Suppose we're given a $k$-long list of rational numbers:

$$q : \{0,\ldots,k-1\} \rightarrow \mathbb{Q}.$$

Then there's a least $n \in \mathbb{Z}_{>0}$ such that $nq_i$ is an integer for all $i$.

Question. What is this integer $n$ called, and does it have any generalizations/related concepts in ring theory?

For what it's worth, I think $n$ can be computed as follows. Given a rational number $q$, lets define that the denominator of $q$ is the least $b \in \mathbb{Z}_{>0}$ such that $q=a/b$ for some integer $a$. This defines a function $$\mathrm{den}:\mathbb{Q} \rightarrow \mathbb{Z}_{>0}.$$ If I'm not mistaken, $n$ can be computed as $$\mathop{\mathrm{lcm}}_{i=0}^{k-1} \mathrm{den}(q_i).$$

Here's a "real" world example of how this shows up. Suppose we have a box with three kinds of chocolate; dark, white, and milk. We're given that if we reach into this box to pull out a random chocolate, there's $1/3$ probability of getting dark, $1/4$ probability of getting white, and hence $5/12$ probability of getting milk. The question is; what's the minimum number of chocolates in this box? The answer can be found by computing $n$ for the aforementioned rational numbers; this gives $12$, so that's the minimum number of chocolates that could be in the box.

  • $\begingroup$ I don't think it has a name, but your formula to find it looks like it ought to be correct. $\endgroup$ – Arthur Jul 24 '17 at 13:03
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    $\begingroup$ LCD $\endgroup$ – Daniel Fischer Jul 24 '17 at 13:08
  • $\begingroup$ @DanielFischer, oh, haha. That makes so much sense. $\endgroup$ – goblin Jul 24 '17 at 13:09
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    $\begingroup$ @DanielFischer As trivial as it is, you should post it as an answer. $\endgroup$ – orlp Jul 24 '17 at 13:43
  • $\begingroup$ It seems that one of my serial downvoters decided to attack this answer. In case something is not clear there, please let me know and I will be happy to elaborate. $\endgroup$ – Bill Dubuque Jul 24 '17 at 22:00

$n\,$ is simply the least common denominator of the rationals $\,q_i.\,$ As you remarked, it is the lcm of the denominators when the fractions are written in least terms, since

$$ n\, a_i/b_i\in\Bbb Z\iff b_i\mid n a_i\overset{(a_i,b_i)\,=\,1}\iff b_i\mid n\iff {\rm lcm}\{b_i\}\mid n$$

For some generalization and related concepts see posts denominator ideals and order ideals.

One common application of denominator ideals is to irrationality proofs. Using Dedekind's notion of conductor (universal denominator) ideal allows one to unify and generalize most of the common descent-based irrationality proofs, e.g. see here. As I show there, they generalize to a one-line proof that PIDs are integrally closed (monic case of Rational Root Test). More generally, the conductor ideal plays a key role when studying rings of algebraic integers.

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  • $\begingroup$ The downvotes surely have notihng to do with math, esp, considering that this and another answer were downvoted within seconds. $\endgroup$ – Bill Dubuque Jul 24 '17 at 20:41
  • $\begingroup$ How lame. I, for one, appreciate the answer. $\endgroup$ – goblin Jul 24 '17 at 22:23

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