
Proving the fundamental theorem of arithmetic - justinhj
https://gowers.wordpress.com/2011/11/18/proving-the-fundamental-theorem-of-arithmetic/
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auggierose
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!

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qmalzp
It helps that the fundamental theorem of arithmetic is a significantly more
approachable topic than perfectoid spaces :)

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justinhj
Follow up to
[https://news.ycombinator.com/item?id=11952290](https://news.ycombinator.com/item?id=11952290)

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gohrt
Which step in the proof succeeds on the integers, but fails when applied to
the ring of {a + b*sqrt(-5)} ?

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Chinjut
Bezout’s theorem fails in that ring.

The particular proof given of Bezout's theorem here uses the fact that modding
out the ambient ring by one of its prime (in the sense of not further
factorable) elements yields a resulting ring with a prime (in the sense of not
further factorable) cardinality, and therefore no nontrivial additive
subgroups.

However, in the ring Z[sqrt(-5)], there are irreducible elements such that
modding out by them produces a ring with non-prime cardinality; for example, 2
is irreducible in this ring, but there are 4 values mod 2 in this ring. 4 is
not a prime cardinal, of course. Thus, the provided proof of Bezout's Theorem
can and does fail in this context. [Indeed, if we look at the range of
multiplication by 2 on this particular additive group of size 4, we see that
it is neither trivially of size 1 nor trivially of size all 4. It is an
intermediate subgroup of size 2, which can exist because 4 has an intermediate
factor of 2]

The provided proof works in the integers because integers can be lined up with
cardinals so that A) integer arithmetic mod n has cardinality corresponding to
n, and B) also, factorizations or lack thereof for an integer are the same as
for the corresponding cardinal. This ensures that integer arithmetic mod a
prime integer has a prime cardinality. This fails in other contexts.

[The proof of Bezout's theorem provided, incidentally, is not my favorite
because it generalizes so very little. A nicer proof of Bezout's theorem, in
my mind, is via the Euclidean algorithm, which then generalizes to all
Euclidean domains (and, with minor modification, slightly beyond those as
well, to principal ideal domains axiomatized in a manner analogous to
Euclidean domains)]

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sukilot
How do you define "prime cardinal" without relying on the definition of prime
numbers? "Prime" (as distinct from irreducible elements) doesn't have a clear
meaning before you prove the Fundamental Theorem of Arithmetic, does it?

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Chinjut
I defined it in the relevant way in the post you're responding to (indeed,
inbetween the very words "prime" and "cardinal" where I first invoke the
concept): the relevant notion is of a cardinal which can't be factored as the
(Cartesian) product of cardinals other than 1 and itself. [Yes, people often
call this concept "irreducible" instead, but I used "prime" for convenience,
and explicitly described what I meant by that.]

This is the concept that is relevant to the provided proof: Gowers argues
that, if a group's size is an unfactorable cardinal, then, by Lagrange's
Theorem (which tells us |G'| divides |G| whenever G' is a subgroup of G), it
has no intermediate subgroups. Thus, if the additive group mod whatever has
size an unfactorable cardinal, then every homomorphism into it is either
constantly identity or surjective (as its range is a subgroup); accordingly,
multiplication by any non-identity element would have to be invertible (modulo
whatever), which is the desired instance of Bezout's Theorem.

~~~
tamana
Thanks for clarifying. I was trying to be careful about understanding meaning
since the OP was all about not taking obvious things for granted.

