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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.

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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 $$u\cdot pMx+vM\cdot Nx=(up+vN)Mx=Mx.$$

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    $\begingroup$ As far as I know, your answer is only correct for algebraic numbers, not for algebraic integers with leading coefficient $1$ and coefficients in $\mathbb Z$... $\endgroup$ – L. Miller Apr 1 at 13:56
  • $\begingroup$ @L.Miller You are absolutely right, I had algebraic numbers in mind somehow. I'll try to answer properly in a bit. $\endgroup$ – Servaes Apr 1 at 14:01
  • $\begingroup$ @L.Miller Actually these types of arguments hold much more generally - see my answer. $\endgroup$ – Bill Dubuque Apr 1 at 17:50
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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.

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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}\, $ $\Rightarrow\, a^{\large M}\equiv 1$

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