Completing Algebraic Integers into Squares

Let $$L/K$$ be an extension of number fields with Galois closure $$E$$, and let $$\theta \in \mathcal{O}_L \setminus \{0\}$$. Let $$\Sigma_E$$ be the set of primes of $$E$$, let $$S' \subset \Sigma_E$$ be a finite set containing all of the ramified primes, and let $$S = \{ \mathfrak{p} \in \Sigma_E \setminus S' : v_{\mathfrak{p}}(\theta) \not\equiv 0 \pmod{2}\}.$$ Notice that $$S$$ is finite, and let $$T$$ be the finite set of primes of $$K$$ lying below the primes in $$S$$. Suppose for each $$p \in T$$ that there exists $$r_p \in K_p^\times$$ such that $$v_{\mathfrak{p}}(r_{p} \cdot \theta) \equiv 0 \pmod{2}$$ for each $$\mathfrak{p} \mid p$$.

Question: Does there exist $$r \in K^\times$$ such that $$v_{\mathfrak{p}}(r\cdot \theta) \equiv 0 \pmod{2}$$ for all primes $$\mathfrak{p} \in \Sigma_E \setminus S'$$?

• The result is obvious when $$L = K$$, because one can simply take $$r = \theta$$.
• Suppose that $$L \neq K$$. Each $$r_p$$ can always be taken to be equal to a $$p$$-adic uniformizer. When $$K$$ has class number $$1$$, these uniformizers can be taken to be prime elements of $$K$$, so $$r = \prod_{p \in T} r_p$$ does the job.
• Suppose that $$L \neq K$$ and $$[L : K]$$ is odd, and let's factorize $$\theta$$ into primes of $$E$$ as follows: $$\theta = \left[\prod_{\mathfrak{p} \in S} \mathfrak{p}\right] \cdot \left[\prod_{\mathfrak{p} \in \Sigma_E \setminus S'} \mathfrak{p}^{e_{\mathfrak{p}}}\right]^2 \cdot \theta',$$ where $$\theta' \in L^\times$$ is a product of primes in $$S'$$. (In the first displayed product above, I'm isolating exactly one copy of each prime in $$S$$, and the second displayed product above contains all remaining prime factors not from $$S'$$). Observe that the existence of the $$r_{p}$$'s implies that $$\prod_{\mathfrak{p} \in S} \mathfrak{p} = \prod_{p \in T} p$$. Letting $$r = \operatorname{N}_{L/K}(\theta)$$, we have that $$r = \left[\prod_{p \in T} \operatorname{N}_{L/K}(p)\right] \cdot \left[\operatorname{N}_{L/K}\left(\prod_{\mathfrak{p} \in \Sigma_E \setminus S'} \mathfrak{p}^{e_{\mathfrak{p}}}\right)\right]^2 \cdot \operatorname{N}_{L/K}(\theta'),$$ so since $$\operatorname{N}_{L/K}(p) = p^{[L: K]}$$, we deduce that $$r \cdot \theta = \left[\prod_{p \in T} p^{[L : K] + 1}\right] \cdot \left[\left(\prod_{\mathfrak{p} \in \Sigma_E \setminus S'} \mathfrak{p}^{e_{\mathfrak{p}}}\right) \cdot \operatorname{N}_{L/K}\left(\prod_{\mathfrak{p} \in \Sigma_E \setminus S'} \mathfrak{p}^{e_{\mathfrak{p}}}\right)\right]^2 \cdot \operatorname{N}_{L/K}(\theta')$$ It is evident from the above expansion that $$r \cdot \theta$$ has the desired property.
• Suppose that $$L \neq K$$ and $$[L : K]$$ is even. Then the answer to the question is not necessarily. For a counterexample, let $$K = \mathbb{Q}(\sqrt{-5})$$, and recall that the class group $$\operatorname{Cl}(K)$$ of $$K$$ is isomorphic to $$\mathbb{Z}/2\mathbb{Z}$$ and is generated by the prime ideal $$(1 + \sqrt{-5}, 1 - \sqrt{-5})$$. Let $$L$$ be the Hilbert class field (maximal unramified abelian extension) of $$K$$, which is a quadratic extension of $$K$$ because the class number of $$K$$ is $$2$$. Recall that every ideal of $$\mathcal{O}_K$$ becomes principal in $$\mathcal{O}_L$$, so there is some $$\theta \in \mathcal{O}_L$$ such that $$(\theta) = (1 + \sqrt{-5}, 1 - \sqrt{-5})$$. Note that for any $$r \in K^\times$$, the class of the product ideal $$I = (r) \cdot (1 + \sqrt{-5}, 1 - \sqrt{-5})$$ in $$\operatorname{Cl}(K)/2\operatorname{Cl}(K)$$ is nontrivial, so the multiplicity of $$(1 + \sqrt{-5}, 1 - \sqrt{-5})$$ in $$I$$ cannot be even. [Thanks to James Tao for this observation.]