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| KRamsay |
 Posted: Wed Dec 24, 2003 4:42 pm |
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Guest
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In article <bscj9e$u27$1@agate.berkeley.edu>, magidin@math.berkeley.edu (Arturo
Magidin) writes:
|> but it
|>takes a bit of work to prove the equivalence (the first proof is
|>apparently due to Gauss, around 1800).
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|That sounds like the proof of the Fundamental Theorem of Arithmetic,
|rather than an explicit proof that all irreducibles have the prime
|divisor property (the prime divisor property, by the way, is that if p
|divides a*b then it divides a or it divides b, and that goes back to
|Euclid!).
Gauss writes in _Disquisitiones Arithmeticae_
Section II. 14. If neither a nor b can be divided by a prime number p,
the product ab cannot be divided by p.
I don't see a definition of "prime number" given, but apparently he's
using the definition as irreducible.
He claims "Euclid had already proved this theorem in his Elements".
The citation is to Book VII, No. 32, but that doesn't seem right. That
proposition reads "Any number either is prime or is measured by a
prime number." I assume the citation has a typo, but I don't see
where the actual proposition is.
This is only a guess, but I would guess that when the concept of
"ideal divisor" entered number theory (now often just called "ideal"),
the "irreducible" definition lost ground relative to the "prime" definition.
An element being prime is the same as the ideal generated by
the element being a prime ideal. The ideal generated by an irreducible
element has no similarly nice property as far as I know.It doesn't
factor into nontrivial principal ideals, but when the element is not
actually prime, this seems to me more like a manifestation of the
fact that the ideal factors happen not to be principal.
I think in early grades, the "irreducible" definition probably would seem
less mysterious and less complicated, though, so I'm guessing that
it will persist.
Keith Ramsay |
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| Arturo Magidin |
| Posted: Thu Dec 25, 2003 1:23 pm |
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In article <yRJGb.13661$d%1.2838844@news20.bellglobal.com>,
curious <cur@curio.us> wrote:
Quote:
KRamsay wrote:
Gauss writes in _Disquisitiones Arithmeticae_
Section II. 14. If neither a nor b can be divided by a prime number p,
the product ab cannot be divided by p.
I don't see a definition of "prime number" given, but apparently he's
using the definition as irreducible.
He claims "Euclid had already proved this theorem in his Elements".
The citation is to Book VII, No. 32, but that doesn't seem right. That
proposition reads "Any number either is prime or is measured by a
prime number." I assume the citation has a typo, but I don't see
where the actual proposition is.
[ ... ]
KRamsay is right, it is a typo (in the English translation, anyway). It
should be Euclid VII.30:
http://aleph0.clarku.edu/~djoyce/java/elements/bookVII/propVII30.html
And Euclid's proof is actually simpler than the one in Gauss.
So Arturo Magidin is right that the proof goes back to Euclid.
The proof that irreducibles in the integers are primes, yes. Euclid
also implicitly uses Unique Factorization (the Fundamental Theorem of
Arithmetic) when he discusses Pythagorean Triples in Book X. An easier
way to spot his use of Unique Factorization is to look at the
explanation in
http://math.nmsu.edu/~history/book/euclidpt.pdf
since at one point he argues that if a product of two relatively prime
integers is a square, then each must be a square. This requires unique
factorization.
The first explicit proof of Unique Factorization is due to Gauss,
however, and that requires the use of the prime divisor property. It
is Art. 16 of the _Disquisitiones Arithmeticae_, and in fact Gauss
starts by saying that it is "clear from elementary considerations that
any composite number can be [factored] as a product of primes", and
then uses the prime divisor property in the usual way to show that any
two factorizations must in fact be equal. The "elementary
considerations" are that you cannot have an infinite descending
sequence of positive integers, and that if a positive number is
composite, you can factor it into two strictly smaller positive
integers. But this argument works for composite numbers, so long as
you can prove that there is no infinite descending sequence of
irreducibles. It is the prime divisor property that is a (possible)
problem.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@math.berkeley.edu |
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| curious |
| Posted: Thu Dec 25, 2003 5:53 pm |
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Guest
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KRamsay wrote:
Quote: In article <bscj9e$u27$1@agate.berkeley.edu>, magidin@math.berkeley.edu (Arturo
Magidin) writes:
|> but it
|>takes a bit of work to prove the equivalence (the first proof is
|>apparently due to Gauss, around 1800).
|
|That sounds like the proof of the Fundamental Theorem of Arithmetic,
|rather than an explicit proof that all irreducibles have the prime
|divisor property (the prime divisor property, by the way, is that if p
|divides a*b then it divides a or it divides b, and that goes back to
|Euclid!).
Gauss writes in _Disquisitiones Arithmeticae_
Section II. 14. If neither a nor b can be divided by a prime number p,
the product ab cannot be divided by p.
I don't see a definition of "prime number" given, but apparently he's
using the definition as irreducible.
He claims "Euclid had already proved this theorem in his Elements".
The citation is to Book VII, No. 32, but that doesn't seem right. That
proposition reads "Any number either is prime or is measured by a
prime number." I assume the citation has a typo, but I don't see
where the actual proposition is.
[ ... ]
KRamsay is right, it is a typo (in the English translation, anyway). It
should be Euclid VII.30:
http://aleph0.clarku.edu/~djoyce/java/elements/bookVII/propVII30.html
And Euclid's proof is actually simpler than the one in Gauss.
So Arturo Magidin is right that the proof goes back to Euclid. |
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| Robin Chapman |
| Posted: Mon Dec 29, 2003 2:40 pm |
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Guest
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David W. Cantrell wrote:
Quote: magidin@math.berkeley.edu (Arturo Magidin) wrote:
In article <XEmFb.1061$d%1.280494@news20.bellglobal.com>,
curious <cur@curio.us> wrote:
[snip]
In his "Commutative Algebra", D. Eisenbud has another definition:
Definition 2. Let R be a commutative ring. An element p of R is prime if
p is not a unit and the following is true: If a and b are elements of
R such that p divides ab, then p divides a or b.
Is Definition 2 common in the commutative algebra community?
It is common in algebraic number theory as well.
Then I must wonder what the common definition of "divides" is. According
to the definition which I typically use, 0 divides 0. And then according
to Definition 2, we would have 0 being a prime, which surely we don't
want.
Not necessarily --- the zero ideal isn't prime in every ring.
Def 2 just says that p is prime iff p generates a prime ideal.
Of course, in algebra, it's prime ideals, not prime elements, that
are of major interest.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"Needless to say, I had the last laugh."
Alan Partridge, _Bouncing Back_ (14 times) |
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