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

Arturo Magidin wrote:

[.snip.]

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),


I suspect that what you are really groping for is the degree of the
algebraic integers. Given an algebraic integer (or in fact, an
algebraic number) a, we define its "degree over Q" to be the degree of
the extension Q(a)/Q, which is equivalent to the degree of the monic
irreducible polynomial for a over Q.

Now let A be a subring of the algebraic integers which is integrally
closed (meaning that A contains all algebraic integers that are
contained in the field of fractions of A). If there is a uniform bound
N for the degrees of all elements of A, that is, if [Q(a):Q]<= N for
all a in A, then it is in fact the case that if F is the field of
fractions of A, then [F:Q] <= N. The converse is of course trivial
(that is, if [F:Q]<=N, then for all a in A we have [Q(a):Q]<=N). For
this direction, note that if there are N+1 elements of F that are
linearly independent over Q, f1,...,f_{N+1}, then let K =
Q(f1,...,f_{N+!}). By definition of the degree of the extension, we
have [K:Q]=N+1, and by the primitive element theorem we know that there
exists an algebraic number r in K such that K=Q(r). Multiplying by an
appropriate integer d>0 we get that dr is an algebraic integer, and
Q(r)=Q(dr), hence dr is an element of K (and so of F, and therefore of
A), with [Q(dr):Q]=N+1>N, contradicting the hypothesis that all degrees
of elements of A are less than or equal to N. Thus, [F:Q]<=N, as
claimed.

Now: if the there is a uniform bound for all elements of A, then you
will necessarily have factorization into irreducibles in A, though it
may not be unique (that is, it may be the case that A is not a UFD,
that not all irreducibles are prime). To show this, define the norm
map from A to Z by mapping each a in A to the product of all its images
under all embeddings of F into the complex numbers (the usual norm
map). This will give you a multiplicative function from A to Z which
can be used to do induction and complete the first part of the FTA:
factorization into irreducibles.

I am not positive that if there is no uniform bound (equivalently, if
the degree of F over Q is infinite) then you will be able to produce an
"infinite descending sequence" of elements of A; that is, a sequence
a1, a2, a3, ..., a_n,... of elements of A, all nonunits, such that
a_{i+1} divides a_i, and is not an associate of a_i, for
i=1,...,.n,...., which will tell you that you may be unable to achieve
a factorization of a1 into irreducibles (though a different
factorization may exist). I am not positive of this, though.

In a sense, you want to think not about their "size" in terms of
embeddings into the complex numbers, but rather their "size" in terms
of how complicated a polynomial you need in order to specify them.
That's what the degree gives you.

Arturo Magidin, sans .sig

.

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