How to precisely define $b/a$ if $a \mid b$ in a general commutative ring?

Question

Let $R$ be a commutative ring and let $a,b \in R$. Suppose that $a \mid b$, which means that there exists some $k \in R$ such that $k\cdot a = b$. I now often see expressions of the form $b/a$ in this context. ( For example the important lcm-gcd-formula $gcd(a,b) = ab / lcm(a,b)$ ). My question:

Is it enough to simply define $b/a := k$ or are there some caveats? (I think one problem here might be that $b/a$ does not have to be unique with this naive definition.)

Could you please explain this to me?

https://en.wikipedia.org/wiki/Localization_(commutative_algebra) — aidangallagher4, May 26 '21 at 19:47
If $R$ is not an integral domain, then yes this can be an issue. However, if it is an integral domain, then $k$ is unique (assuming $a \not= 0$). Can you prove this for yourself? — J. Loreaux, May 26 '21 at 19:48
@aidangallagher4 Beware that overrings of fractions need not be localizations. That is true only for Bezout domains — Bill Dubuque, May 26 '21 at 20:20

ramhuw · Accepted Answer · 2021-05-28T00:33:52.873

1

For the case $[a, b] = ab / (a, b)$, the quotient exists in GCD domain and is unique.

For an arbitrary commutative ring, it's possible to define 'quotient', see Localization.

edited May 28 '21 at 00:33

answered May 28 '21 at 00:14

ramhuw

64

How to precisely define $b/a$ if $a \mid b$ in a general commutative ring?

1 Answers1