Let $K/\mathbb Q$ be a cubic extension and $(p) = \mathfrak p_1 \mathfrak p_2 \mathfrak p_3 $ be a factorization into distinct prime ideals. Suppose that $\alpha \in \mathcal O_K$ is an integer such that $\text{T}_{K/\mathbb Q}(\alpha) = 0$. Show that if $\mathfrak p_1 \mathfrak p_2$ divides $(\alpha)$, then $\mathfrak p_3$ also divides $(\alpha)$.
Here are some immediate findings from the premise:
1. For each $i = 1 \text{ to } 3$, the ideal norm $\text{N}(\mathfrak p_i) = q_i^{e_i}$, a prime power.
2. Thus, $\text{N}((p)) = q_1^{e_1}q_2^{e_2}q_3^{e_3}$, and $\text{N}(\mathfrak p_1)\text{N}(\mathfrak p_2)$ divides $\text{N}((\alpha)) = \left| \text{N}_{K/\mathbb Q}(\alpha) \right |$.
I attempted to find a prime number, $p_0$, such that $\text{N}((\alpha))$ is divisible by $p_0$, and that $(p_0)$ contains $\mathfrak p_3$, and I am getting stuck in the problem because I am having a hard time keeping track between prime number and prime ideals. Also, I am not sure how the condition that $\text{Tr}_{K/\mathbb Q}(\alpha) = 0$ fits into the solution; that suggests that this is not the correct approach.
Here’s my take on the problem, but I’m not fully sure that it’s sound.
Your hypothesis that $p$ splits completely in $K$ implies that there are three embeddings $\rho_1,\rho_2,\rho_3$ of $\Bbb Q$ into $\Bbb Q_p$, equivalently that $K\otimes_{\Bbb Q}\Bbb Q_p\cong\Bbb Q_p^3$. The hypothesis that $\mathfrak p_1\mathfrak p_2\,\vert\,(\alpha)$, in other words $\alpha\in\mathfrak p_1\mathfrak p_2=\mathfrak p_1\cap\mathfrak p_2$ says that $v_p(\rho_1(\alpha))\ge1$, similarly for $\rho_2(\alpha)$. Now, for $z\in K$, $\text{Tr}^K_{\Bbb Q}(z)$ is the trace of $z$ when considered in $\Bbb Q_p^3$, which is just $\sum_i\rho_i(z)$, and your hypothesis is that for $z=\alpha$, this sum is zero. Since $p\big|\big(\rho_1(\alpha)+\rho_2(\alpha)\big)$, it must also be that $p|\rho_3(\alpha)$, and you take this back to conclude that $\alpha\in\mathfrak p_3$.