Trace of algebraic integer $\alpha$ that is in every prime ideal of $\mathcal{O}_K$ lying over $(p)$ Let $K$ be a number field with ring of integers $\mathcal{O}_K$ and let $p$ be a rational prime. Let $(p) = \mathfrak{p}_1^{e_1}\ldots\mathfrak{p}_r^{e_r}$ be the prime factorisation of (p) over $\mathcal{O}_K$, and suppose that $\alpha \in \mathfrak{a} = \mathfrak p_1\ldots \mathfrak{p}_r$. Then show that $\text{Tr}_{K/\mathbb{Q}}(\alpha) \equiv 0$ (mod $p$). 
I'd really appreciate any help in proving this. Thanks for reading! 

Special Case
I am able to prove the result in the case where $K/\mathbb{Q}$ is a Galois extension. In that case, each embedding $\sigma$ of $K$ is actually a $\mathbb{Q}$-automorphism, and $\sigma$ permutes the $\mathfrak{p}_i$, hence $\sigma(\alpha) \in \mathfrak{a}$, so clearly $\text{Tr}_{K/\mathbb{Q}}(\alpha) = \sum_\sigma \sigma(\alpha) \in \mathfrak{a} \cap \mathbb{Z} \subseteq p\mathbb{Z}$.
However, if $K/\mathbb{Q}$ is not Galois, the argument fails because the embeddings no longer permute the $\mathfrak{p}_i$, so the conjugates of $\alpha$ are no longer in $\mathfrak{a}$. 

Other Ideas
I'm aware that $\text{Tr}_{K/\mathbb{Q}}$ is the trace of the $\mathbb{Q}$-linear transformation of $K$ given by $v \mapsto \alpha v$, so I have thought about the matrix of this linear transformation with respect to an arbitrary integral basis, but haven't been able to make much headway with that.
We also have that $\alpha^e \in (p)$, where $e = \max_i \{e_i\}$, so that $\text{Tr}_{K/\mathbb{Q}}(\alpha^e)\in (p) \cap \mathbb{Z} = p\mathbb{Z}$, so I've thought about trying to relate the trace of $\alpha$ to the trace of $\alpha^e$, but also to no avail. 
 A: One can prove the general case by essentially reducing to the Galois case (though see a remark at the end). Let $L/\mathbb Q$ be a Galois extension containing $K$. Any prime $\mathfrak q$ lying above $p$ in $L$ contains one of the primes $\mathfrak p_i$ from $K$, and hence $\mathfrak q$ contains $\alpha$. By your own argument, any conjugate of $\alpha$ is also in $\mathfrak q$, and so the sum $A$ of conjugates of $\alpha$ is in $\mathfrak q\cap\mathbb Z=p\mathbb Z$. We have $A=\text{Tr}_{\mathbb Q(\alpha)/\mathbb{Q}}(\alpha)$, and since $K$ contains $\mathbb Q(\alpha)$, we have that $\text{Tr}_{K/\mathbb{Q}}(\alpha)=[K:\mathbb Q(\alpha)]\text{Tr}_{\mathbb Q(\alpha)/\mathbb{Q}}(\alpha)\in p\mathbb Z$.
Now for the promised remark, my first proof attempt trying to literally reduce to the Galois case, which would tell us (in notation above) that $\text{Tr}_{L/\mathbb{Q}}(\alpha)\in p\mathbb Z$ and try to deduce that $\text{Tr}_{K/\mathbb{Q}}(\alpha)\in p\mathbb Z$. However, the two differ by a factor of $[L:K]$ which itself might be divisible by $p$, which is a problem. My proof avoids this by directly looking at $A$, which is the sum of conjugates without any unnecessary repetitions.
A: It is known that the different ideal $D_K$ is divisible by $\mathfrak{p}_1^{e_1-1}\cdots \mathfrak{p}_r^{e_r-1}$, hence contained in $\mathfrak{p}_1^{e_1-1}\cdots \mathfrak{p}_r^{e_r-1}$ . Therefore $\mathfrak{p}_1^{1-e_1}\cdots \mathfrak{p}_r^{1-e_r}\subset D_K^{-1}.$
Now recall that for a fractional ideal $I$, we have $Tr_{K/\mathbb{Q}}(I)\subset \mathbb{Z}\iff I\subset D_K^{-1}$.
Now, $p^{-1} \mathfrak{a}=\mathfrak{p}_1^{1-e_1}\cdots \mathfrak{p}_r^{1-e_r}\subset D_K^{-1}$, so $Tr_{K/\mathbb{Q}}(p^{-1}\mathfrak{a})\subset \mathbb{Z}$, which is equivalent to $Tr_{K/\mathbb{Q}}(\mathfrak{a})\subset p\mathbb{Z}$, as required.
