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Let $K/ \mathbb Q$ be a number field with $[K:\mathbb Q]=n$. Using that there exists a prime $p\in \mathbb Z$ which splits completely, that is $p\mathcal O_K=P_1...P_n$ for some distinct primes $P_i$ in $\mathcal O_K$, show that there exists an element $\alpha \in K\setminus \mathcal O_K$ such that

$$\text{Norm}_{K/ \mathbb Q}(\alpha)=\pm 1 $$

Here is what I tried to do. Let $p$ and $P_i$ as above. By the finiteness of the class group, we have that $(P_1...P_n)^k$ is principal for some $k\in \mathbb N$, say $(P_1...P_n)^k=(\beta)$ with $\beta \in K$. Now take $t=\dfrac{\beta}{p^k}$. Then the ideal norm of the fractional ideal generated by $t$ is 1. Can this help me get an element outside $\mathcal O_K$ of norm $\pm1$ ?

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I would have used a different strategy, as follows.
Use finiteness of class number, as you did, to find $z_i\in\mathcal O_K$ with $P_i^{m_i}=(z_i)$, just for $i=1,2$. That is, $v_{P_i}(z_i)=m_i>0$ and $v_Q(z_i)=0$ for all primes $Q$ of $\mathcal O_K$ different from $P_i$. Notice that the norm of $P_i$ is $p$, so that the norm of $P_i^{m_i}$ is $p^{m_i}$, and thus the norm of $z_i$ is also $\pm p^{m_i}$. Now take $z_1^{m_2}/z_2^{m_1}$.

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