How to eliminate prime factors from algebraic integers? I'm trying to eliminate prime factors from algebraic integers.
Is the following true without further restrictions? And how can I prove it?
Let $p,N,M\in\mathbb Z$, let $p$ be prime such that $gcd(p;N)=1$.
Let $x\in\mathbb C$ such that $p\cdot M\cdot x$ and $N\cdot x$ are algebraic integers.
Then $M\cdot x$ is also an algebraic integer.
 A: Because $\gcd(p,N)=1$ there exist integers $u,v\in\Bbb{Z}$ such that
$$up+vN=1.$$
By assumption $pMx$ and $Nx$ are algebraic integers, hence so is
$$u\cdot pMx+vM\cdot Nx=(up+vN)Mx=Mx.$$
A: This is true. Let $ap + bN = 1$, where $a,b \in \mathbb{Z}$. $NMx$ is also an algebraic integer, so $a(pM)x + b(NM)x = Mx$ is an algebraic integer.
A: More conceptually: the set $I = \{n\in \Bbb Z\ :\  n x \in \Bbb I\}$ is easily verified to be a (denominator) ideal.
Since $I$ contains $pM,N$ it contains their gcd $\,(pM,N) = (M,N),\,$ so also its multiple $M$.
Remark $ $ Likely you are familiar with more elementary manifestations such as the well-known fact that if a fraction can be written with denominators $\,c,d\,$ then it can be written with denominator $\,\gcd(c,d),\,$ or analogous results about orders of elements, e.g. here.. Conceptually it is better to avoid direct Bezout-based proofs and instead think of these results in terms of denominator or order ideals..
I used an exact order analog here, where I mention the fractional form below (with your notation)
Lemma $ $ If a fraction is writable with denominator $\,N\,$ and also with denominator $\,pM\,$ where $(N,p)=1$ then the fraction can  be written with denominator $\,M$.
The order form used there is as follows.
Lemma $\,\ $ If $\ (N,p)=1\,$ then $\, a^{\large N}\equiv 1\equiv a^{\large pM}\, $ $\Rightarrow\, a^{\large M}\equiv 1$
