Direct sums and prime numbers In my linear algebra class today, I was reminded of an episode of Code Lyoko, where Aelita gives a rather odd definition for a pair of prime numbers.
http://youtu.be/wvwSaKaz4is?t=9m41s

A pair of prime numbers occurs when their highest common denominator
  is one. In a principal ring, it's equivalent to the fact that their
  direct sum is equal to one, but of course, with a non principal
  factorial ring, that is not the case.

What does this even mean?  Can I perform a direct sum on two integers yielding another integer that is different from regular addition?
 A: The cited phrase does not makes sense, and just show once again that makers of such films couldn't care less that the scene is (extra) ridiculous to anyone who knows what this is (supposed to be) about. In any case prime numbers are plain (positive) integers (the corresponding notion is called "irreducible element" in a more general context of rings), and there is no reason for being posh about pairs of prime numbers, which are just pairs of integers that happen to be prime numbers.
There is a vague allusion to the fact that in a factorial ring (or more generally a gcd ring) two elements $a,b$ are relatively prime (or coprime; by definition this means thay have no non-invertible common divisors) if and only if their greatest common divisor is $1$, and that (only) in a principal ideal domain this is equivalent to being able to write $1$ as $1=sa+tb$, a sum of multiples of $a~$and$~b$. "Highest common denominator" is nonsense, born of confusion between "highest common factor" (a synonym for gretest common divisor) and "lowest common denominator".
It takes some effort to find any link with direct sums at all. It is a fact however that in modules over a principal ideal domain, the submodule of elements with $ab$-torsion (those becoming zero when multiplied by $ab$) can be decomposed as a direct sum of the submodules of elements with $a$-torsion and with $b$-torsion if $a$ and $b$ are relatively prime, and that a Bezout relation $1=sa+tb$ is instrumental in proving this.
