Product of "Fake"-Galois Conjugates My apologies if this question ends up being a duplicate; I did my best to search for an answer, but I have no idea what to call this stuff I'm working with, so I couldn't really find much. There is a rather long wind up, so I've done my best to put in line breaks between the hypotheses.
Let $L$ be a normal field extension of $\mathbb{Q}$ of finite degree, let $\mathcal{O}_L$ be its ring of integers, let $G=\text{Aut}(L/\mathbb{Q})$, and let $\lambda\in\mathbb{C}$ be a transcendental number. 
For each map $g\in G$, extend that map to a ring automorphism $\hat{g}:L[\lambda]\rightarrow L[\lambda]$ by requiring $\hat{g}(\lambda)=\lambda$, and call the collection of such extensions $\hat{G}$. 
Let $p(\lambda)\in\mathcal{O}_L[\lambda]$, and define $\displaystyle \hat{p}(\lambda)=\prod_{\hat{g}\in\hat{G}}\hat{g}(p(\lambda))$. 
Is it true that $\hat{p}(\lambda)\in\mathbb{Z}[\lambda]$?
It's pretty easy to show that the leading coefficient and the constant coefficient of $\hat{p}(\lambda)$ are integers, and I think that it is believable that all of the coefficients are integers; the problem is that I don't see a good way of proving that all of them are. My most recent attempt kept running into big combinatorial messes, and I feel like there should be a slick proof of this out there somewhere. Does anyone know if there is a name for this construction, and does anyone know how to come up with a nice proof (or heaven forbid, disproof) of the claim? 
 A: The key thing to notice is that by your construction, each $\hat g$ fixes $\mathbb Q[\lambda]$. So rather than just extending each $g\in G$ to a map $\hat {g}:L[\lambda]\to L[\lambda]$, it seems logical to extend this map further to a field automorphism $\tilde g:L(\lambda)\to L(\lambda)$ by setting $$\tilde g\left(\frac{1}{p(\lambda)}\right) = \frac{1}{\hat g(p(\lambda))}.$$
In this way $\tilde g$ becomes a field automorphism of $L(\lambda)/\mathbb Q(\lambda)$, so that we can hope to apply Galois theoretic techniques. Note that $\tilde g$ and $\hat g$ agree on $L[\lambda]$, so that the results proven for $\tilde g$ will still hold for $\hat g$ when we restrict to $L[\lambda]$.
Letting $\tilde G$ be the collection of all such extensions, we see that actually, $$\tilde G = \mathrm{Gal}(L(\lambda)/\mathbb Q(\lambda)).$$
Indeed, every field automorphism of $L(\lambda)/\mathbb Q(\lambda)$ must fix $\lambda$, so is determined entirely by its action on $L$.
Let $p(\lambda)\in\mathcal O_L[\lambda]$ and define $$\tilde p(\lambda) =\hat p(\lambda) = \prod_{\tilde g\in\tilde G}\tilde g(p(\lambda)).$$
Then for every $\tilde h\in\tilde G$, we see that $$\tilde h(\tilde p(\lambda))=\tilde p(\lambda),$$so $\tilde p(\lambda)\in \mathbb Q(\lambda)$. Since $\tilde p(\lambda) \in \mathcal O_L[\lambda]$, the result now follows.
