Trace of algebraic integer $\alpha$ that is in every prime ideal of $\mathcal{O}_K$ lying over $(p)$

Question

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!

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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 $\mathfrak{a}$.

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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.

Answer 1

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.

Answer 2

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.

Answer 3

Here is a nice idea that doesn't assume $K/\mathbb{Q}$ is Galois: If the $\sigma_i$ are the embeddings into $ \mathbb{C},$ note that $$ \mathrm{Tr}(\alpha^p) = \sum \sigma(\alpha^p) = \sum \sigma(\alpha)^p = \mathrm{Tr}(\alpha) \pmod p.$$

This means we can take repeated $p$-th powers within the trace and it won't change the residue class modulo $p.$ Picking $N$ such that $p^N \geq \max{(e_1, \ldots, e_r)}$ guarantees that $\alpha^{p^N} \in (p).$

Now we are done, as $\alpha^{p^N} = p\beta$ for some $\beta \in O_K,$ and we see $$ \mathrm{Tr}(\alpha) = \mathrm{Tr}(\alpha^{p^N}) = \mathrm{Tr}(p\beta) = 0 \pmod p.$$