Unique Factorization Integers Definition

Question

I'm currently reading on integers, specifically for unique factorization for my abstract algebra course and I'm already stuck.

The text reads:

Write d|m if d divides m. It is easy to prove, from the definition, that if d|x and d|y then d|(ax + by) for any integers x, y, a, b.

It's not immediately obvious to me why you wouldn't just write d|(x+y) or d|(ax+ay) instead of d|(ax+by). My initial thought was to show that d|(ax+by) won't be a fraction, but I can't think of an example that would make d|(x+y) be a fraction.

Thank you for helping!

Answer 1

Writing $d|(ax+by)$ is more general. It allows you to pick any $a,b$ you want and apply the theorem. If we just wrote $d|(x+y)$ we wouldn't know we could multiply each of the terms by any integer we want and still have the divisibility. You could prove a second theorem that $d|x \implies d|(ax)$ and then use the two together plus the commutativity of addition to get what you want, but it is more work. It is better to make the theorem more general and save the work later.

Answer 2

They will soon invent in the text book the notion of an "ideal", in our case $(d)$ (in standard notation), the set of all elements divisible by $d$, is such an ideal. An "ideal" $I$ of some commutative ring $R$ can be defined in the following two obviously equivalent ways:

• $I$ is an ideal of $R$, iff it is closed w.r.t. sum of its elements, and w.r.t multiplication with arbitrary elements of $R$. Using letters, for all $x,y\in I$, all $a$ in $R$ the elements $x+y$ and $ax$ are in $I$.

• $I$ is an ideal of $R$, iff for all $x,y\in I$, all $a,b$ in $R$ the element $ax+by$ is in $I$.

It is just matter of taste which of the relations should be extracted and considered as definition. Usually, when we show that some $I$ is an ideal, we use the first definition, which wants "less". When we use the property of being an ideal, we use often the second definition, thus possibly having the economy of only one sentence.