Dgdrxrt

I'm trying to eliminate prime factors from algebraic integers.
Is the following true without further restrictions? And how can I prove it?

Let $p,N,Minmathbb Z$, let $p$ be prime such that $gcd(p;N)=1$.

Let $xinmathbb C$ such that $pcdot Mcdot x$ and $Ncdot x$ are algebraic integers.

Then $Mcdot x$ is also an algebraic integer.

Because $gcd(p,N)=1$ there exist integers $u,vinBbbZ$ such that
$$up+vN=1.$$
By assumption $pMx$ and $Nx$ are algebraic integers, hence so is
$$ucdot pMx+vMcdot Nx=(up+vN)Mx=Mx.$$

This is true. Let $ap + bN = 1$, where $a,b in mathbbZ$. $NMx$ is also an algebraic integer, so $a(pM)x + b(NM)x = Mx$ is an algebraic integer.

More conceptually: the set $I = nin Bbb Z : n x in Bbb I$ 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 Nequiv 1equiv a^large pM, $ $Rightarrow, a^large Mequiv 1$