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

In sci.math, Arturo Magidin
<magidin@xxxxxxxxxxxxxxxxx>
wrote
on 1 Jan 2007 12:57:49 -0800
<1167685069.819648.73530@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:

The Ghost In The Machine wrote:
In sci.math, Arturo Magidin
<magidin@xxxxxxxxxxxxxxxxx>
wrote
on Sun, 31 Dec 2006 14:14:57 +0000 (UTC)
<en8gl1$qso$1@xxxxxxxxxxxxxxxxxx>:
In article <870j64-nsg.ln1@xxxxxxxxxxxxxxxxxxxxxxx>,
The Ghost In The Machine <ewill@xxxxxxxxxxxxxxxxxxxxxxx> wrote:
I have a dumb question, and am not sure precisely how to
phrase it to Google.

As everyone in this forum should know, for every positive
integer N, one can find a unique decomposition into primes
such that

N = 2^e_2 * 3^e_3 * 5^e_5 * 7^e_7 * ... * p ^ e_p

where e_2, e_3, e_5, etc. are nonnegative integers,
and p are special positive integers usually called primes,
divisible by only themselves and 1.

Usually the zero exponents are omitted, so that one get
things such as

33 = 3 * 11
42 = 2 * 3 * 7
54 = 2 * 3^3
65 = 5 * 13
109 = 109^1
etc.

This is of course the Fundamental Theorem of Arithmetic,
proven long ago by either Euclid or Gauss. I'll admit to
not being familiar with Ernst Kummer's work but am curious
as to why this factorization fails entirely in the ring
of algebraic integers, which are, of course, those roots
(real or complex) for irreducible members of Z[x] whose
highest power term has a coefficient of +/- 1.

In the full ring of algebraic integers, there are no "primes" (there
are no irreducible elements). So you cannot have unique factorization
into primes; in fact, you have NO factorization into irreducibles for
any algebraic integer other than units (which have the empty
factorization).


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?

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). 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), 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.


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.


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.


where everything involved is a non-unit integer.\
Apparently, a variant of infinite integer factorization
is what breaks the Fundamental Theorem of Arithmetic therein.

I cannot make heads or tails of what you are trying to say here. There
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. :-)
He's yet to find the *cat*.


Arturo Magidin


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