# 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.

## 3 Answers

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

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

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.

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$$