Re: algebra with field of quotients.


"William Elliot" <marsh@xxxxxxxxxxxxxxxxxx> wrote in message
news:Pine.BSI.4.58.0609102041000.18881@xxxxxxxxxxxxxxxxxxxx
On Mon, 11 Sep 2006, mina_world wrote:

i know that Q is a field of quotients of Z.

for a pair (4,6) of Z,

No. (4,6) in Z^2

we can't say that (4,6) is a element of "field of quotients of Z".

we can say that equivalent pair [(4,6)] is a element of
"field of quotients of Z".

so, (4,6) is not a element of "field of quotients of Z".
but, it is fact that 4/6 is a element of Q.

Indeed, it's defined as [ (4,6) ] or equivalently as [ (2,3) ]
or [ (8,12) ]. So what's new? 4/6 = 2/3 = 8/12 ? ;-)

i think....
if Q' only contain irreducible fractions,
we can say that Q' is a field of quotients of Z.

is this a foolish thinking ?

Yes. [ (4,6) ] = [ (2,3) ]. So where is irreducible?
Just as with fractions 4/6 = 2/3, where is irreducible?

Irreducible isn't in the rationals, it's in the representation
of the quotient class or in the representation of the fraction.

Let q = n/m be a rational with coprime n and m.
q then is representated as an irreducible fraction.

q = 2n/2m isn't representated as an irreducible fraction
even tho it's the same number.


if (4,6) = [(4,6)], my thinking is needless.
but it's false.
it's true that (4,6) in [(4,6)].

if we say that Q is a field of quotients of Z,
4/6 is a element of "field of quotients of Z".
but it's false.
if [4/6] = {x | x = 2/3}={2/3, 4/6, 6/9,.....},
it's true that [4/6] is a element of "field of quotients of Z".

thus, if (4,6) = [(4,6)], my thinking is needless.
but it's false.
how do you think about it ?
i need your advice.
because, sci-math's members are genius.


.

Relevant Pages

  • Re: algebra with field of quotients.
    ... so, is not a element of "field of quotients of Z". ... of the quotient class or in the representation of the fraction. ... q then is representated as an irreducible fraction. ...
    (sci.math)
  • Re: algebra with field of quotients.
    ... You can define addition and multiplication on the equivalence classes, ... The irreducible fraction you speak of is a representative member of the ... we can't say that is a element of "field of quotients of Z". ...
    (sci.math)