*}^2$. The condition $m\alpha \in {K^*}^2$ can be translated as : $\alpha {K^*}^2 \in Im \phi_{K/\mathbf Q}$ (*). In other words, the problem amounts to the determination of $Im \phi_{K/\mathbf Q}$ in terms of parameters in $K$. The minimal polynomial of $\alpha$ over $\mathbf Q$ can be classically reduced to the form $X^3 + pX + q$, with discriminant $D= -4p^3 - 27q^2$. Introduce "the" normal closure $N$ of $K$, with Galois group $G$ isomorphic to $S_3=D_6$ or $A_3 = C_3$. The distinction between the two cases depends on $D$ being or not in ${\mathbf Q^*}^2$ :
1) If $D$ is a square, then $N=K$ and $G\cong C_3$, and $Im \phi_{K/\mathbf Q}$ can be conveniently determined using Kummer theory and Galois cohomology. The exact sequence of $G$-modules $1 \to (\pm 1) \to K^* \to {K^*}^{2} \to 1$  (1) gives rise to the cohomology exact sequence $ H^1(G, K^*)\to H^1(G, {K^*}^2)\to H^2(G,(\pm 1))$. But $H^1(G, K^*)=1$ by Hilbert's thm. 90 and $H^1(G,(\pm 1))=1$ because $G$ and $(\pm 1)$ have coprime orders. Finally $H^1(G, {K^*}^2)=1$. Besides, the exact sequence $1\to {K^*}^2 \to K^* \to K^*/{K^*}^2\to 1$  (2) produces an exact sequence
$1\to ({K^*}^2)^G \to {K^*}^G=\mathbf Q^*\to (K^*/{K^*}^2)^G \to H^1(G, {K^*}^2)=1$, so $Im \phi_{K/\mathbf Q}= (K^*/{K^*}^2)^G $. As for $Ker \phi_{K/\mathbf Q}$, note that $({K^*}^2)^G={K^*}^2 \cap \mathbf Q^*$ and the latter group is just ${\mathbf Q^*}^2$ because the square root of each of its element gives rise to an extension of degree 1 or 2 inside $K$ which has degree 3 over $\mathbf Q$ ; in other words, $\phi_{K/\mathbf Q}$ is injective. In conclusion, $\mathbf Q^*/{\mathbf Q^*}^2 \cong (K^*/{K^*}^2)^G $. Note that $K^*/{K^*}^2$ parametrizes the quadratic extensions of $K$ whereas $(K^*/{K^*}^2)^G $ parametrizes those among them which are normal over $\mathbf Q$ (easy Galois exercise) . The isomorphism just obtained means that those particular quadratic extensions of $K$ are obtained by composing $K$ with quadratic extensions of $\mathbf Q$, and this due essentially to $2\neq 3$.
2) The more complicated case is when $N\neq K$ and $G=Gal(N/\mathbf Q) \cong S_3$, which occurs when $D$ is not a square. The subextensions of $N/\mathbf Q$ are the following : $L=\mathbf Q(\sqrt D)$ is the fixed field of $A_3$ and the unique quadratic field contained in $N$; the three conjugate roots $\alpha, \alpha ', \alpha ''$ generate three cubic subextensions $K, K', K''$ fixed by the transpositions of $S_3$, and $N=K(\sqrt D)$. Put $H=Gal(N/L)$ and $J=G/H=Gal(L/\mathbf Q)\cong Gal(N/K)$. We want to determine the image of $\phi_{K/\mathbf Q}$ using the same cohomological machinery as in 1). To circumvent the inconvenient that $K/\mathbf Q$ is not Galois, just exploit the transition formula $\phi_{N/\mathbf Q}=\phi_{N/K}. \phi_{K/\mathbf Q}=\phi_{N/L}.\phi_{L/\mathbf Q}$. I don't give all the details, only the key steps :
a) To compute $\phi_{L/\mathbf Q}$, take as before the cohomology of the analog of the exact sequence (1) relative to $L/\mathbf Q$. The point is that $[L:\mathbf Q]=2$ so we can no longer conclude that $H^1(J, {L^*}^2)=1$. Going on to $H^2$, we get instead $H^1(J,{L^*}^2)=\epsilon$, where $\epsilon = (1)$ or ($\pm 1$), and $\epsilon=1$ iff $-1$ is a norm in $L/\mathbf Q$ (this condition is a kind of Pell-Fermat equation). Then the cohomology of the analog of the exact sequence (2) gives that $Ker \phi_{L/\mathbf Q}= D.{\mathbf Q^*}^2/{\mathbf Q^*}^2$ and $(L^*/{L^*}^2)^J/Im\phi_{L/\mathbf Q}=\epsilon$.
b) Replacing $\mathbf Q$ by $L$ in 1), the computation of $\phi_{N/L}$ goes on exactly as in 1) to yield  $Im \phi_{N/L}= (N^*/{N^*}^2)^H $ . In conclusion, $Im \phi_{N/\mathbf Q}$ "is $(N^*/{N^*}^2)^G $ up to $\epsilon$". The same kind of interpretation in terms of quadratic extensions as at the end of 1) is still valid here. The quadratic field $L$ is ruled out because it’s already contained in $N$.
Remarks.
(i) All the calculations above can be carried on over any field of characteristic $\neq 3$ instead of $\mathbf Q$.
(ii) Part 1) remains valid for any odd prime $p$ in place of $3$. So do some pieces of part 2).
