Let $p$ be an odd prime number. Then there are three quadratic extensions of $\mathbb{Q}_p$. Assume that $t$ is a nonresidue module $p$, then these three extensions can be obtained by adjoining $\sqrt{t}$, $\sqrt{p}$ and $\sqrt{pt}$. Let $K/\mathbb{Q}$ be a Galois extension, for which the ramification index of $p$ is equal to $e=1$, and the inertia degree is equal to $f=2$. Let $\mathfrak{p}$ be a prime above $p$, then $K_{\mathfrak{p}}/\mathbb{Q}_p$ is an extension of degree $2$. But I can not decide to which of those three extensions it corresponds to. How can I find the corresponding extension? Which of these three extensions $\mathbb{Q}_p(\sqrt{t}), \mathbb{Q}_p(\sqrt{p}), \mathbb{Q}_p(\sqrt{pt})$ is obtained?
Edit: The comment by reuns answered my question. Now I have another similar question. How can we determine the corresponding extension, when $e=2$ and $f=1$? (The new question in detail: I mean Let $K/\mathbb{Q}$ be a Galois extension, for which the ramification index of $p$ is equal to $e=2$, and the inertia degree is equal to $f=1$. Let $\mathfrak{p}$ be a prime above $p$, then $K_{\mathfrak{p}}/\mathbb{Q}_p$ is an extension of degree $2$. But I can not decide to which of those three extensions it corresponds to. How can I find the corresponding extension? Which of these three extensions $\mathbb{Q}_p(\sqrt{t}), \mathbb{Q}_p(\sqrt{p}), \mathbb{Q}_p(\sqrt{pt})$ is obtained?)
@NeoTheComputer For $p$ odd, $\mathbf Q_p^*/{\mathbf Q_p^*}^{2}\cong {C_2}^2$. This comes from the general canonical decomposition $\mathbf Q_p^* \cong p^{\mathbf Z} \times \mathbf \mu_{p-1} \times (1+p\mathbf Z_p)$, which implies that $\mathbf Q_p^*/{\mathbf Q_p^*}^{2}$ is as announced, with representatives ${1, p, u, up}$, where $u$ is a unit s.t. $(u/p)=-1$ . See e.g. Serre's "A course in arithmetic", chap. I, §3.2. I misread completely your question.
NB: It seems to me that your question has been answered by Reuns. What does still bother you ?