Re: On The Fundamental Theorem of Arithmetic and Why it Breaks Down for the Algebraics

In article <ulro64-h2n.ln1@xxxxxxxxxxxxxxxxxxxxxxx>,
The Ghost In The Machine <ewill@xxxxxxxxxxxxxxxxxxxxxxx> wrote:

[.snip.]

That's fine as far as it goes; the rest of it was
an attempt why. I wanted to know if that attempted
explanation was correct and/or if there was an alternate
explanation which makes more sense.


Your "explanation" read to me like a lot of confused double-talk.

Maybe. :-)

What
"breakdown" points? The theorem fails because there are no primes.

And why are there no primes?

Because there are no irreducibles. Every prime has to be irreducible
(though irreducibility does not imply primeness).

The
theorem states that every number can be written as a product of
finitely many primes, in a unique way up to units and order. If there
are no primes, you cannot have a factorization, let alone a finite and
unique one. THAT's the "breakdown". Period.

Actually, I'm not sure if that's the way to look at it, or that one
cannot break down a number into a finite product (which in the positive
integers is obviously always possible).

Utter and absolute nonense. Perhaps you meant to say something else,
but what you said is patently false.

Of COURSE you can break down an algebraic integer into a finite
product. The problem is that you can ALWAYS break down a non-unit
algebraic integer into a product of two non-unit algebraic integers.

In the algebraics one can have
non-unit integers arbitrarily close to 1 (a^(1/n) for any integer a and
any positive integer n),

You are already going in the wrong track. You are introducing the
notion of "closeness", which requires a norm. Thus, you are already
trying to introduce analytic notions.

But in any case, this is completely irrelevant. There are rings of
algebraic integers (number rings), in which you can have numbers
arbitrarily close to 1 in their embeddings into the real numbers. For
example, take a quadratic real number field, and look at its rings of
integers. The group of units has a free part of rank 1, so there is a
number u, an algebraic integer, such that for every integer (positive
or negative), u^n is also an algebraic integer.

If |u|>1, then you get a bunch of numbers, u^{-n} with n positive,
which get "smaller and smaller", arbitrarily close to 0. Add 1 to them
to get nonunits which are arbitrarily close to 1. If |u|<1, then take
the positive powers instead.

Yet in these rings, you DO have irreducibles. Some even have unique
factorization. So this is not what you want.

but in the regular positive integers one has
'1' (which doesn't count), and '2' as the smallest possible factors.

Call it double-talk if one wishes but it's clear that some assumption
that is relevant for the integers is not relevant for the algebraics,
JSH et al notwithstanding. :-) I get the feeling that the two are
roughly equivalent:

[1] there are no primes,
[2] there is no guarantee that a factorization within the given ring
will terminate finitely.

No. These are again not equivalent. Take Z and adjoin all rational
powers of an indeterminate x. Then there are primes and irreducibles,
so [1] fails, and [2] holds.

I don't see why you think this does not make "enough" sense.

I don't think the proof for the Fundamental Theorem of Arithmetic starts
off with the primes, though I'd frankly have to look.

Sigh.

If you are trying to figure out how the proof of the FTA fails for the
algebraic integers, don't you think you should LOOK AT THE PROO FOF
THE FTA?

I mean, really. Isn't that, like, the first thing to do>

The usual proof for the integers works by induction on n. It assumes
you have shown that every irreducible is prime.

First one establishes the existence of the factorization:

n=1 has a factorization into primes: the empty one.

Assume all k<n have a factorization into primes. If n is prime, then
n itself is its own factorization. If n is not prime, then n is not
irreducible. Therefore, n=ab, with 1<a,b<n. By induction, each of a
and b have a factorization, and concatenating them gives you a
factorization for n. This establishes existence.

To prove uniqueness. assume

n= p1*...*p_m = q1 * ... * q_s

are two factorizations, p_i, q_j primes, m<=s. Proceed by induction on
m.

If m=0, then n=1, so s=0, and we are done: the two are the same.

Otherwise, p1 | n, so p1 divides one of q1,...,q_s. By reordering,
we may assume p1|q1. Since q1 and p1 are both primes, p1=q1 or
p1=-q1. Thus, p1=q1 up to a unit. Cancelling, we get

(n/p1) = p2* ...*p_m = q2*...*q_{s}

By induction, m-1 = s-1, so m=s, and up to a reordering and a unit,
p2=q2, ..., p_m=q_m. QED



The proof thus proceeds along two steps: FIRST, one uses a
"finiteness" of the integers, which is perhaps what you are groping
for. For the second, one uses the prime divisor property.

For number rings, one can rescue the "finiteness" by considering the
norm of the number, which gives you a multiplicative map into the
integers, which thus allows you to talk about how "big" an algebraic
integer is. But this "bigness" is not the usual one from the embedding
into the reals. For example, in Z[sqrt(-2)], the norm of a+b*sqrt(-2) is
defined to by a^2 + 2b^2.


You talk about "the implicit assumption that there is a smallest prime
dividing the origina number." There is nothing "implicit" about that,
and there is no such assumption in the proof of the fundamental theorem
(at least, not in the proofs I know). This assumption certainly breaks
down because there are NO primes, and therefore certainly no such thing
as a "smallest prime dividing the original number." How much more sense
do you want than that?


However, in algebraics, one gets nasty things such as

7 = 7^(1/2) * 7^(1/4) * 7^(1/8) * 7^(1/16) * ...

This, as written, is nonsense. Infinite products, like infinite sums,
do not follow readily from the finite nature of products. You would
need to DEFINE what you mean by an infinite product, which would of
course rely on some sort of limiting process. That will almost
certainly take you out of the standard concerns of algebra and into
those of analysis.

If you prefer one can generate an infinite number of finite products.
Same problem.

No, same problem FOR YOU. To talk about infinite products, or
"infinite number of finite products", you need it to MAKE SENSE. And
for it to make sense, you usually need some sort of limiting
process. That will take you out of the concerns of algebra and into
those of analysis. You are quite simply looking at the wrong
thing. And what you wrote is not valid in the ring of algebraic
integers; and the "infinite number of finite products", NONE is equal
to 7, and in order to talk about them approaching 7 you would need
some sort of metric, which is by no means given.


are no PRIMES, there are no IRREDUCIBLES, in the context of the full
ring of algebraic integers. The very statement of the FTA is nonsense
when N is replaced with that ring. In Spanish, I would say that you are
looking for the cat's three legs, meaning that you are engaged in a
search for the wrong thing and confusing yourself along the way.

Entirely possible. But if so, I'm farther along than JSH. :-)

Quite a compliment, there.


--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
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Arturo Magidin
magidin-at-member-ams-org

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