Proving the fundamental theorem of arithmetic 53 points by justinhj on July 18, 2016 | hide | past | web | favorite | 10 comments

 Just saw Gowers yesterday in the first row of a conference talk given by Peter Scholze about perfectoid spaces. I left after 10 minutes as I was lost after 5 minutes :-)In my opinion conference talks are just not the right way to spread mathematical knowledge, except if you are deep into the subject matter. That can be seen because after the talk, there is usually just 1 question (maybe 2 questions), out of politeness. Blog articles like this have a much bigger impact!
 It helps that the fundamental theorem of arithmetic is a significantly more approachable topic than perfectoid spaces :)
 Denoting {a + b sqrt(-5) for a, b integers} by Z[sqrt(-5)], it's worth pointing out that you can actually use the norm to get a bit of the FTA--namely that every nonzero nonunit factors into irreducibles. By x being "irreducible", I mean that x is a nonzero nonunit element of Z[sqrt(-5)] and whenever x = yz for y,z in Z[sqrt(-5)] then either y or z is a unit.So, denoting x = a + b sqrt(-5) in Z[sqrt(-5)], we define the norm[1] of x as follows:`````` N(x) = a^2 + 5b^2 `````` Note that, in this case, N is a function from Z[sqrt(-5)] to the natural numbers.For x, y in Z[sqrt(-5)], it turns out that the following properties of the norm hold (proving them is fairly straightforward, as it's essentially "plug-and-chug" combined with a bit of reasoning about how things work in the natural numbers):1. N(xy) = N(x)N(y)2. N(x) = 0 if and only if x = 03. N(x) = 1 if and only if x = 1 or -1 (ie x is a unit).So, if x is a nonzero nonunit that isn't irreducible, then we can, by definition, write x = y*z where y and z are also nonzero nonunits. Applying strong induction via the norm, we can show that x can be written as a product of irreducibles. Of course this product is, in general, not uniquely determined.